# Proof

(related to Proposition: Discovery of Irrational Numbers)

• According to the Pythagorean theorem, given a square with the side $$q$$ for some integer $$q\in\mathbb N$$, the length of the diagonal $$p$$ is given by $p=\sqrt 2 q,~\quad\text{or}\quad\sqrt 2=\frac pq.$
• Assume that $$\frac pq$$ is a rational number $$\frac pq\in\mathbb Q$$ and that the positive integers $$p$$ (numerator) and $$q$$ (denominator) have no common prime factors (otherwise cancel out these prime factors to get the reduced fraction).
• Given these numbers, we can construct a smaller triangle (shaded) with the sides $$2q-p$$, $$p-q$$, $$p-q$$, like shown in the below picture.

• Since both triangles are similar, we conclude that $\sqrt 2=\frac pq=\frac {2q-p}{p-q}.$
• Since the numerator and the denominator of $$\frac {2q-p}{p-q}$$ are smaller then the numerator and the denominator of $$\frac pq$$, the natural numbers $$2q-p$$ and $$p$$ as well as $$p-q$$ and $$q$$ have to have different factorizations.
• But this is a conctradiction to how the fraction $$\frac pq$$ was constructed (i.e. there were no more common prime factors we could cancel out from $$p$$ and $$q$$).
• On the other hand, only by such a cancellation, we would preserve the ratio of $$\sqrt 2$$. Therefore, our assumption that $$p$$ is a whole number must be wrong.
• Consequently, $$\sqrt 2$$ must be irrational.

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### References

#### Bibliography

1. Alsina, Claudi; Nelsen, Roger B.: "Bezaubernde Beweise", Springer Spektrum, 2013