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Topic: Sloping Squares (Read 488 times) |
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Sir Col
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A sloping square is drawn on the diagonal of a right angle triangle with integral length sides, a and b. A second sloping square is drawn on the diagonal of the first sloping square.
Let c be the length of the first sloping square and d be the length of the second sloping square.
Prove that c and d can never both be rational.
For which values of a and b is either c or d rational?
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Guest
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Well, I have a solution, but I think you are looking for something more elegant.
part 1: prove that "c" and "d" cannot both be rational numbers
Given : SQRT(2) is an irrational number.
Given : Irrational number * Rational number = Irrational number
For any square with sides of length "c" the diagonal "d" is always going to be SQRT(2)c.
c[sup2] + c[sup2] = d[sup2]
2c[sup2] = d[sup2]
SQRT(2)c = d
therefore :
d = [Sqrt(2) * c] = [irrational number * rational number] = [irrational number]
So, if c is a rational number then d must be an irrational number.
part 2: when is "c" a rational number
c[sup2] = a[sup2] + b[sup2]
c = SQRT(a[sup2] + b[sup2])
therefore,
if SQRT(a[sup2] + b[sup2]) is a rational number, then c = rational number
part 3: when is "d" a rational number
d^2 = (a+b)[sup2] + (a-b)[sup2]
d^2 = (a[sup2] + 2ab + b[sup2]) + (a[sup2] -2ab + b[sup2])
d^2 = 2a[sup2] + 2b[sup2]
d = sqrt(2a[sup2] + 2b[sup2])
if SQRT((a[sup2] + b[sup2])/2) is a rational number, then d = rational number
Perhaps there is another way of simplifying the criteria for parts 2 & 3.
test x[sup2] = x^2
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Icarus
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Re: Sloping Squares
« Reply #2 on: Nov 10th, 2003, 5:57pm » |
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Triples (a, b, c) of integers which satisfy the pythagorean equation a2 + b2 = c2 are called pythagorean triples, and as Guest has guessed, there is a nice way of generating them. But since I learned this a long time ago, I will leave it to see if someone unfamiliar with the result can come with the answer for themselves.
I will say that it is fairly easy to see that since a and b are restricted to integers, if c is rational, it will be an integer. Similarly, if d is rational it must be also be an integer (this one is harder to show than for c, but only slightly).
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