A calculator display shows that  = 1.414213562, and  This suggests that  is a rational num...

A calculator display shows that  = 1.414213562, and  This suggests that  is a rational number, which contradicts Theorem. Explain the discrepancy.

Theorem

Irrationality of

 is irrational.

Proof:

[We take the negation and suppose it to be true.] Suppose not. That is, suppose  is rational. Then there are integers m and n with no common factors such that

Equation 1

[by dividing m and n by any common factors if necessary]. [We must derive a contradiction.]

Squaring both sides of equation 1 gives

Equation 2

Or, equivalently,

m2 = 2n2.

Note that equation 2 implies that m2 is even (by definition of even). It follows that m is even (by Proposition). We file this fact away for future reference and also deduce (by definition of even) that

Equation 3

m = 2k for some integer k.

Substituting equation 3 into equation 2, we see that

m2 = (2k)2 = 4k2 = 2n2.

Dividing both sides of the right-most equation by 2 gives

n2 = 2k2.

Consequently, n2 is even, and so n is even (by Proposition). But we also know that m is even. [This is the fact we filed away.] Hence both m and n have a common factor of 2. But this contradicts the supposition that m and n have no common factors. [Hence the supposition is false and so the theorem is true.]

Proposition

For all integers n, if n2 is even then n is even.

Proof (by contraposition):

Suppose n is any odd integer. [We must show that n2 is odd.] By definition of odd, n = 2k + 1 for some integer k. By substitution and algebra,

n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k)+ 1.

But 2k2 + 2k is an integer because products and sums of integers are integers. So n2 = 2•(an integer) + 1, and thus, by definition of odd, n2 is odd [as was to be shown].