The area of a right triangle whose sides are integers in not a square implies ===>

• Jul 1st 2010, 05:38 PM
Dark Sun
The area of a right triangle whose sides are integers in not a square implies ===>
Hello,

I have been trying to show that the fact that S1="The area of a right triangle whose sides are integers in not a square" implies that S2="There is no triangle whose sides are rational and whose area is 1".

So far I have that if we assume S1, and also consider an arbitrary triangle whose sides are rational and whose area is 1, then we have that:

$\frac{1}{2}\frac{x}{y}\cdot h=1\Rightarrow h=\frac{2y}{x},$ which shows that the height of one of these such triangles must be rational. Then, if we let $h=a/b$, we have that:

$\frac{1}{2}\frac{x}{y}\frac{a}{b}=1\Rightarrow \frac{1}{2}ax=by$

I have also considered the fact that our triangle is two right triangles put together, and that our triangle may be enclosed within a rectangle, but so far nothing has panned out.

The book treats this implication as trivial, saying that "if such a triangle existed, we would be able to obtain, by multiplying all three sides by a suitable integer, a triangle whose sides are integers and whose area is a square..." And I can see this, but I cannot see that this is a right triangle, as is needed to contradict S1.
• Jul 1st 2010, 07:15 PM
tonio
Quote:

Originally Posted by Dark Sun
Hello,

I have been trying to show that the fact that S1="The area of a right triangle whose sides are integers in not a square" implies that S2="There is no triangle whose sides are rational and whose area is 1".

So far I have that if we assume S1, and also consider an arbitrary triangle whose sides are rational and whose area is 1, then we have that:

$\frac{1}{2}\frac{x}{y}\cdot h=1\Rightarrow h=\frac{2y}{x},$ which shows that the height of one of these such triangles must be rational. Then, if we let $h=a/b$, we have that:

$\frac{1}{2}\frac{x}{y}\frac{a}{b}=1\Rightarrow \frac{1}{2}ax=by$

I have also considered the fact that our triangle is two right triangles put together, and that our triangle may be enclosed within a rectangle, but so far nothing has panned out.

The book treats this implication as trivial, saying that "if such a triangle existed, we would be able to obtain, by multiplying all three sides by a suitable integer, a triangle whose sides are integers and whose area is a square..." And I can see this, but I cannot see that this is a right triangle, as is needed to contradict S1.

So you've already a triangle ABC whose sides are integers and whose area is a square; now form the right angle with legs A,H (H = the height to A in ABC) and voila!

Tonio
• Jul 1st 2010, 07:39 PM
Dark Sun
Hi Tonio,

Assuming that ABC has Area=x^2, for x an integer, How do we know that the right triangle formed in this way (with the height of ABC as one of the legs) also has area that is a square?
• Jul 1st 2010, 08:31 PM
tonio
Quote:

Originally Posted by Dark Sun
Hi Tonio,

Assuming that ABC has Area=x^2, for x an integer, How do we know that the right triangle formed in this way (with the height of ABC as one of the legs) also has area that is a square?

Because you're forming the right triangle with one of the triangle's sides and its height as legs...! The area of a right triangle is leg times leg times 1/2, and this makes it sure the right triangle has the same area as the first one, and there you've your contradiction.

Tonio
• Jul 7th 2010, 03:56 PM
Dark Sun
I'm sorry, perhaps I am missing something. I drew some figures to illustrate my confusion:

http://i942.photobucket.com/albums/a...s/SDC10902.jpg

Since in order to have a contradiction, we need that all three sides of our triangle be integers.
• Jul 22nd 2010, 03:35 PM
Dark Sun
I talked to my professor, and after much deliberating, he found a counter-example on Wikipedia, confirming that there was a typo in the book. This theorem only applies to right triangles. The word "right" is missing from the book.