Sunday, February 17, 2008

That √2 is irrational

Rudin's text on mathematical analysis provides a proof that √2 is irrational by showing that the set of rational numbers A consisting of all rational numbers whose square is greater than 2 has no least element and the set B of all rational numbers whose square is lesser than 2 has no greatest element. He considers A first and shows that for any p in A, we can find
a number q that is lesser than p and yet in A. He conjures up from apparently nowhere the rational number

q = p – (p^2 - 2)/(p + 2) ……….(1)

The number q is immediately seen to satisfy the needs of the proof for the set A.

How did he arrive at (1)? Some algebra shows that it does not matter what denominator is chosen in (1) as long as it is a rational greater than p + √2. So this could have been 2p for example. This it turns out is sufficient for us to work up a number such as q in (1) ourselves.

Since p > √2, we expect 2/p < √2. So √2 and hence the rational q we are searching for lies between p and 2/p, and of course q should be greater than √2. But first let us find out by what ratio the number √2 divides the segment (2/p,p). Let us say the ratio is µ. Then


√2 = μ 2/p + (1 - μ) p = μ (2/p-p) + p……….(2)

From which, we get = p / (√2 + p). The above equation suggests that as μ varies between 0 and p / (√2 + p), the number μ 2/p + (1 - μ) p spans the segment (√2,p), which is exactly the interval in which we are looking for rational numbers q. We now can find q, by choosing a appropriate value for μ. Now since p >
√n, we can choose μ = p / (p + p) < p/(√2 + p), so we have q = p/2 + 2/2p = p/2 + 1/p.

How close is this number to the in (1)?. As I said earlier in the post this would be the same number if we replaced the denominator in (1) by (p+p) instead of p + 2.


 


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