Compilation of elementary results related to permutations and combinations: Pre RMO, RMO, IITJEE math

1. Disjunctive or Sum Rule:

If an event can occur in m ways and another event can occur in n ways and if these two events cannot occur simultaneously, then one of the two events can occur in $m+n$ ways. More generally, if $E_{i}$ ($i=1,2,\ldots,k$) are k events such that no two of them can occur at the same time, and if $E_{i}$ can occur in $n_{i}$ ways, then one of the k events can occur in $n_{1}+n_{2}+\ldots+n_{k}$ ways.

2. Sequential or Product Rule:

If an event can occur in m ways and a second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, $if $E_{1}$ can occur in $n_{1}$, $E_{2}$ can occur in $n_{2}$ ways (no matter how $E_{1}$ occurs), $E_{3}$ can occur in $n_{3}$ ways (no matter how $E_{1}$ and $E_{2}$ occur), $\ldots$, $E_{k}$ can occur in $n_{k}$ ways (no matter how the previous $k-1$ events occur), then the k events can occur simultaneously in $n_{1}n_{2}n_{3}\ldots n_{k}$ ways. )3. Definitions and some basic relations: Suppose X is a collection of n distinct objects and r is a nonnegative integer less than or equal to n. An r-permutation of X is a selection of r out of the n objects but the selections are ordered. An n-permutation of X is called a simply a permutation of X. The number of r-permutations of a collection of n distinct objects is denoted by $P(n,r)$; this number is evaluated as follows: A member of X can be chosen to occupy the first of the r positions in n ways. After that, an object from the remaining collections of $(n-1)$ objects can be chosen to occupy the second position in $(n-1)$ ways. Notice that the number of ways of placing the second object does not depend upon how the first object was placed or chosen. Thus, by the product rule, the first two positions can be filled in $n(n-1)$ ways,….and all r positions can be filled in $P(n,r) = n(n-1)\ldots (n-r+1) = \frac{n!}{(n-r)}!$ ways. In particular, $P(n,n) = n!$ Note: An unordered selection of r out of the n elements of X is called an r-combination of X. In other words, any subset of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects is denoted by $n \choose r$ (read as ” n ‘choose’ r). For each r-subset of X there is a unique complementary $(n-r)$-subset, whence the important relation ${n \choose r}$ = $n \choose {n-r}$. To evaluate $n \choose r$, note that an r-permutation of an n-set X is necessarily a permutation of some r-subset of X. Moreover, distinct r-subsets generate r-permutations each. Hence, by the sum rule: $P(n,r)=P(r,r)+P(r,r)+\ldots + P(r,r)$ The number of terms on the right is the number of r-subsets of X. That is, $n \choose r$. Thus, $P(n,r)=P(r,r) \times {n \choose r}$=$r! \times {n \choose r}$. The following is our summary: 1. $P(n,r) = \frac{n!}{(n-r)!}$ 2. ${n \choose r}=\frac{P(n,r)}{r!}=\frac{n!}{r! (n-r)!}$=$n \choose {n-r}$ 4. The Pigeonhole Principle: Basic Version: If n pigeonholes (or mailboxes) shelter n+1 or more pigeons (or letters), at least 1 pigeonhole (or mailbox) shelters at least 2 pigeons (or letters). 5. The number of ways in which $m+n$ things can be divided into two groups containing m and n equal things respectively is given by : $\frac{(m+n)!}{m!n!}$ Note: If $m=n$, the groups are equal (and hence, indistinguishable), and in this case the number of different ways of subdivision is $\frac{(2m)!}{2!m!m!}$ 6. The number of ways in which m+n+p things can be divided into three groups containing m, n, p things severally is given by: $\frac{(m+n+p)!}{m!n!p!}$ Note: If we put $m=n=p$, we obtain $\frac{(3m)!}{m!m!m !}$ but this formula regards as different all the possible orders in which the three groups can occur in any one mode of subdivision. And, since there are 3! such orders corresponding to each mode of subdivision, the number of different ways in which subdivision into three equal groups can be made in $\frac{(3m)!}{m!m!m!3!}$ ways. 7. The number of ways in which n things can be arranged amongst themselves, taking them all at a time, when p of the things are exactly alike of one kind, q of them are exactly alike of a another kind, r of them are exactly alike of a third kind, and the rest are all different is as follows: $\frac{n!}{p!q!r!}$ 8. The number of permutations of n things r at a time, when such things may be repeated once, twice, thrice…up to r times in any arrangement is given by: $n^{r}$. Cute quiz: In how many ways, can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? (Compare your answers with your friends’ answers :-)) 9. The total number of ways in which it is possible to make a selection by taking some or all of n things is given by : $2^{n}-1$ 10. The total number of ways in which it is possible to make a selection by taking some or all out of $p+q+r+\ldots$ things, whereof p are alike of one kind, q alike of a second kind, r alike of a third kind, and so on is given by : $(p+1)(q+1)(r+1)\ldots-1$. Regards, Nalin Pithwa. Tutorial problems. I. Symmetric and Alternating functions. RMO/IITJEE Math Exercises: 1. Show that $(bc-ad)(ca-bd)(ab-cd)$ is symmetric with respect to a, b, c, d. 2. Show that the following expressions are cyclic with respect to a, b, c, d, taken in this order: $(a-b+c-d)^{2}$ and $(a-b)(c-d)+(b-c)(d-a)$ 3. Expand the expression using $\Sigma$ notation: $(y+z-2x)(z+x-2y)(x+y-2z)$ 4. Expand the expression using $\Sigma$ notation: $(x+y+z)^{2}+(y+z-x)^{2}+(z+x-y)^{2}+(x+y-z)^{2}$ 5. Prove that $(\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+\alpha^{2}\beta^{2})(\alpha+\beta+\gamma)= \Sigma\alpha^{2}\beta^{2}+\alpha\beta\gamma\Sigma\alpha\beta$ 6. Prove that $(\alpha-\beta)(\alpha-\gamma)+(\beta-\gamma)(\beta-\alpha)+(\gamma-\alpha)(\gamma-\beta)=\Sigma{\alpha^{2}}-\Sigma{\alpha}{\beta}$ 7. Prove that $(\beta-\gamma)(\beta+\gamma-\alpha)+(\gamma-\alpha)(\gamma+\alpha-\beta)+(\alpha-\beta)(\alpha+\beta-\gamma)=0$ 8. Prove that : $\alpha(\beta-\gamma)^{2}+\beta(\gamma-\alpha)^{2}+\gamma(\alpha-\beta)^{2}=\Sigma{\alpha^{2}}{\beta}-6\alpha\beta\gamma$ 9. Prove that: $(\beta^{2}\gamma+\beta\gamma^{2}+\gamma^{2}\alpha+\gamma\alpha^{2}+\alpha^{2}\beta+\alpha\beta^{2})(\alpha+\beta+\gamma)=\Sigma{\alpha^{2}}\beta+2\Sigma{\alpha^{2}}{\beta^{2}}+2\alpha\beta\gamma\Sigma{\alpha}$ 10. Prove that : $a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+abc(a+b+c)=\Sigma{a^{2}}.\Sigma{ab}$. 11. Prove that: $(a+b-c)(a^{2}+b^{2}-c^{2})+(b+c-a)(b^{2}+c^{2}-a^{2})+(c+a-b)(c^{2}+a^{2}-b^{2})=3\Sigma{a^{3}}-\Sigma{a^{2}{b}}$ 12. Prove that: $(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})=(ax+by+cz)^{2}+(bz-cy)^{2}+(cx-az)^{2}+(ay-bz)^{2}$ 13. Prove that: $(b^{2}-ac)(c^{2}-ab)+(c^{2}-ab)(a^{2}-bc)+(a^{2}-bc)(b^{2}-ac)=-(bc+ca+ab)(a^{2}+b^{2}+c^{2}-bc-ca-ab)$ 14. Prove that: $(a^{2}-bc)(b^{2}-ac)(c^{2}-ab)=abc(a^{2}+b^{2}+c^{2})-(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2})$ 15. If one of the numbers a, b, and c is the geometric mean of the other two, use the previous problem to prove the following: $abc(a^{2}+b^{2}+c^{2})=b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}$ 16. If the numbers x, y, z taken in some order or other form an AP, use problem 3 to prove that $2(x+y+z)^{2}+27xyz=9(x+y+z)(yz+zx+xy)$ 17. Express $2(a-b)(a-c)+2(b-c)(b-a)+2(c-a)(c-b)$ as the sum of three squares. Hence, show that $(b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c)$ is negative for all real values of a, b, c except when $a=b=c$. Hint: Put $b-c=x$, $c-a=y$, $a-b=z$, and notice that $x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=(x+y+z)^{2}=0$. 18. If $x+y+z=0$, show that (i) $2yz=x^{2}-y^{2}-z^{2}$; (ii) $(y^{2}+z^{2}-x^{2})(z^{2}+z^{2}-y^{2})(x^{2}+y^{2}-z^{2})+8x^{2}y^{2}z^{2}=0$ (iii) $ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy$ can be expressed in the form $px^{2}+qy^{2}+rz^{2}$; and, find p, q, r in terms of a, b, c, f, g, h. Cheers, Nalin Pithwa. Symmetric Functions. Alternating Functions. Algebra for RMO/IITJEE Math Reference: Higher Algebra by Bernard and Child. I. Symmetric Functions. A function which is unaltered by the interchange of any two of the variables which it contains is said to be symmetric with respect to (wrt) these two variables. Thus, $yz+zx+xy$ and $(x^{2}y+y^{2}z+z^{2}x)(x^{2}z+y^{2}z+z^{2}y)$ are symmetrical w.r.t. x, y, z. The interchange of any two letters, x, y, z is called the transposition (xy). Terms of an expression which are such that one can be changed into the other by one or more transpositions are said to be of the same type. Thus, all the terms of $x^{2}y+x^{2}z+y^{2}z+y^{2}x+z^{2}x+z^{2}y$ are of the same type, and the expression is symmetric with respect to x, y, z. A symmetric function which is the sum of a number of terms of the same type is often written in an abbreviated form thus: Choose any one of the terms and place the letter $\Sigma$ (sigma) before it. For instance: $x+y+z$ is represented by $\Sigma{x}$ and $xy+yz+zx$ by $\Sigma{xy}$. Again, $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=\Sigma{x^{2}}+2\Sigma{xy}$. It is obvious that (i) if a term of some particular type occurs in a symmetric function, then all terms of the same type will also occur. (ii) The sum, difference, product and quotient of two symmetric functions are also symmetric functions. (PS: there is no need for a grand proof of the above; just apply the definitions of symmetric functions…check with some examples). Considerations of symmetry greatly facilitate many algebraical processes as illustrated in the following examples: Example 1: Expand $(y+z-x)(z+x-y)(x+y-z)$. Solution 1: This expression is symmetric, homogeneous and of the third degree in x, y, z We may therefore assume that $(y+z-x)(z+x-y)(x+y-z)= a.\Sigma{x^{3}}+b.\Sigma{x^{2}y}+cxyz$, where a, b, c are independent of x, y, z. In this assumed identity: (i) put $x=1, y=0, z=0$, then $-1=a$ (ii) put $x=1, y=1, z=0$, then $0=2a+2b$, and so $b=1$. (iii) put $x=1, y=1, z=1$, then $1=3a+6b+c$ and so $c=-2$. Hence, the required product is $-x^{2}-y^{2}-z^{2}+y^{2}z+yz^{2}+z^{2}x+zx^{2}+x^{2}y+xy^{2}-2xyz$. Example 2: Expand $(a+b+c+d)(ab+ac+ad+bc+bd+cd)$. Test the result by putting $a=b=c=d=1$. Solution 2: the product is the sum of all terms of the product obtained by multiplying any one term of the first expression by any other term of the second expression. Hence, the result will have terms of the type : $a^{2}b, abc$, The coefficient of $a^{2}b$ in the product is 1; because this term is obtained as product of a and ab and in no other way. The coefficient of abc is 3; because this term is obtained in each of the three ways $a(bc), b(ca), c(ab)$. Hence, the required answer is $\Sigma{a^{2}b}+3\Sigma{abc}$ Test: The number of terms of the type $a^{2}b$ is 12 and the number of terms of the type abc is 4; hence, if $a=b=c=d=1$, then $\Sigma{a}.\Sigma{ab}=4.6=24$ and $\Sigma{a^{2}b}+3.\Sigma{abc}=12+3.4=24$ so that the test is satisfied. Example 3: Factorize $(x+y+z)^{5}-x^{5}-y^{5}-z^{5}$. Solution 3: Method I: Brute force is really difficult (I did give it a shot …:-)) Method II; Some of you might try the binomial theorem for positive integral index, but to extract the factors is still ..a little bit like brute force method only. Method III: Check whether the given expression is symmetric wrt any two variables (namely, x & y; y & z; z & x; ) and whether it is homogeneous and if so, what is the degree. Also from observations of past solved problems, we need to check how many terms of each type are there: Observations are as follows: the degree of the expression is five only; and the expression is also homogeneous with each term being of degree five; to check for symmetry, let us proceed as follows: $E_{1}=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}$ and switching x and y gives us $E_{2}=(y+x+z)^{5}-y^{5}-x^{5}-z^{5}$. Quite clearly, the expression is symmetric w.r.t. x and y; y and z; and, z and x. To factorize it, we use fundamental theorem of algebra or factor or remainder theorem. Substitute $x=-y$ so that the expression is equal to $(z)^{5}-x^{5}-(-x)^{5}-z^{5}=z^{5}-x^{5}+x^{5}-z^{5}=0$ so that $(x+y)$ is a factor of the expression. Similarly, the other factors are $(y+z)$ and $(z+x)$. By the fundamental theorem of algebra, we still need a quadratic factor of x, y and z. This factor should be homogeneous also. Hence, let $E=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}=(x+y)(y+z)(z+x){A(x^{2}+y^{2}+z^{2})+B(xy+yz+zx)}$, where A and B are pure numeric coefficients independent of x, y and z. So, put $x=1,y=1, z=0$, then $2A+B=15$ and put $x=1, y=1, z=1$, then $a+b=10$ so that $A=B=5$. So, $E=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}=5(x+y)(y+z)(z+x)(x^{2}+y^{2}+z^{2}+xy+yz+zx)$. II Alternating Functions: If a function E of x, y, z …is transformed into -E by the interchange of any two of the set x, y, z, …, then E is called an alternating function of x, y, z…(Note that just as in the case of symmetric functions, we talk of alternating functions w.r.t. a pair of variables at a time.) ((PS: At this juncture, it behooves you to recall the definitions of even and odd functions, and also to recall the fact that every function can be expressed as a sum of an even function and an odd function. Compare all three now: symmetric, alternating and even/odd functions. )) Such an alternating function is $x^{n}(y-z)+y^{n}(z-x)+z^{n}(x-y)$; for, the interchange of any two letters, say x and y, transforms it into $y^{n}(x--z)+x^{n}(x-y)+z^{n}(y-z)=-E$. Observe that the product and the quotient of two alternating functions are symmetric functions. (here, again, it does not require any grand proof…just pore over the definitions in your head…) Thus, $\frac{x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)}{(y-z)(z-x)(x-y)}$ is symmetric w.r.t. x, y, and z. (PS: please do some scribbling and verify this little observation/fact). Example 1: Factorize : $x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$. Solution 1: Let $E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$ We know that $E=0$ when $x=y$, $y=z$ and $z=y$. Hence, the following is a factor of E: $(x-y)(y-z)(z-x)$. As the given expression E is homogeneous of degree 4, it should have one more homogeneous linear factor. The only such factor possible is $K(x+y+z)$. So, now, $E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)=K(x-y)(y-z)(z-x)(x+y+z)$. To find K, the numerical coefficient independent of x, y, z, let us equate the coefficient of $x^{3}y$ on each side; then, $K=-1$. (Alternatively, we could have substituted some numerical values for x, y, z and found K as it is an identity.) Hence, $E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)=-(x-y)(y-z)(z-x)(x+y+z)$. III. Cyclic Expressions: An algebraic expression is said to be cyclic with respect to the letters a, b, c, d, …, h, k arranged in this order when it remains the same if we replace a by b, b by c, c by d, …., h by k, and k by a. This “cycle of interchange of letters” is called the cyclic substitution denoted by $(abcd\ldots hk)$. (PS this reminds you of the the right hand unit vectors i, j, k and their cross products). Thus, the expression $a^{2}b+b^{2}c+c^{2}d+d^{2}a$ is cyclic with respect to a, b, c, and d (in this order only) because the cyclic substitution $(abcd)$ changes the first term to the second term, the second term to the third term and the fourth term to the first term. It is clear that: (i) If a term of some particular type occurs in a cyclic expression, then the term which can be derived from this by the cyclic interchange, must also occur; and, the coefficients of these terms must be equal. (ii) The sum, difference, product and quotient of two cyclic expressions is also cyclic. In writing a cyclic expression, it is unnecessary to write the whole expression or all the terms explicitly. Thus, instead of writing the full $x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)$ it suffices just to abbreviate it as $\Sigma{x^{2}(y-z)}$. (Please note that the use of $\Sigma$ here has a different meaning than earlier.) Sometimes, it is also written in short as $x^{2}(y-z)+\ldots+\ldots$. We need to be familiar with the following important basic cyclic identities: 1. $(b-c)+(c-a)+(a-b)=0$ 2. $a(b-c)+b(c-a)+c(a-b)=0$ 3. $a^{2}(b-c)+b^{2}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)$. 4. $bc(b-c)+ca(c-a)+ab(a-b)=-(b-c)(c-a)(a-b)$. 5. $a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=-(b-c)(c-a)(a-b)$. 6. $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)$. 7. $(a+b+c)(ab+bc+ca)=a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})+3abc$ 8. $(b+c)(c+a)(a+b)=a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})+2abc$ 9. $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$. 10. $(b-c)^{2}+(c-a)^{2}+(a-b)^{2}=2(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ 11. $(a+b+c)(b+c-a)(c+a-b)(a+b-c)=-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}$. Note that identity 9 can be proved by at least three non-trivial ways 🙂 PS again :-)) It helps to prove the above identities from LHS to RHS and also from RHS to LHS !! it will be proved later that i) Any symmetric function of $\alpha, \beta, \gamma$ can be expressed in terms of $\Sigma{\alpha}$, $\Sigma{\alpha\beta}$ and $\alpha\beta\gamma$. ii) Any symmetric function of $\alpha, \beta, \gamma, \delta$ can be expressed in terms of $\Sigma{\alpha}$, $\Sigma{\alpha\beta}$, $\Sigma{\alpha\beta\gamma}$, and $\alpha\beta\gamma\delta$. In the above two cases, the notation $\Sigma$ is used in the sense of a symmetric function. This mode of expression is extremely useful in factorizing symmetric functions, and in proving identities: Example 1: Factorize $a(1-b^{2})(1-c^{2}) + b(1-c^{2})(1-a^{2}) + c(1-a^{2})(1-b^{2})-4abc$. Solution 1: PS: Comment: this is not easy. But, go through it and there is ample scope to improve via exercises in the next blog 🙂 Denoting the given expression by E, we have $E=a(1-(b^{2}+c^{2})+b^{2}c^{2})+\ldots+\ldots-4abc$ $E=\Sigma{a}-\Sigma{ab^{2}}+abc\Sigma{ab}-4abc$, but from identity 7 above, we see that $\Sigma{ab^{2}}=\Sigma{a}.\Sigma{ab}-3abc$; Hence, we get $E = \Sigma{a}-\Sigma{a}.\Sigma{ab}-abc+abc\Sigma{ab}$ So, $E=\Sigma{a}.(1-\Sigma{ab})-abc(1-\Sigma{ab})=(1-bc-ca-ab)(a+b+c-abc)$. IV. Substitutiions: We consider processes by which one arrangement (permutation) of a set of elements may be transformed into another: Taking the permutations $cdba, bdac$ of a, b, c, d, the first is changed into the second by replacing a by c, b by a, c by b and leaving a unaltered. This process is represented by the operator $\left(\begin{array}{cccc}abcd\\cabd \end{array}\right)$ or $\left(\begin{array}{ccc}abc\\cab\end{array}\right)$ and, we write $\left(\begin{array}{ccc}abc\\cab\end{array}\right)cdba=bdac$. Such a process and also the operator which affects it is called a substitution. As previously stated, the interchange of two elements a, b is called the transposition $(ab)$. Also, a substitution such as $\left(\begin{array}{cccc}abcd\\bcda\end{array}\right)$ in which each letter is replaced by the one immediately following it and the last by the first, is called a cyclic substitution or cycle, and is denoted by $(abcd)$. If two operators are connected by the sign =, the meaning is that one is equivalent to the other, thus $(abcd)=(bcda)$. Two or more substitutioins may be applied successively. This is indicated as follows, the order of operations being from right to left. Let $S=(ab)$, $T=(ba)$, then $STabcd=Sacbd=bcad$ and $TSabcd=Tbacd=cabd$. Thus, $ST=\left(\begin{array}{cccc}abcd\\bcad\end{array}\right)$ and $TS=\left(\begin{array}{cccc}abcd\\cabd\end{array}\right)$ This process is called multiplication of substitutions, and the resulting substitution is called the product. Multiplication of this kind is not necessarily commutative, but if the substitutions have no common letter, it is commutative. The operation indicated by $(ab)(ab)$, in which $(ab)$ is performed twice, produces no change in the order of the letters, and is called an identical substitution. Any substituion is cyclic or is the product of two or more cyclic substitutions which have no common element. As an instance, consider the substitution $S = \left(\begin{array}{ccccccccc}abcdefghk\\chfbgaedk\end{array}\right)$ Here, a is changed to c, c to f, f to a, thus completing the cycle $(acf)$. Also, b is changed to h, h to d, d to b, making the cycle $(bhd)$. Next, c is changed to g, and g to e, giving the cycle $(eg)$. The element k is unchanged, and we write $S=(acf)(bhd)(eg)(k)$ or $S=(acf)(bhd)(eg)$. This expression for S in unique, and the order of the factors is indifferent. Moreover, the method applies universally, for in effecting any substitution, we must arrive at a stage when some letter is replaced by the first, thus completing a cycle. The same argument applies to the set of letters not contained in the cycle. A cyclic substitution of n elements is the product of $(n-1)$ transpositions: Consider $(abc)=(ab)(bc)$, $(abcd)=(abc)(cd)=(ab)(bc)(cd)$, $(abcde)=(abcd)(de)=(ab)(bc)(cd)(de)$, and so on. We also have equalities such as : $(ae)(ad)(ac)(ab)=(abcde)$ and $(ab)(ac)(ad)(ae)=(edcba)$. A substitution which deranges n letters and which is the product of r cycles is equivalent to $(n-r)$ transpositions. This follows at once from our previous work. Thus, if $S = \left(\begin{array}{cccccccc}abcdefgh\\chfbgaed\end{array}\right)$, then $S=(acf)(bhd)(eg)=(ac)(cf)(bh)(hd)(eg)$. If we introduce the product $(ab)(ab)$, S is unaltered and the number of transpositions is increased by 2. Thus, if a given substitutition is equivalent to j transpositions, the number j is not unique. We shall prove that : $j=n-r+2s$ where r is a positive integer or zero. This is a very important theorem, and to prove it we introduce the notion of “inversions.” *** Taking the elements a, b, c, d, e choose some arrangement, as $abcde$, and call it a normal arrangement. Consider the arrangement $bdeac$. Here b precedes a, but follows it in the normal arrangement. On this account, we say that the pair ba constitutes an inversion. Thus, $bdeac$ contains five inversions, namely, $ba, da, dc, ea, ec$. Theorem 1: If i is the number of inversions which are introduced or removed by a substitution which is equivalent to j transpositions, then i and j are both even or both odd. Proof of theorem 1: Consider the effect of a single transposition $(fg)$. If f, g are consecutive elements, the transposition $(fg)$ does not alter the position of f or g relative to the other elements. It therefore introduces or removes a single inversion due to the interchange of f, g. If f, g are separated by n elements p, q, r, …, x, then f can be moved to the place occupied by g by $n+1$ interchanges of consecutive elements, and then g can be moved to the place originally occupied by f by n such interchanges. Thus, the transposition $(fg)$ can be effected by $2n+1$ interchanges of consecutive elements. Therefore, any transposition introduces or removes an odd number of inversions, and the theorem follows. QED. Again, for a given substitution, i is a fixed number, and therefore whatever value j may have, it must be even or odd, according as i is even or odd. Hence, we get the following: Theorem 2: If one arrangement A of a given set of elements is changed into another B by j transpositions, then j is always even or always odd. In other words: the number of transpositions which are equivalent to a given substitution is not unique, but is always even or always odd. The minimum value of j is $n-r$. Thus, substitutions may be divided into two distinct classes. We say that a substitution is even or odd according as it is equivalent to an even or an odd number of transpositions. Rule: To determine the class of a substitution S we may express it as the product of cycles, and count the number of cycles with an even number of elements: then S is even or odd according as this number is even or odd. Or, we can settle the question by counting the number of inversions, but this generally takes longer. The tutorial exercises follow this blog. Cheers, Nalin Pithwa Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems) Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications. Problem 1: Prove that if $A \bigcup B=A$ and $A \bigcap B=A$, then $A=B$. Problem 2: Show that in general $(A-B)\bigcup B \neq A$. Problem 3: Let $A = \{ 2,4, \ldots, 2n, \ldots\}$ and $B= \{ 3,6,\ldots, 3n, \ldots\}$. Find $A \bigcap B$ and $A - B$. Problem 4: Prove that (a) $(A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)$ Prove that (b) $A \Delta B = (A \bigcup B)-(A \bigcap B)$ Problem 5: Prove that $\bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})$ Problem 6: Let $A_{n}$ be the set of all positive integers divisible by $n$. Find the sets (i) $\bigcup_{n=2}^{\infty}A_{n}$ (ii) $\bigcap_{n=2}^{\infty}A_{n}$. Problem 7: Find (i) $\bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}]$ (ii) $\bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$ Problem 8: Let $A_{\alpha}$ be the set of points lying on the curve $y=\frac{1}{x^{\alpha}}$ where $(0. What is $\bigcap_{\alpha \geq 1}A_{\alpha}$? Problem 9: Let $y=f(x) = $ for all real x, where $$ is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval $\frac{1}{4} \leq y \leq \frac{3}{4}$? Partition the real line into classes of points with the same image. Problem 10: Given a set M, let $\mathscr{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a \in M$, and let $aRb$ if and only if $(a,b) \in \mathscr{R}$. Interpret the relation R. Problem 11: Give an example of a binary relation which is: • Reflexive and symmetric, but not transitive. • Reflexive, but neither symmetric nor transitive. • Symmetric, but neither reflexive nor transitive. • Transitive, but neither reflexive nor symmetric. We will continue later, 🙂 🙂 🙂 PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also. Nalin Pithwa Set Theory, Relations, Functions: Preliminaries: part VIIIA (We continue from part VII of the same blog article series with same reference text). Theorem 4: A set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M. Proof of Theorem 4: Every partition of M determines a binary relation on M, where $aRb$ means that “a belongs to the same class as b.” It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M. Conversely, let R be an equivalence relation on M, and let $K_{a}$ be the set of all elements $x \in M$ such that $xRa$ (clearly, $a \in K_{a}$, since R is reflexive). Then, two classes $K_{a}$ and $K_{b}$ are either identical or disjoint. In fact, suppose that an element c belongs to both $K_{a}$ and $K_{b}$, so that $cRa$ and $cRb$. But by symmetry of R, being an equivalence relation, we can infer that $aRc$ also and, further by transitivity, we say that $aRb$. If now, $x \in K_{a}$ then we have $xRa$ and hence, $xRb$ (since we already have $aRb$ and using transitivity). Similarly, we can prove that $x \in K_{b}$ implies that $x \in K_{a}$. Therefore, $K_{a}=K_{b}$ if $K_{a}$ and $K_{b}$ have an element in common. Therefore, the distinct sets $K_{a}$ form a partition of M into classes. QED. Remark: Because of theorem 4, one often talks about the decomposition of a set M into equivalence classes. There is an intimate connection between mappings and partitions into classes, as illustrated by the following examples: Example 1: Let f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image $b=f(a) \in B$. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point $(x,y)$ into the point $(x,0)$. Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f. Example 2: Given any partition of a set A into classes, let B be the set of these classes and associate each element $a \in A$ with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line $[0, \infty)$ each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line $[0,\infty)$. Example 3: Suppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of “winding” the real line onto a circle of unit circumference. (Note: The largest integer $\leq x$ is called the integral part of x, denoted by [x], and the quantity $x -[x]$ is called the fractional part of x). In the next blog article, let us consider a tutorial problem set based on last two blogs of this series. 🙂 🙂 🙂 Nalin Pithwa Set Theory, Relations, Functions: Preliminaries: Part VIII SETS and FUNCTIONS: Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books. Operations on sets: Let A and B be any two sets. Then, by the sum or union of A and B, denoted by $A \bigcup B$, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets $A_{\alpha}$ (indexed by some parameter $\alpha$), we mean the set, denoted by $\bigcup_{\alpha}A_{\alpha}$ of all elements belonging to at least one of the sets $A_{\alpha}$. By the intersection $A \bigcap B$ of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets $A_{\alpha}$, we mean the set, denoted by $\bigcap_{\alpha}A_{\alpha}$ of all elements belonging to every one of the sets $A_{\alpha}$. Two sets A and B are said to be disjoint if $A \bigcap B=\phi$, that is, if they have no elements in common. More generally, let $\mathscr{F}$ be a family of sets such that $A \bigcap B=\phi$ for every pair of sets A, B in $\mathscr{F}$. Then, the sets in $\mathscr{F}$ are said to be pairwise disjoint. It is an immediate consequence of the above definitions that the operations $\bigcup$ and $\bigcap$ are commutative and associative, that is, $A \bigcup B=B \bigcup A$ and $(A \bigcup B) \bigcup C= A \bigcup (B \bigcup C)$; $A \bigcap B=B \bigcap A$; and, $(A \bigcap B)\bigcap C = A \bigcap (B \bigcap C)$. Moreover, the operations $\bigcup$ and $\bigcap$ obey the following distributive laws: $(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$….call this I $(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$….call this II. For example, suppose that $x \in (A \bigcup B)\bigcap C$, so that x belongs to the left-hand side of I. Then, x belongs to both C and $A \bigcup B$, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets $A \bigcap C$ and $B \bigcap C$, that is, $x \in (A \bigcap C)\bigcup (B \bigcap C)$, so that x belongs to the right hand side of I. Conversely, suppose that $x \in (A \bigcap C)\bigcup (B \bigcap C)$. Then, x belongs to at least one of the two sets $A \bigcap C$ and $B \bigcap C$. It follows that x belongs to both C and at least one of the two sets A and B, that is, $x \in C$ and $x \in A \bigcup B$, or equivalently $x \in (A \bigcup B)\bigcap C$. This proves I, and II is proved similarly. By the difference of two sets, $A - B$ between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that $A \supset B$. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by $A \delta B$ and defined as the union of the two differences $A-B$ and $B-A$: $A \delta B=(A-B)\bigcup (B-A)$. We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference $R-A$ is called the complement of A, denoted by $A^{'}$. An important role is played in set theory and its applications by the following duality principle: $R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})$…call this III $R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})$…call this IV. In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that $x \in R - \bigcup_{\alpha}A_{\alpha}$….call this V. Then, x does not belong to the union $\bigcup_{\alpha}A_{\alpha}$. …call this VI. That is, x does not belong to any of the sets $A_{\alpha}$. It follows that x belongs to each of the complements $R - \bigcup_{\alpha}A_{\alpha}$, and hence, $x \bigcap_{\alpha}(R-A_{\alpha})$….call this VII. Conversely, suppose that VII holds, so that x belongs to every set $R - A_{\alpha}$. Then, x does not belong to any of the sets $A_{\alpha}$, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3. Homework: Prove IV similarly. Remark: The designation “symmetric difference” for the set $A \Delta B$ is not too apt, since $A \Delta B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the two statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in $A \Delta B$ the same two statements are joined by “or” used in the ordinary “either…or…” sense (as in “to be or not to be”). In other words, x belongs to $A \bigcup B$ if and only if x belongs to either A or B or both, while x belongs to $A \Delta B$ if and only if x belongs to either A or B but not both. The set $A \Delta B$ can be regarded as a kind of “modulo-two sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B). Functions and mappings. Images and preimages: (Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) 🙂 🙂 🙂 A rule associating a unique real number $y=f(x)$ with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers $f(x)$ such that $x \in X$ is called the range of f. More generally, let M and N be two arbitrary sets. Then a rule associating a unique element $b=f(a) \in N$ with each element $a \in M$ is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b). If a is an element of M, the corresponding element $b=f(a)$ is called the image of a (under the mapping f). Every element of M with a given element $b \in N$ as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by $f^{-1}(b)$. If A is a subset of M, the set of all elements $f(a) \in N$ such that $a \in A$ is called the image of A, denoted by $f(A)$. The set of all elements of M whose images belong to a given set $B \subset N$ is called the preimage of B, denoted by $f^{-1}(B)$. If no element of B has a preimage, then $f^{-1}(B)=\phi$. A function f is said to map M into N if $f(M)=N$, as is always the case, and onto N if $f(M)=N$. Thus, every “onto” mapping is an “into” mapping, but converse is not true. Suppose f maps M onto N. Then, f is said to be one-to-one if each element $b \in N$ has a unique preimage $f^{-1}(b)$. In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping $f^{-1}$ associating $f^{-1}(b)$ is called the inverse of f. Theorem I: The preimage of the union of two sets is the union of the preimages of the sets $f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B)$. Proof of Theorem I: If $x \in f^{-1}(A \bigcup B)$, then $f(x) \in A \bigcup B$ so that $f(x)$ belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcup f^{-1}(B)$. Conversely, if $x \in f^{-1}(A) \bigcup f^{-1}(B)$, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$. Therefore, $f(x)$ belongs to at least one of the sets A and B, that is, $f(x) \in A \bigcup B$. But, then $x \in f^{-1}(A \bigcup B)$. QED. Theorem 2: The preimage of the intersection of two sets is the intersection of the preimages of the sets: $f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B)$. Proof of Theorem 2: If $x \in f^{-1}(A \bigcap B)$, then $f(x) \in A \bigcap B$, so that $f(x) \in A$ and (meaning, simultaneously) $f(x) \in B$. But, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcap f^{-1}(B)$. Conversely, if $x \in f^{-1}(A) \bigcap f^{-1}(B)$, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. Therefore, $f(x) \in A$ and $f(x) \in B$, that is, $f(x) \in A \bigcap B$. But, then $x \in f^{-1}(A \bigcap B)$. QED. Theorem 3: The image of the union of two sets equals the union of the images of the sets $f(A \bigcup B) = f(A) \bigcup f(B)$. Proof of theorem 3: If $y \in f(A \bigcup B)$, then $y=f(x)$ where x belongs to at least one of the sets A and B. Therefore, $y=f(x)$ belongs to at least one of the sets $f(A)$ and $f(B)$. That is, $x \in f(A) \bigcup f(B)$. Conversely, if $y \in f(A) \bigcup f(B)$, then $y=f(x)$. where x belongs to at least one of the sets A and B, that is, $x \in A \bigcup B$ and hence, $y=f(x) \in f(A \bigcup B)$. QED. Remark I: Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point $(x,y)$ into the $(x,0)$. Then, the segments $0 \leq x \leq 1$ with $y=0$, and $0 \leq x \leq 1$ with $y=1$ do not intersect, although their images coincide. Remark 2: In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f is well-defined. That is, unambiguously defined. For example, if the set A is the union of two subsets $A_{1}$ and $A_{2}$, and we are considering a function $f: A \longrightarrow B$, then one can try to specify a function from set A to the set B $\{ 0,1\}$ by declaring that f is to map everything in $A_{1}$ to 0 and is to map everything in $A_{2}$ to 1. This unambiguously defines f unless $A_{1}$ and $A_{2}$ have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that $A_{1}$ and $A_{2}$ have no intersection. Remark 3: Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets $A_{\alpha}$: $f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})$ $f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})$ $f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})$ Decomposition of a set into classes. Equivalence relations. Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes. A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if $b>a$. In fact, if $b>a$, b must be assigned to same class as a but then a cannot be assigned to same class as b, since $a. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1! These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If $(a,b)$ is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, $aRb$ might mean $a, while if a and b are triangles, $aRb$ might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair $(a,b)$ for every $a \in M$. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions: • Reflexivity: $aRa$ for every $a \in M$. • Symmetry: If $aRb$, then $bRa$. • Transititivity: If $aRb$ and $bRc$, then $aRc$. 🙂 🙂 🙂 Nalin Pithwa, more later Set theory, relations, functions: preliminaries: Part VII FINITE, COUNTABLE AND UNCOUNTABLE SETS: (Reference: Principles of Mathematical Analysis, Third Edition, Walter Rudin:) We begin with the function concept. Definition: Consider two sets A and B whose elements may be any objects whatsoever, and suppose that with each element x of A there is associated in some manner, an element of B, which we denote by $f(x)$. Then, f is said to be a function from A to B (or a mapping of A into B). The set A is called the domain of f (we also say f is defined on A), and the elements f(x) are called the values of f. The set of all values of f is called the range of f. Definition: Let A and B be two sets and let f be a mapping of A into B. If $E \subset A$, then $f(E)$ is defined to be the set of all elements f(x), for $x \in E$. We call $f(E)$ the image of E under f. In this notation, $f(A)$ is the range of f. It is clear that $f(A) \subset B$. If $f(A)=B$, we say that f maps A onto B. (Note that, according to this usage, onto is more specific than into). If $E \subset B$, $f^{-1}(E)$ denotes the set of all $x \in A$ such that $f(x) \in E$. We call $f^{-1}(E)$ the inverse image of E under f. If $y \in B$, then $f^{-1}(y)$ is the set of all$latex  \in A\$ such that $f(x)=y$. If, for each $y \in B$, $f^{-1}(y)$ consists of at most one element of A, then f is said to be a 1-1 (one-one) mapping of A into B. This may also be expressed as follows: f is a 1-1 mapping of A into B provided that $f(x_{1}) \neq f(x_{2})$ whenever $x_{1} \neq x_{2}$ and $x_{1} \in A$ and $x_{2} \in A$.

(The notation $x_{1} \neq x_{2}$ , means that $x_{1}$ and $x_{2}$ are distinct elements, otherwise we write $x_{1} = x_{2}$).

Definition:

If there exists a 1-1 mapping A of onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or briefly, that A and B are equivalent, and we write $A \sim B$. This relation clearly has the following properties:

It is reflexive: $A \sim A$

It is symmetric: If $A \sim B$, then $B \sim A$.

It is transitive: If $A \sim B$, and $B \sim C$, then $A \sim C$.

Any relation with these three properties is called an equivalence relation.

Definition:

For any positive integer n, let $J_{n}$ be the set whose elements are the integers $1, 2, \ldots, n$. And, let J be the set consisting of all positive integers. For any set A, we say:

(a) A is finite if $A \sim J_{n}$ for some n (the empty set is also considered to be finite).

(b) A is infinite, if A is not finite.

(c) A is countable if $A \sim J$.

(d) A is uncountable if A is neither finite nor countable.

(e) A is at most countable if A is finite or countable.

Countable sets are sometimes called enumerable, or denumerable.

For two finite sets A and B, we evidently have $A \sim B$ if and only if A and B contain the same number of elements. For infinite sets, however, the idea of “having the same number of elements” becomes quite vague, whereas the notion of 1-1 correspondence retains its clarity.

Example:

Let A be the set of all integers. Then, A is countable. For, consider the following arrangement of the sets A and J:

A: 0,1,-1,2,-2,3,-3,…

J:1,2,3,4,5,6,7,…

We can, in this example, even give an explicit formula for a function f from J to A which sets up a 1-1 correspondence:

When n is even, $f(n)=\frac{n}{2}$

When n is odd, $f(n)=-\frac{n-1}{2}$

Remark:

A finite set cannot be equivalent to one of its proper subsets. That this is, however, possible for infinite sets, is shown by the above example, in which J is a proper subset of A.

In fact, we can define infinite sets as follows: A set A is infinite if A is equivalent to one of its proper subsets.

Definition: SEQUENCES:

By a sequence, we mean a function f defined on the set J of all positive integers. If $f(n)=x_{n}$, for $n \in J_{n}$, it is customary to denote the sequence f by the symbol $\{ x_{n}\}$, or sometimes by $x_{1}, x_{2}, x_{3}, \ldots$. The values of f, that is, the elements $x_{n}$ are called the terms of the sequence. If A is a set and if $x_{n} \in A$, for all $n \in J$, then $\{ x_{n}\}$ is said to be a sequence in A, or a sequence of elements of A.

Note that the terms $x_{1}, x_{2}, x_{3}, \ldots$ of a sequence need not be distinct.

Since every countable set is the range of a 1-1 function defined on J, we may regard every countable set as the range of a sequence of distinct terms. Speaking more loosely, we may say that the elements of any countable set can be “arranged” in a sequence.

Sometimes, it is convenient to replace J in this definition by the set of all non-negative integers, that is, to start with 0 rather than with 1.

Regards,

Nalin Pithwa

PS: the above exposition will be extremely useful to RMO, INMO and IITJEE advanced maths students and students preparing for Chennai Mathematics Institute Entrance Examination.