Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 litres of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that is also a prime number. Then, the first number theorist would pipette out 1/p litres of chemical and the second litres. How many times do they have to play this game to empty the flask completely?
A bit of real analysis is required.
I will publish the reference when I post the solution. So that all students/readers can curb their impulse to see the solution immediately!!!
I hope you will be hooked to the problem in a second….!!! Here is a beautiful utility of pure math! 🙂
PS: I do not know if the above problem did (or, will?? )appear as RMO question. It is just my wild fun with math to kindle the intellect of students in analysis !! 🙂
In the Feb 23 2018 blog problem, we posed the following question:
Sum the following infinite series:
The sum can be written as:
, where .
Thus, . This is the answer.
If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.
Young philosopher Leibnitz went to Huygens for training in mathematics. Huygens’ asked Leibnitz to find the sum of the following infinite series:
Question: Calculate the sum.
Kindly share your ideas, even partial solutions are welcome.
We call a sequence a Cauchy sequence if for all there exists an such that for all m, .
Every Cauchy sequence is a bounded sequence and is convergent.
By definition, for all there is an such that
for all m, .
So, in particular, for all , that is,
for all .
Let and .
It is clear that , for all .
We now prove that such a sequence is convergent. Let and . Since any Cauchy sequence is bounded,
But since is Cauchy, for every there is an such that
for all .
which implies that . Thus, for all . This is possible only if .
Thus, we have established that the Cauchy criterion is both a necessary and sufficient criterion of convergence of a sequence. We state a few more results without proofs (exercises).
For sequences and .
(i) If and , then too is convergent and .
(ii) If , then , .
(v) If and are both convergent, then , , and are convergent and we have , and .
(vi) If , are convergent and , then is convergent and .
Reference: Understanding Mathematics by Sinha, Karandikar et al. I have used this reference for all the previous articles on series and sequences.
Sequences of integers are a favourite of olympiad problem writers since such sequences involve several different mathematical concepts, including for example, algebraic techniques, recursive relations, divisibility and primality.
Consider the sequence defined by , and
for all . Prove that all the terms of the sequence are positive integers.
There is no magic or sure shot or short cut formula to such problems. All I say is the more you read, the more rich your imagination, the more you try to solve on your own.
Replacing n by yields, and we obtain
This is equivalent to
or for all . If n is even, we obtain
while if n is odd,
It follows that , if n is even,
and , if n is odd.
An inductive argument shows that all are positive integers.
Given and , we can find such that .
Let . If , then and hence . On the other hand, if , then . Thus, A is non-empty. Next, observe that if , then v is an upper bound of A. In particular, is an upper bound of A.
By the least upper bound property, A admits a least upper bound. Let us denote it by y. We will rule out the possibilities and implying that .
If , let and . It can be checked that so that . But, , contradicting the fact that y is the least upper bound of A. (we have used the inequalities 7 and 8 in the previous blog).
On the other hand, if , let and . Again, it can be verified that and hence w is an upper bound of A. But, , contradicting the fact that y is the least upper bound of A. Hence, . QED.
In particular, we see that there is an element such that and hence also which means that the equation has two solutions. The positive one of those two solutions is . In fact, the above theorem has guaranteed its extraction of the square root, cube root, even nth root of any positive number. You could ask at this stage, if this guarantees us extraction of square root of a negative number. The answer is clearly no. Indeed, we have
We can further extend to include numbers whose squares are negative, which you know leads to the idea of complex numbers.
We have shown that Q is a subset of . One can also show that between any two distinct real numbers there is a rational number. This naturally leads to the decimal representation of real number: Given any real number x and any , we can get a unique and unique such that and
You are invited to try to prove this familiar decimal representation.
If we have a terminating decimal representation of a real number, then surely it is rational.But, we know that rationals like 1/3, 1/7, 1/11, do not have a terminating decimal expansion.
It is clear that the decimal representation of cannot terminate as it is not rational. There are many elements of which are not in Q. Any element of which is not in Q is called an irrational number, and irrational numbers cannot have terminating decimal representation.