Hi. I'm trying to prove something, and I don't really know if this form is considered valid.
Suppose you have a right triangle, with hypotenuse z and the other two legs designated as x and y. Using the fact that the area A=(z^2)/4 , I want to prove that x=y, aka that the triangle is isosceles. I will also end up using the fact that the area A=(1/2)xy and the Pythagorean theorem, z^2 = x^2 + y^2.
My problem is that in my proof, i take the identity (1/2)xy=(z^2)/4, and do a handful of operations to either side of it, which eventually transform it to x=y. I am a first year undergraduate, and don't have much experience with proofs, but I'm taking a course on discrete math that's supposed to serve as an introduction to proofs, among other things, and all the direct proofs I've seen, and the explanation in my text, are such that you take one expression (not an equality) and transform it to another. So in this case, I would take x, and transform it to y. Im having trouble finding a way to do this, so is what i have something that is considered correct? (btw this is not homework, it's just me wandering through problems in the text, for my own edification)
1. (z^2)/4 = (1/2)xy
2. Z^2 = 2xy (multiply both sides by 4)
3. x^2 + y^2 = 2xy (substitute by the pythagorean theorem)
4. (x^2 + y^2)/(xy) = 2 (multiply both sides by xy)
5. (y/x) + (x/y) = 2 (decompose the fraction on the left side)
6. Y/x = 2  x/y (subtract x/y from both sides)
7. Y = 2x  (x^2)/y (multiply both sides by x)
8. Y = x(2  x/y) (factor right side)
9. y^2 = x(2y  x) (multiply both sundes by y)
10. (d/dx)(y^2) = (d/dx)[x(2y  x)] (multiply both sides by d/dx)
11. 2y(dy/dx) = x[2(dy/dx)  1] + (2y  x) (carry out derivetives)
12. 2y(dy/dx) = 2x(dy/dx)  x  2y  x (distribute on right side)
13. 2y(dy/dx)  2y = 2x(dy/dx) 2x (subtract 2y from both sides)
14. 2y[(dy/dx)  1] = 2x[(dy/dx)  1] (factor both sides)
15. 2y = 2x (divide both sides by (dy/dx)  1 )
16. y = x (guess)
Forming a proof about a triangle
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Re: Forming a proof about a triangle
Atmosck wrote: but I'm taking a course on discrete math that's supposed to serve as an introduction to proofs, among other things, and all the direct proofs I've seen, and the explanation in my text, are such that you take one expression (not an equality) and transform it to another. So in this case, I would take x, and transform it to y.
That's not the only way to prove two things are equal. There's no rule that says you have to start with x and transform it to y. You're allowed to have a proof that almost "magically" results in x=y as a consequence  provided, of course, that you start with results that are taken to be true, and you use correct reasoning.
By the way, your proof looks fine to me at first glance, though there could be a clever "shortcut" at some point that might result in a shorter proof.
Edit: If you're using derivatives, that introduces subtleties. You might be better off avoiding that if possible.
 phlip
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Re: Forming a proof about a triangle
One thing that stands out is that you divide by dy/dx1, without checking that it's nonzero... and it will be zero for y=x. For completeness, you should say that it implies that "2x = 2y or dy/dx1 = 0", and then show that both options of that imply that y=x, which they do... 2y=2x does pretty easily, dy/dx=1 implies y=x+C, and you can probably substitute back into an earlier line to find C has to be 0. A similar thing applies for step 4, where you divide by xy (though you say multiply in the explanation)... though the proof that xy can't be 0 is pretty simple.
However, what I would've done is starting at your step 3:
3. x^{2} + y^{2} = 2xy
4. x^{2}  2xy + y^{2} = 0 (rearranged)
5. (x  y)^{2} = 0 (factorised)
6. x  y = 0 (square root... technically should be "plusorminus 0", but that's pretty trivial)
7. x = y (rearrange again)
All this messing around with derivatives and whatnot still works, it's just a bit overcomplicated. Your main overcomplicating step is step 4, where you divide both sides to make a rational function... normally you'd be going the other way, if you're given an equation with a rational function (ie a fraction of two polynomials) you'll multiply both sides by the divisor, so that you get just a normal polynomial, 'cause they're much easier to work with.
Also note that all your messing around in steps 49, dividing by xy, rearranging, multiplying by x and y, rearranging some more, ultimately doesn't achieve much... if you expand the factorisation on the RHS, you're back at x^{2}  2xy + y^{2} = 0 again.
That's not to say that the way that you did it is wrong, just... more work than necessary.
As an aside, the other proof style for equality you mentioned, where you start with the lefthand side, and manipulate it a bunch until it equals the right side, is just a shorthand for:
1. a = b (some reasoning)
2. b = c (some reasoning)
3. c = d (some reasoning)
4. therefore a = d (transitive property)
which is the same style as what you've got here... listing a bunch of facts, and deriving things from them, and ending up with "a=d" as the final thing.
However, what I would've done is starting at your step 3:
3. x^{2} + y^{2} = 2xy
4. x^{2}  2xy + y^{2} = 0 (rearranged)
5. (x  y)^{2} = 0 (factorised)
6. x  y = 0 (square root... technically should be "plusorminus 0", but that's pretty trivial)
7. x = y (rearrange again)
All this messing around with derivatives and whatnot still works, it's just a bit overcomplicated. Your main overcomplicating step is step 4, where you divide both sides to make a rational function... normally you'd be going the other way, if you're given an equation with a rational function (ie a fraction of two polynomials) you'll multiply both sides by the divisor, so that you get just a normal polynomial, 'cause they're much easier to work with.
Also note that all your messing around in steps 49, dividing by xy, rearranging, multiplying by x and y, rearranging some more, ultimately doesn't achieve much... if you expand the factorisation on the RHS, you're back at x^{2}  2xy + y^{2} = 0 again.
That's not to say that the way that you did it is wrong, just... more work than necessary.
As an aside, the other proof style for equality you mentioned, where you start with the lefthand side, and manipulate it a bunch until it equals the right side, is just a shorthand for:
1. a = b (some reasoning)
2. b = c (some reasoning)
3. c = d (some reasoning)
4. therefore a = d (transitive property)
which is the same style as what you've got here... listing a bunch of facts, and deriving things from them, and ending up with "a=d" as the final thing.
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 Yakk
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Re: Forming a proof about a triangle
Critique of things that don't matter much.
Don't change case of variables. A and a are often different!
Don't you mean divide both sides by xy? This is invalid if x = 0 or y = 0.
It is better to "presume" something like "if x=0", then "if y=0, similar conclusion", then "if neither x nor y is zero".
You can do that in the other order (putting the meat of the proof earlier, because the zero cases are probably trivial). Another method is to take the meat, lemma it out of the proof, present the lemma prior to the proof (maybe just mentioning it), then proving it elsewhere.
Again, watch case. It is a good habit to get into. Being able to use y and Y for different related things (like an element and the set it comes from) is useful, but if you cannot maintain case, you cannot use it in your proofs. Plus, in more complex proofs, readers can be confused by the introduction of a new case that means nothing.
In many cases, having bidirectional proofs is useful  each step both being implied, and implying, the previous one. This breaks that. Sometimes you want to note that. (if x=0, this becomes rather boringly true, but does not imply line 6, unlike most of your steps). This is an aside, and is not a flaw in this particular proof, as yours is a forward chain of implication.
Taking the derivative is not multiplication, at least it isn't commonplace multiplication.
Remember that not all operations on both sides of an equality are friendly. In this case, if two differentiable functions equal each other everywhere, their derivative also agrees.
Wait, are you implying that y is a function of x now? This is a relatively strong statement to pop in. If y is a simple function of x, then you aren't a circle or spiral  cases have been ruled out that where not ruled out before.
As noted, this is invalid if dy/dx = 1.
Atmosck wrote:1. (z^2)/4 = (1/2)xy
2. Z^2 = 2xy (multiply both sides by 4)
Don't change case of variables. A and a are often different!
3. x^2 + y^2 = 2xy (substitute by the pythagorean theorem)
4. (x^2 + y^2)/(xy) = 2 (multiply both sides by xy)
Don't you mean divide both sides by xy? This is invalid if x = 0 or y = 0.
It is better to "presume" something like "if x=0", then "if y=0, similar conclusion", then "if neither x nor y is zero".
You can do that in the other order (putting the meat of the proof earlier, because the zero cases are probably trivial). Another method is to take the meat, lemma it out of the proof, present the lemma prior to the proof (maybe just mentioning it), then proving it elsewhere.
5. (y/x) + (x/y) = 2 (decompose the fraction on the left side)
6. Y/x = 2  x/y (subtract x/y from both sides)
Again, watch case. It is a good habit to get into. Being able to use y and Y for different related things (like an element and the set it comes from) is useful, but if you cannot maintain case, you cannot use it in your proofs. Plus, in more complex proofs, readers can be confused by the introduction of a new case that means nothing.
7. Y = 2x  (x^2)/y (multiply both sides by x)
In many cases, having bidirectional proofs is useful  each step both being implied, and implying, the previous one. This breaks that. Sometimes you want to note that. (if x=0, this becomes rather boringly true, but does not imply line 6, unlike most of your steps). This is an aside, and is not a flaw in this particular proof, as yours is a forward chain of implication.
8. Y = x(2  x/y) (factor right side)
9. y^2 = x(2y  x) (multiply both sundes by y)
10. (d/dx)(y^2) = (d/dx)[x(2y  x)] (multiply both sides by d/dx)
Taking the derivative is not multiplication, at least it isn't commonplace multiplication.
Remember that not all operations on both sides of an equality are friendly. In this case, if two differentiable functions equal each other everywhere, their derivative also agrees.
11. 2y(dy/dx) = x[2(dy/dx)  1] + (2y  x) (carry out derivetives)
Wait, are you implying that y is a function of x now? This is a relatively strong statement to pop in. If y is a simple function of x, then you aren't a circle or spiral  cases have been ruled out that where not ruled out before.
12. 2y(dy/dx) = 2x(dy/dx)  x  2y  x (distribute on right side)
13. 2y(dy/dx)  2y = 2x(dy/dx) 2x (subtract 2y from both sides)
14. 2y[(dy/dx)  1] = 2x[(dy/dx)  1] (factor both sides)
15. 2y = 2x (divide both sides by (dy/dx)  1 )
As noted, this is invalid if dy/dx = 1.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Forming a proof about a triangle
Yakk wrote:Don't change case of variables. A and a are often different!3. x^2 + y^2 = 2xy (substitute by the pythagorean theorem)
4. (x^2 + y^2)/(xy) = 2 (multiply both sides by xy)
Yeah, sorry about the cases. I understand the point of being consistent about it, but this isn't a very formal setting, and i typed the origional message on an iPad, which likes to automatically capitalize what it thinks is the beginning of a sentence.
[/quote]11. 2y(dy/dx) = x[2(dy/dx)  1] + (2y  x) (carry out derivetives)
Wait, are you implying that y is a function of x now? This is a relatively strong statement to pop in. If y is a simple function of x, then you aren't a circle or spiral  cases have been ruled out that where not ruled out before.12. 2y(dy/dx) = 2x(dy/dx)  x  2y  x (distribute on right side)
13. 2y(dy/dx)  2y = 2x(dy/dx) 2x (subtract 2y from both sides)
14. 2y[(dy/dx)  1] = 2x[(dy/dx)  1] (factor both sides)
15. 2y = 2x (divide both sides by (dy/dx)  1 )
As noted, this is invalid if dy/dx = 1.
I guess I should have differentiated with respect to t in stead of x. Like I said in the first post, this is really about a static geometric figure, so nothing is actually changing, so the differentiation is purely abstract.

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Re: Forming a proof about a triangle
Yakk wrote:Critique of things that don't matter much.Atmosck wrote:1. (z^2)/4 = (1/2)xy
2. Z^2 = 2xy (multiply both sides by 4)
Don't change case of variables. A and a are often different!3. x^2 + y^2 = 2xy (substitute by the pythagorean theorem)
4. (x^2 + y^2)/(xy) = 2 (multiply both sides by xy)
Don't you mean divide both sides by xy? This is invalid if x = 0 or y = 0.
x and y are the side lengths of a right triangle
take into account the triangle inequality theorem.
it states that the sum of two sides of a triangle are always greater than the remaining side. Which means triangles can't have side lengths of 0.
i think.
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Re: Forming a proof about a triangle
It is not a strict inequality, just because there can be sides of 0 length.. But then, when you state $$ \frac{xy}{2} = \frac{z^2}{4}$$, and it's a given that one of the sides is 0, it follows quite trivially that the other 2 sides are zero too, and therefore the equality x=y holds true.
Re: Forming a proof about a triangle
You reached this result in a really roundabout way.
Start with the formula for area:
A=(xy)/2
Using the assumption that x=y:
A=(x^2)/2
But, using that same assumption on the Pythagorean Theorem:
z^2 = x^2 + y^2
z^2 = 2x^2
(z^2)/2 = x^2
So by substitution:
A = ((z^2)/2)/2
Simplify and you have your result:
A = (z^2)/4
You were looking to prove that a right triangle with an area of (z^2)/4 must be isosceles, but I thought that direction was a little strange, so instead I showed that an isosceles right triangle must have an area of (z^2)/4. Clearly those imply each other so it was a mere aesthetic choice.
So now what about any isosceles triangle? Does the same hold true?
Start with the formula for area:
A=(xy)/2
Using the assumption that x=y:
A=(x^2)/2
But, using that same assumption on the Pythagorean Theorem:
z^2 = x^2 + y^2
z^2 = 2x^2
(z^2)/2 = x^2
So by substitution:
A = ((z^2)/2)/2
Simplify and you have your result:
A = (z^2)/4
You were looking to prove that a right triangle with an area of (z^2)/4 must be isosceles, but I thought that direction was a little strange, so instead I showed that an isosceles right triangle must have an area of (z^2)/4. Clearly those imply each other so it was a mere aesthetic choice.
So now what about any isosceles triangle? Does the same hold true?
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Re: Forming a proof about a triangle
Eastwinn wrote:You were looking to prove that a right triangle with an area of (z^2)/4 must be isosceles, but I thought that direction was a little strange, so instead I showed that an isosceles right triangle must have an area of (z^2)/4. Clearly those imply each other so it was a mere aesthetic choice.
Um... no. You've just proved the converse. The converse does not imply the original problem.
What would be an equivalent thing to show is the contrapositive, which in this case would be to show that if a right triangle is not isosceles, then its area is not z^2/4.
Re: Forming a proof about a triangle
letterX wrote: Um... no. You've just proved the converse. The converse does not imply the original problem.
What would be an equivalent thing to show is the contrapositive, which in this case would be to show that if a right triangle is not isosceles, then its area is not z^2/4.
Right, right. But I thought it was obvious that the converse must also be true. On second thought, however, it's not so obvious, and I'm not totally sure, nor do I have the brain power to figure it out right now
Edit: To below: Absolutely.
Last edited by Eastwinn on Mon Dec 06, 2010 12:46 am UTC, edited 1 time in total.
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 phlip
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Re: Forming a proof about a triangle
You can only simply convert a proof of A>B into a proof of B>A if every step is of the "ifandonlyif" variety... so you can simply reverse the order of the steps, and you're done. But that's not the case with your proof... substitution isn't an ifandonlyif thing. You can go from "a=f(b) and b=g(c)" to "a=f(g(c))", but you can't go from "a=f(g(c))" to "a=f(b) and b=g(c)".
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