Hey guys [smile]. I have some questions about proofs. 10th grade in the USA, please bear with me because I don't have anyone else as knowledgeable as you guys to ask. Recently I took a trig test, and one of the questions said: choose one of the following identities and prove it. The identities in the question weren't the standard identities; they were just something one can come up with by messing around with an equation while still keeping it an equation. Then I had to take it and show that the equation is, in fact, true. To keep things simple, let's just do something like (cos x)/(cos² x) = 1/(cos x) (the real questions had multiple steps to them and were therefore much more complex) Now, according to my teacher, the correct answer would be:

(cos x)/(cos² x) = 1/(cos x) <carry out division on left side>
1/(cos x) = 1/(cos x) <Q.E.D.>

I did this:

(cos x)((cos x) / (cos² x)) = (1 / (cos x))(cos x) <multiply both sides by cos x>
1 = 1

He said that's wrong because you're not allowed to change the sides like that. Now, I can see why that wouldn't work:

1 = 2
(0)(1) = (2)(0) <multiply both sides by 0>
0 = 0 therefore 1 must equal 2

Of course, that's wrong. However, is my teacher correct in his method of "proving"? I thought that to "prove" something, you had to start with a trivially true statement, then work up to the original statement, showing that it must be true. In our example:

cos x = cos x <by identity>
1 / (cos x) = 1 / (cos x) <by identity>
((cos x)/(cos x)) 1/(cos x) = 1/(cos x) <multiply left side by equivalent of 1/1>
(cos x)/(cos² x) = 1/(cos x) <Q.E.D.>

Thus, you can't just start out with the original and simplify it, because that's what you're trying to prove in the first place. Am I right or wrong? If we were to start out with the original statement, we would need to use proof by contradiction:

(cos x)/(cos² x) =/= 1/(cos x)
1/(cos x) =/= 1/(cos x) <contradiction, therefore the original statement must be true>

Right? Basically, I understand why my method was wrong, but I'm confused as to whether my teacher's method is actually correct.

Quote:
Now, I can see why that wouldn't work:


1 = 2

(0)(1) = (2)(0) <multiply both sides by 0>

0 = 0 therefore 1 must equal 2

His proof contains the same situation, by carrying out division he has assumed that cos(x) != 0. Other than that I'm not sure what you mean by "because you're not allowed to change the sides like that". Is he suggesting that you cannot manipulate both sides of the equation to find a satisfactory proof?

Its been a while, but generally such proofs have two paths: the normal path for when the denominators are reasonable, and an alternate path where the denominators are troublesome. This is usually expressed as a constraint on the proof (which here means that the identity wasn't fully specified to begin with).

(cos x)/(cos2 x) = 1/(cos x)            // carry out division on left side1/(cos x) = 1/(cos x) [cos(x) != 0]     // Q.E.D.

With such a qualifier, I would consider your proof as valid.

Quote:
His proof contains the same situation, by carrying out division he has assumed that cos(x) != 0.

That wasn't the exact problem, but his way of doing things contains similar tactics.

Quote: Other than that I'm not sure what you mean by "because you're not allowed to change the sides like that". Is he suggesting that you cannot manipulate both sides of the equation to find a satisfactory proof?

His exact statement was that "in a proof, you can't change either side by anything other than 1" (for example, you can multiply one side by the equivalent of 1/1 to simplify something, but you can't multiply both sides by 2)

Let's say we're proving that sin^2 θ + cos^2 θ = 1, using the right-triangle definition of sin and cos and the Pythagorean theorem.

sin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2
sin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2
sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2
sin^2 θ + cos^2 θ = H^2 / H^2
sin^2 θ + cos^2 θ = 1

In this formulation, each line is satisfactorily proven if and only if the previous line is satisfactorily proven. The middle steps aren't necessarily important on their own, but have nevertheless been conclusively proven.

Now, at any point in that proof, if you felt like it, you could multiply both sides by zero. No problem with that, you're still proving statement after statement. HOWEVER, once you do that, all you're proving is that 0=0, which nobody was going to argue against in the first place. You've essentially discarded your previous structure of proof in favor of proving a triviality. The burden of a proof is not simply to show that something is true, but to show that the statement in question is true. That's why "QED" is there. "Quod erat demonstrandum"; "That which was to be shown."

So when your teacher says "you can't change both sides", he's simply reminding you that changing both sides leads you down an unproductive path of proving things that won't help you prove what you're trying to prove, and that proving something is not the same as proving your thing.

Quote:
So when your teacher says "you can't change both sides", he's simply reminding you that changing both sides leads you down an unproductive path of proving things that won't help you prove what you're trying to prove, and that proving something is not the same as proving your thing.

Right, I get that part.

Quote: Original post by Sneftel
Let's say we're proving that sin^2 θ + cos^2 θ = 1, using the right-triangle definition of sin and cos and the Pythagorean theorem.

sin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2
sin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2
sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2
sin^2 θ + cos^2 θ = H^2 / H^2
sin^2 θ + cos^2 θ = 1

You started with a trivially true statement, basically using the exact definition of the left-hand side of the equation. His way of proving it would essentially be the opposite, by starting out with the identity and then manipulating the right side, like this:

sin^2 θ + cos^2 θ = 1sin^2 θ + cos^2 θ = H^2 / H^2 // multiply by the equivalent of 1/1sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2 // expand H^2 based by Pythagorean theoremsin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2 // distributesin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2 // rearrangesin^2 θ + cos^2 θ = sin^2 θ + cos^2 θ // substitute

That's what I'm confused by, since I thought you can't start with the statement itself, regardless of the issue with multiplying both sides by something.

Quote: Original post by nullsquared

Quote: Original post by Sneftel
Let's say we're proving that sin^2 θ + cos^2 θ = 1, using the right-triangle definition of sin and cos and the Pythagorean theorem.

sin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2
sin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2
sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2
sin^2 θ + cos^2 θ = H^2 / H^2
sin^2 θ + cos^2 θ = 1

You started with a trivially true statement, basically using the exact definition of the left-hand side of the equation. His way of proving it would essentially be the opposite, by starting out with the identity and then manipulating the right side, like this:

sin^2 θ + cos^2 θ = 1sin^2 θ + cos^2 θ = H^2 / H^2 // multiply by the equivalent of 1/1sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2 // expand H^2 based by Pythagorean theoremsin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2 // distributesin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2 // rearrangesin^2 θ + cos^2 θ = sin^2 θ + cos^2 θ // substitute

That's what I'm confused by, since I thought you can't start with the statement itself, regardless of the issue with multiplying both sides by something.

Right, you definitely don't start by assuming what you have is true. That makes for quite the meaningless argument. What you've shown is fine if reversed, but to start your proof with what you're trying to prove would mean being marked wrong by anyone in the [pure] math business. This is how I would perhaps write things:

1 = H^2 / H^2  = (O^2 + A^2) / H^2 // expand H^2 based by Pythagorean theorem  = O^2 / H^2 + A^2 / H^2 // distribute  = (O/H)^2 + (A/H)^2 // rearrange  = sin^2 θ + cos^2 θ // substitute

This clearly shows how you manipulated one side to obtain the opposite side. You began with something that was true (1 = H^2 / H^2) and manipulated it to draw a conclusion.

[EDIT]
And just to add on to the original post and rip-off's comments: (cos x)/(cos² x) = 1/(cos x), this equality itself implies that cos(x) != 0 since if it were, we would have began our proof with an equality that makes no sense. Hence, I don't think a qualifier is even necessary because of this implicit assumption. Nevertheless, there's no harm in being explicit and/or verbose to avoid any ambiguities. What I'm leading to is that whether you divide or multiply (in this case) doesn't really matter, as long as one realizes that there is an assumption that cos(x) != 0.

[Edited by - aryx on March 12, 2010 5:12:32 PM]

Okay, great.

Another question: once we start with a trivially true statement, are we allowed to do the same thing to both sides in order to end up with the final statement, thus proving it?

Quote: Original post by nullsquared
You started with a trivially true statement, basically using the exact definition of the left-hand side of the equation. His way of proving it would essentially be the opposite, by starting out with the identity and then manipulating the right side, like this:

sin^2 θ + cos^2 θ = 1sin^2 θ + cos^2 θ = H^2 / H^2 // multiply by the equivalent of 1/1sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2 // expand H^2 based by Pythagorean theoremsin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2 // distributesin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2 // rearrangesin^2 θ + cos^2 θ = sin^2 θ + cos^2 θ // substitute

That's a perfectly acceptable proof, because it's exactly the same thing; it's just written the other way around. Just as in my formulation each statement is supported by the statement before it, in this formulation each statement is supported by the statement after it. In my formulation each statement is provably true as soon as it's written down; in this formulation nothing is proven until you get to the last line, at which point justification ripples back through the proof to the beginning. So because it's written this way, each line has to lead logically to the previous line. If you put in 0=0 somewhere, that would no longer be a valid proof, because the line before it would not be adequately justified. But as long as we don't change the left hand side, we're guaranteed that statement n is justified by statement n+1, due to transitive equality. In fact we could change the right hand side and still have this property, as long as we didn't change both sides at the same time.

Quote: Original post by nullsquared
Okay, great.

Another question: once we start with a trivially true statement, are we allowed to do the same thing to both sides in order to end up with the final statement, thus proving it?

Yes, in so far as you account for all cases. For instance, you can take the square root of both sides as long as you are careful not to lose track of the fact that square roots are not unique (positive and negative roots) and do not always exist (square roots of negative numbers are imaginary). Squaring a number, however, doesn't have the same problem, even though taking square root is the "inverse" of squaring a number. The same thing arises in a lot of different cases.

Typically a teacher will have you draw implication arrows to make sure you're aware of what follows from what. This way you can "start" with the statement you want to prove and end up with a trivial statement, as long as it's true that the statements all actually follow from the "trivially true statement" (EDIT: this is basically what Sneftel said right above me).

Quote: Original post by Sneftel

Quote: Original post by nullsquared
You started with a trivially true statement, basically using the exact definition of the left-hand side of the equation. His way of proving it would essentially be the opposite, by starting out with the identity and then manipulating the right side, like this:

sin^2 θ + cos^2 θ = 1sin^2 θ + cos^2 θ = H^2 / H^2 // multiply by the equivalent of 1/1sin^2 θ + cos^2 θ = (O^2 + A^2) / H^2 // expand H^2 based by Pythagorean theoremsin^2 θ + cos^2 θ = O^2 / H^2 + A^2 / H^2 // distributesin^2 θ + cos^2 θ = (O/H)^2 + (A/H)^2 // rearrangesin^2 θ + cos^2 θ = sin^2 θ + cos^2 θ // substitute

That's a perfectly acceptable proof, because it's exactly the same thing; it's just written the other way around. Just as in my formulation each statement is supported by the statement before it, in this formulation each statement is supported by the statement after it. In my formulation each statement is provably true as soon as it's written down; in this formulation nothing is proven until you get to the last line, at which point justification ripples back through the proof to the beginning. So because it's written this way, each line has to lead logically to the previous line. If you put in 0=0 somewhere, that would no longer be a valid proof, because the line before it would not be adequately justified. But as long as we don't change the left hand side, we're guaranteed that statement n is justified by statement n+1, due to transitive equality. In fact we could change the right hand side and still have this property, as long as we didn't change both sides at the same time.

Hm, I see what you're saying. The only problem now is that we have two contradicting views in the thread [grin]; I hope a few more people come in to confirm one way or the other.