discrete mathematics problem

To prove: The square root of 2 is irrational. In other words, there
is no rational number whose square is 2.

Proof by contradiction: Begin by assuming that the thesis is false,
that is, that there does exist a rational number whose square is 2.

By definition of a rational number, that number can be expressed in
the form c/d, where c and d are integers, and d is not zero.
Moreover, those integers, c and d, have a greatest common divisor,
and by dividing each by that GCD, we obtain an equivalent fraction
a/b that is in lowest terms: a and b are integers, b is not zero,
and a and b are relatively prime (their GCD is 1).

Now we have

[1] (a/b)^2 = 2

Multiplying both sides by b^2, we have

[2] a^2 = 2b^2

The right side is even (2 times an integer), therefore a^2 is even.
But in order for the square of a number to be even, the number
itself must be even. Therefore we can write

[3] a = 2f

Using this to replace a in [2], we obtain

[4] (2f)^2 = 2b^2
[5] 4f^2 = 2b^2
[6] 2f^2 = b^2

The left side is even, therefore b^2 must be even, and by the same
reasoning as before, b must be even. But now we have found that both
a and b are even, contradicting the assumption that a and b are
relatively prime. Therefore the assumption is incorrect, and there
must NOT be a rational number whose square is 2.
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That's my proof. Admittedly it still leaves out some steps, for
instance a proof that if the square of a number is even then the
number itself must be even. But I filled in the part that had you
confused.

-Doctor Rick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/