My math teacher berated me for doing something in comp. sci. involving math that I hadn't proved (Lummox eventually proved it on the forums), even though I was rather sure it was true (and new it was in the scale we were working). Out of spite, I tried to prove it myself, and this is what I came up with:
a2 + (ax)2 = c2
a2 + a2x2 = c2
a2 + a2x2 - c2 = 0
a2x2 - c2 = a2
x2 - c2 = -1
c2 - 1 = x2
c and x will never be whole unless c = 1. Reasoning:
f(x) = (x + 1)2 - x2 = 2x - 1
As x increases, f(x) > f(x - 1) where x > 0, meaning there are no perfect squares where x2 - 1 is also a perfect squiare, save for 1.
Therefore, in a2 + (ax)2 = c2, a, c, and x can never be whole numbers, unless one side of the triangle has a length of 0, meaning it is not a triangle. This means there will never be a Pythagorean triple with one leg as a multiple of the other.
So, where did I screw up my math?
ID:182556
 Apr 28 2008, 6:05 pm (Edited on Apr 28 2008, 6:10 pm)
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Er, whoops, yea. The question is, is there ever a Pythagorean triple where one leg is a multiple of the other? x, in this case, would be the multiple. This is also the first time I have ever tried to write a proof, and I suck at math, so it's no surprise I screwed up there so easily.
Well, back to trying to figure out a way to prove it. | |
Alright, here's a second attempt:
a2 + (ax)2 = c2 a2 + a2x2 = c2 a2x2 = c2 - a2 x2 = (c2 - a2) / a2 x = √((c2 - a2) / a2) The fraction under the radical can never be a whole number, as the numerator will always be less than the denominator, and c2 will always be greater than a2 (because the hypotenuse is always longer than either of the legs). x will never be a whole number here because the square root of a number is always less than the number, unless x = 1 or 0. If x = 0, then the shape isn't a triangle. If x = 1, then triangle will be a 45°-45°-90° triangle, and the hypotenuse will be a multiple of √2, meaning it is not whole. Therefore, a Pythagorean triple with one leg as a multiple of the other can not exist. Did I screw up at all this time? | |
Popisfizzy wrote:
The fraction under the radical can never be a whole number, as the numerator will always be less than the denominator, and c2 will always be greater than a2 (because the hypotenuse is always longer than either of the legs). The numerator will not always be less than the denominator. Whenever x>2y, x-y>y and (x-y)/y > 1. Consider the situation where a=1 and c=3. x = sqrt((c2-a2)/a2) = (9-1)/1 = 8. That does not mean b=xa=sqrt(8)*1 is an integer, since it's obviously not, but it does contradict part of your attempted proof. x will never be a whole number here because the square root of a number is always less than the number, The square root of an integer is always less than the original number. Just because x is an integer and y is an integer does not mean that x/y is an integer. Where 0 < x < 1, sqrt(x) > x, which will be the situation in your case when c2<2a2, though in such a case the square root of the number will itself be between 0 and 1 as well, meaning that x is not an integer and therefor not appropriate to your search for multiples; however, you still need to show this as a case in your proof for it to be valid. I got points taken off in many of my math classes for not taking my work far enough. | |
Alright, thanks. The two cases you showed can be handled in it, as the first Pythagorean triple is 3-4-5, meaning anything divided by it one of the legs will not be whole (for what I stated in my previous post), and, as you showed for the second case, it won't work. I'll include both in this. As neither of what you pointed out nullifies what I already have, is there no fundamental errors that make the proof invalid?
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If a2+b2=c2 is your triangle, and b is the smaller leg, then the question is whether, if a and b are integers, c is also an integer. It cannot be.
The proof falls out right from the beginning, where you need to reduce the triangle to its smallest form by taking out any common factors. That is, if some number n divides all of a, b, and c, you can divide the whole equation by n2. Obviously if there's a triple a,b,c satisfying the relationship we want, then na,nb,nc would also. At this point, a, b, and c are relatively prime. If a and b had any common factors, those would also be expressed in c, so they would have to be part of n and already eliminated. For the same reason, no common factors can exist between a and c, or b and c. Since these numbers are relatively prime, then the only way c can be a multiple of b--since it shares no factors--is if b=1. For this to be true, a must be some number such that a2 and a2+1 are both square. Since a is positive, a2<a2+1≤(a+1)2. Since there are no squares between a2 and (a+1)2 because a is an integer, a2+1=(a+1)2 which solves to a=0. This is illegal since at the outset it was established by convention that a>b, and also this is a degenerate case. Lummox JR | |
(a2x2 - c2) / a2 = x2 - c2 / a2.
I don't know what the hell you're trying to prove with all this, though, so I'm having trouble following what the hell you're talking about. What the crap is x in this case?
Oh:
f(x) = (x + 1)2 - x2 = 2x - 1
This is also wrong. That should be 2x + 1.