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Date: 09/25/97 at 21:40:52
From: Sara Price
Subject: Geometric Proofs

Prove that no isoceles right triangle exists which has all three
sides integers.

I have tried to prove it by determining that this is a 45-45-90
triangle, allowing the legs to = x and the hypotenuse to equal
(x)(square root of 2), then proving that the square root of 2 is not
an integer, and that there is never a time when x is an integer and
can be multiplied by the square root of two and equal an integer.

How do you write this as a proof though?
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Date: 09/25/97 at 22:36:25
From: Doctor Pete
Subject: Re: Geometric Proofs

Hi,

Well, it seems that you've done essentially what is a sketch of the
will understand the general principles of the proof, but will not
necessarily be able to fill in the details.

To write a proof, you need to be rigorous and actually provide some
sort of mathematics to support your claims; for example, you say that
you have determined that Sqrt[2] is not an integer, but how?  Also,
you must do more than prove Sqrt[2] is not an integer; more precisely,
you must prove Sqrt[2] is *irrational*, because while 5/7 is not an
integer, multiplying it by 7 makes it an integer. Therefore, proving
Sqrt[2] is irrational automatically proves that no integer multiple of
Sqrt[2] is an integer.

Now, I would advise that you go about determining that the length of
the hypotenuse is Sqrt[2] times the length of a leg in a different
way; namely, by the Pythagorean Theorem. Since the triangle is
isoceles, it follows that the length of the two legs must be equal
(why must it be the two legs that are equal, and not a leg and
hypotenuse?). Therefore, if we let a leg have length x, the
Pythagorean Theorem states that c^2 = x^2 + x^2 = 2x^2; hence
c = Sqrt[2]x.  This way you need not deal with angles.

The rest of the proof, then, relies on showing Sqrt[2] is irrational.
The proof, which is a classic, is by contradiction. Suppose Sqrt[2]
is rational; therefore, it is expressible as some fraction p/q,
where p and q are relatively prime (greatest common divisor is 1).
In other words, Sqrt[2] = p/q, or squaring both sides, 2 = p^2/q^2.
Hence 2q^2 = p^2.

Now, since p and q are integers, p must be divisible by 2, because the
lefthand side is divisible by 2. Therefore p = 2r, for some integer r.
So 2q^2 = (2r)^2 = 4r^2, or q^2 = 2r^2. Then similarly, q must be
divisible by 2, for the righthand side is now an integer multiple
of 2. But this contradicts our assumption that p and q are relatively
prime. Therefore no such p and q exist, and Sqrt[2] is irrational.

Since Sqrt[2] is irrational, the hypotenuse Sqrt[2]x can never be an
integer for any integer value of x; therefore, no such triangle
exists.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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