Proving that the ratio of the hypotenuse of an isosceles right triangle to the leg is irrational

Question

In an isosceles right angled triangle, vertex opposite to the hypotenuse which has $90^\circ$ is always on the perpendicular bisector of the hypotenuse.

So my question is, if we consider a sides of length $a$ and hypotenuse as $c$, then how to prove that length of segment $c$ is not a rational multiple of length of segment $a$.

Sum of lengths of two sides of a traingle is always greater than third side. So we have $2a>c$ so $c/a<2$ and $a<c$ so $1<c/a$, or $1<c/a<2$. But I unable to go any further.

Answer 1

Ancient Greeks were well aware that certain length ratios they could easily construct had to be irrational. One such ratio was, indeed, the hypotenuse/side ratio in an isosceles right triangle.

This ratio can be proved irrational by contradicting the existence of a fraction in lowest terms. For suppose such a fraction $p/q$ were to exist. 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 $2q/p$. Therefore $p/q=2q/p$, and both are then equal to $(2q-p)/(p-q)$.

But $p>q$ 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. 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.

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. That such quantities could be rendered beyond our ability to measure them exactly is a remarkable feature of elementary geometry.

Answer 2

By the Pythagorean Theorem, $c^2=a^2+a^2=2a^2$. Simplifying gives $c=\sqrt2 a$. Hence, we have the conclusion, since $\sqrt2$ is irrational.

Answer 3

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". 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: https://www.youtube.com/watch?v=sDfzCIWpS7Q&t=712s&ab_channel=Mathologer.

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:

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.

Now copy this picture and rotate it $45^\circ$ so the long diagonal is horizontal:

We can now superimpose (put $F_1'$ on top of $A$, and make the points $F_1', K_1, G_1'$ form a horizontal line like they already do in the above picture; using (1) above we know that then $K_1$ and $G_1'$ lie on integer lattice points):

By construction (we rotated the left picture above to make it into the right picture, so everything is congruent) the triangles $\triangle I_3' F_1 G_1$ and $\triangle DAC$ are congruent (to each other, and also to $T$), as are the 2 smaller (similar to the original $T$) triangles $\triangle C'B'E$ and $\triangle G'' G_1' G_1$. $\color{red}{\text{IF we can show $G''=C$, then}}$ this smaller similar triangle is an isosceles right triangle that can be put on the integer 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, exactly like the original triangle $T$ above!!!

This means we can repeat the entire process, and end up with smaller and smaller triangles whose vertices are all on the integer lattice grid, which results in a contradiction because the points in the integer lattice grid don't become arbitrarily close together.

$\color{red}{\text{It remains to show the red "IF" above.}}$ First observe that $A=F_1', F_1, D, I_3'$ are all co-linear (all on the $45^\circ$ diagonal), and that line is parallel to line segment $\overline{G'' G_1'}$. So, "sliding" the triangle $\triangle I_3' F_1 G_1$ down the "ramp" $\overline{G'' G_1'}$ (more rigorously, translating it by the vector $\vec{G_1' G''}$), we get that $G_1'$ lands on $G''$, and because the quadrilateral $\square F_1' F_1 G_1' G''$ is a parallelogram (look at 2nd picture above to avoid the clutter of the 3rd and last picture), $F_1$ lands on $A=F_1'$; and moreover the line segment $\overline{F_1 G_1}$ remains horizontal (overlaying the line $\overline{AC}$) and the line segment $\overline{F_1 I_3'}$ remains $45^\circ$ (overlaying the line $\overline{AD}$).

So, thinking of $A=F_1'$ as the origin, the previous paragraph shows that the "slid-down" version of the triangle $\triangle I_3' F_1 G_1$ (let's call it $\triangle ?F_1'G''$ since we know $F_1 \mapsto F_1', G_1 \mapsto G''$, and $I_3'\mapsto ?$) must be a dilation of $\triangle DAC$. But $\triangle I_3' F_1 G_1$ and $\triangle DAC$ are both congruent to the original $T$, hence to each other, and the only way dilates can be congruent is if they are literally, exactly equal (point for point) $\triangle ?F_1'G'' = \triangle DAC$, $\color{red}{\text{so indeed $C=G''$ as desired}}$.

P.S. the interactive Geogebra link for the illustrations is https://www.geogebra.org/geometry/a4b9xnwg.