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The proof that square root of 2 is irrational:

 

Let's suppose √2 were a rational number.  Then we can write it √2  = a/b where a, b are whole numbers, b not zero.  We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction.

 

It follows that 2 = a²/b²,  or  a² = 2 * b².  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 * a would be odd too.  Odd number times odd number is always odd.  Check if you don't believe that!

 

Okay, 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 = a²/b², this is what we get:

2 = (2k)²/b²

2 = 4k²/b²

2*b² = 4k²

b² = 2k².

This means b² 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/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.