SOLUTION: prove that, square root of 2 is not a rational number.

Question 485329: prove that, square root of 2 is not a rational number.
Found 2 solutions by richard1234, MathLover1:
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Assume that sqrt(2) is a rational number, i.e.

where a and b are integers and a/b is irreducible. Squaring both sides,

this implies that since the LHS is even, then the RHS is also even, and a is a multiple of 2. We can write 2k instead of a:

Similarly, we can write b as 2m for some integer m. This contradicts our original statements, because this would imply a = 2k, b = 2m and they are not in simplest form (plus, we can apply this technique infinitely many times -- also not good). Hence we have a contradiction and sqrt(2) is irrational.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

The proof that is :

Let's were a number. Then we can write it where , are numbers, .
We additionally make it so that this is simplified to the lowest terms, since that can obviously be done with any fraction.
It follows that , or . So the square of is an since it is .
From this we can know that itself is also an number. Why? Because it can't be ; if itself was , then would be . Odd number times odd number is always odd.
-if itself is an number, then is some other whole number, or where is this other number. We don't need to know exactly what is; it won't matter. Soon is coming the :
If we substitute into the original equation , this is what we get:

This means is , from which follows again that itself is an !
WHY is that a ? Because we started the whole process saying that is simplified to the lowest terms, and now it turns out that and would be . So be .
conclusion: to be .
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