Proof by ContradictionDate: 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? 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 proof. In other words, any math-literate person reading your message 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/