Can area be irrational?

Question

I'm stuck in a question of my book which says:

If in an equilateral triangle the coordinates of two vertices are integral then what can we say about the coordinates of the third?

The answer is that at least one of the coordinate of third vertex is irrational.

Now I don't know how to prove this although there is an explanation also in my book that I may not agree to. But first here is my attempt:

I have assumed an equilateral triangle with coordinates of 2 vertices as $(0,0)$ and $(2,2)$.

Now, let the 3rd vertex be $(x,y)$.

Using the fact that all sides are equal in equilateral triangle and using distance formula I get

$x^2+y^2=(x-2)^2+(y-2)^2$

Then we get $x+y=2$ so $x$ and $y$ are not irrational for sure !!!!!!

P. S= i know that the above line is wrong now but its better that is there so as to know how i attempted it...

Coming to what my book says... It says that lets assume the coordinates of 3rd vertex to be rational so when we use the area of a triangle formula if 3 vertices are given we get a rational area. But we know area of equilateral triangle is $\frac{\sqrt{3}}{4}side^2$ so the area is irrational that's not possible so one coordinate has to be irrational.

Now why can't area be irrational? Isn't it the case with a circle also? And why am I proving the wrong thing?

Please help.

Answer 1

If $ABC$ is an equilateral triangle and $A=(0,0),B=(x,y)$, then $C$ lies in: $$ C=\left(\frac{x\mp \sqrt{3}y}{2},\frac{y\pm \sqrt{3}x}{2}\right),$$ so assuming that $B$ is a point with rational coordinates, $C$ is not.

Moreover, the area of an equilateral triangle is just $\frac{\sqrt{3}}{4}l^2$ where $l$ is the length of a side.

In the same way, if the side length is a rational number, the area is not.

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With an alternative approach, if all the three vertices of an equilateral triangle have rational coordinates then the area is a rational number by the shoelace formula. But in such a case the squared length of the side is also a rational number, hence the area is at the same time a rational number and $\sqrt{3}$ times a rational number, contradiction.

Answer 2

Your counterexample doesn't work, because $x+y=2$ doesn't imply that $x$ and $y$ are rational. In fact, we have also that $x^2+y^2=8$. By substituting $y=2-x$ into the latter equation, you will get a quadratic equation with two irrational roots.

Answer 3

You should re-read the book's proof...

Briefly,

1. Assume that the third point has rational co-ordinates
2. Then by the formula you cite, the area is rational
3. But the area is an irrational number times the length of a side squared
4. But by the original assumptions, two vertices have rational coordinates, so the side length squared is also rational
5. A rational times an irrational is irrational; therefore in this case, the area is irrational
6. Thus the assumption (Point 1) leads to a contradiction (Point 2 vs Point 5)

It's Points 4 and 5 that the book fails to state explicitly...

Answer 4

Confronting the actual question asked by you, instead of that asked by your book:

Area in general CAN be irrational. The Area of that particular triangle cannot.

According to your question:

It says that lets assume the coordinates of 3rd vertex to be rational so when we use the area of a triangle formula if 3 vertices are given we get a rational area. But we know area of equilateral triangle is 3 √ 4 side 2 so the area is irrational that's not possible so one coordinate has to be irrational.

This quote mentions that a triangle with 3 rational vertices will have a rational area.

If it has a rational area, then it cannot also have an irrational area.

Thus, a triangle with 3 rational vertices cannot have an irrational area.

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(Returning to your book's given answer, it argues that an equilateral triangle with two rational vertices must have an irrational area, therefore the third vertex cannot be completely rational since that could not produce an irrational area.)

Answer 5

Let the vertices of the triangle be $A=(0,0),B=(a,b),C=(c,d)$, and label the midpoint of $AB$ as $D=(a/2,b/2)$. Then the vectors $\overset{\to}{AD}=(a/2,b/2)$ and $\overset{\to}{DC}=(c-a/2,d-b/2)$ are perpendicular, being the altitude and (half-)base of the triangle, and $|DC|=\sqrt 3|AD|$. Rotating the vector $DC$ by $90^\circ$ to get $v=(b/2-d,c-a/2)$, we see that $v=\pm\sqrt 3\overset{\to}{AD}$, while both $v$ and $AD$ are rational vectors, so $\sqrt 3=\left|\dfrac{b/2-d}{a/2}\right|=\left|\dfrac{c-a/2}{b/2}\right|$ are two rational decompositions of $\sqrt 3$, either one of which is a contradiction. Similarly, the area of the triangle is $\rho=\frac{\sqrt 3}2|(a,b)|^2$, which yields the rational decomposition $\sqrt 3=\frac{2\rho}{|(a,b)|^2}$ under the assumption that $A$ is rational.