prove that there is no rational number whose square is 6
Answers
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To prove that there is no rational number whose square is 6, we can prove that √6 is not a rational number.
On the contradiction, Let's suppose √6 is a rational number. Then we can write it √6 = 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 √6 = a/b it follows that 6 = a^2/b^2, or a^2 = 6 · b^2 or a^2 = 2. 3 · 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.
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 6 = a^2/b^2, this is what we get:
6 = (2k)^2/b^2
6 = 4k^2/b^2
2*3*b^2 = 4k^2
3* 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!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √6 is rational) is not correct. Therefore √6 cannot be rational.