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Math and science::Analysis::Tao::05. The real numbers

# The irrationality of $$\sqrt(2)$$

Consider a rational number fully reduced, $$\frac{p}{q}$$ such that $$\frac{p^2}{q^2} = 2$$. Investigate whether $$p$$ or $$q$$ are odd or even numbers leads to a contradiction which shows that $$2$$ cannot be rational.

• $$p$$ and $$q$$ can't both be even, as otherwise they could be simplified.
• $$p$$ and $$q$$ must both be odd, as an odd $$\times$$ odd is odd, so if $$p$$ is odd, then $$q$$ must be odd also, as $$p^2$$ (odd) will divide by $$q^2$$ to produce 2 only if $$q^2$$ is also odd. Similarly, if \)q\) is odd, so too must $$p$$ be.
• $$p$$ is even as $$p^2$$ is even. This can be seen by rearranging the original expression to get $$p^2 = 2q^2$$.