How can we conclude that "root"2 is irrational

Asked by Mani Bharathi | 19th Apr, 2013, 07:26: AM

Expert Answer:

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 assume that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in its simplest terms, both a andb must be not be even. One or both must be odd. Otherwise, you could simplify.

From the equality ?2  = a/b   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.

Answered by  | 19th Apr, 2013, 08:03: AM

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