prove that root 2 is an irrational number
Answers
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.
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 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.
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