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Prove that root 2 is a irrational number. |
| Answer» 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 = a2/b2, or a2 = 2 · b2. 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 = a2/b2, this is what we get:2\t=\t(2k)2/b22\t=\t4k2/b22*b2\t=\t4k2b2\t=\t2k2This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!! | |