Prove that the square root of 2 is irrational.

We will use the well ordering principle for this proof and try to reach a contradiction.

Let A be a set in the natural numbers so that there exists b and a in the natural numbers so that sqrt(2)=(a)/(b)

thus 2 =(a^2)/(b^2)

and (a^2)= 2(b^2)

From the formula above we see that a must be an even number because it is equal to 2(b^2). If we substitute 2k for a to represent this we get:

(2k^2) = 2 (b^2)

(b^2) = 4(k^2) / 2 = 2(k^2)

so

b = 2k

And so b also must be even, however if both a and b are smallest elements in the set A, then one must be even and one must be odd so we have a contradiction and the square root of 2 cannot be expressed as the ratio of two numbers, thus it is irrational.

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