![]() ![]() Now copy this picture and rotate it $45^\circ$ so the long diagonal is horizontal: Oscar Lanzi then sees that there is a triangle that can be formed with vertices all on the integer lattice points, similar to the original one, but smaller: the triangle $\triangle C'B'E$ above. More explicit explanation of Oscar Lanzi's proof in terms of geometry and lattice points: First you suppose that an isosceles right triangle $T$ can be put on the integer lattice grid in 2 ways: (1) where the legs are vertical and horizontal and all the vertices are on the integer lattice points, and (2) where the diagonal is horizontal and the endpoints of the diagonal are on the integer lattice points: in the image below, the points $A, B, B', C, C'$ are all on integer lattice points: ![]() See this beautiful Mathologer video "What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented" for other examples of this proof technique. Just a fun note: the heart of user Oscar Lanzi's answer can be phrased as "a square integer lattice grid can not have infinitely shrinking shapes". That such quantities could be rendered beyond our ability to measure them exactly is a remarkable feature of elementary geometry. ![]() Similar arguments could be applied to what we now call the golden ratio (which can be constructed from a right triangle whose legs are in the ratio $2:1$) and the ratio of the legs in the right triangle obtained by dividing an equiliateral triangle along its mirror plane. So the denominator $p-q$ is positive and less than $p$, while the numerator $2q-p$ is also a whole number, contradicting the requirement that $p/q$ was to be in lowest terms. Therefore $p/q=2q/p$, and both are then equal to $(2q-p)/(p-q)$.īut $p>q$ since $p$ is the hypotenuse of the right triangle and $q$ is a leg, and $p<2q$ from the hypotenuse being shorter than the path between its endpoints via the legs. Then by adjoining the original triangle with a congruent one sharing a leg, forming a similar larger right triangle, the same ratio would be rendered as $2q/p$. For suppose such a fraction $p/q$ were to exist. This ratio can be proved irrational by contradicting the existence of a fraction in lowest terms. One such ratio was, indeed, the hypotenuse/side ratio in an isosceles right triangle. Ancient Greeks were well aware that certain length ratios they could easily construct had to be irrational. ![]()
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