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

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

The proof that sqrt%282%29 is irrational:

Let's suppose sqrt%282%29 were a rational number. Then we can write it sqrt%282%29=a%2Fb+ where a, b are whole numbers, b+not+zero.
We additionally make it so that this a%2Fb is simplified to the lowest terms, since that can obviously be done with any fraction.
It follows that 2+=+a%5E2%2Fb%5E2, or +a%5E2+=+2+%2A+b%5E2. So the square of a is an even number since it is two times+something.
From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then+a+%2A+a would be odd+too. Odd number times odd number is always odd.
-if a itself is an even number, then a is 2+times some other whole number, or a+=+2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:
If we substitute a+=+2k into the original equation 2+=+a%5E2%2Fb%5E2, this is what we get:
2=+%282k%29%5E2%2Fb%5E2
2=+%094k%5E2%2Fb%5E2
2%2Ab%5E2=+4k%5E2
b%5E2%09=+2k%5E2
This means b%5E2 is even, from which follows again that b itself is an even+number!
WHY is that a contradiction? Because we started the whole process saying that a%2Fb is simplified to the lowest terms, and now it turns out that a and b would both be even. So sqrt%282%29 CANNOT be rational.
conclusion: sqrt%282%29 HAS to be irrational.