Question

prove by the method of controdiction that √2 is irrational _____HELP!!!

Answer 1

Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a , b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a 2 / b 2 , or a 2 = 2 · b 2 . So the square of a is an even number since it is two times something.
From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 =
a 2 /b 2 , this is what we get:
2 = (2k) 2 /b 2
2 = 4k 2 /b 2
2*b 2 = 4k 2
b 2 = 2k 2
This means that b 2 is even, from which follows again that b itself is even. And that is a contradiction!!!

Answer 2

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