37
$\begingroup$

Here is what I had to prove.

Question: For positive reals $a$ and $b$ prove that $a^2+b^2 \geq 2ab$.

Here is how my teacher did it:

First assume that it is in fact, true that $a^2+b^2 \geq 2ab$. Therefore $a^2+b^2-2ab \geq 0$ . We have $(a-b)^2$ is greater than or equal to zero which is true. Hence what was assumed originally is true.

Why does this method work?

I cannot understand how you can first assume that it is true and then prove that it is.

$\endgroup$
  • 25
    $\begingroup$ You cannot assume that to start. However did your teacher use equivalency of each statement, Ie that the first statement is true if and only if the second is and so on, then it is right. $\endgroup$ – Macavity Apr 12 '16 at 17:06
  • 2
    $\begingroup$ A proof is often "A is true, A implies B, hence B is true". But you start with B. You note that B implies A. Then you see that quite independently of that, you can prove A true. That first step (noting that B implies A) is necessary, because you have to identify a plausible B to start your proof with. $\endgroup$ – almagest Apr 12 '16 at 17:11
  • 4
    $\begingroup$ The teacher's approach might be a thought process you use while trying to solve the problem in order to come up with ideas, but it is not rigorous and when writing a proof you must only use valid steps. $\endgroup$ – M.M Apr 13 '16 at 1:37
  • 7
    $\begingroup$ You ask, "Why does this method work?". But this method does not work: your teacher has not yet proved that $a^2+b^2\ge 2ab$! $\endgroup$ – Daniel Wagner Apr 13 '16 at 6:08
  • 1
    $\begingroup$ I think the professor may have misspoke what he actually meant to say. Often in scratch work it is helpful to start with the conclusion and do some algebraic manipulations until you reach an obviously true statement (like $x^2 \geq 0$), and then use that true statement in your actual proof, which is usually a backwards process of your scratch work. This is one strategy that I was introduced to in my proofs class. $\endgroup$ – Benedict Voltaire Apr 13 '16 at 21:57

10 Answers 10

61
$\begingroup$

As written your teacher's proof is incorrect. The correct way to write your teacher's proof would be to write $$a^2+b^2 \geq 2ab\iff a^2+b^2-2ab \geq 0\iff(a-b)^2\geq 0$$ where each of the statements are equivalent to each other. The final statement is true, hence so is the first. It is much cleaner to write the proof in the reverse direction i.e. $$(a-b)^2\geq 0\implies a^2+b^2-2ab \geq 0\implies a^2+b^2 \geq 2ab$$

$\endgroup$
  • 6
    $\begingroup$ I'm failing to see how this is not equivalent to what the teacher did. Remember that using this kind of notation while speaking is a bit hard. $\endgroup$ – T. Sar Apr 13 '16 at 11:32
  • 9
    $\begingroup$ @ThalesPereira The difference is that your teacher starts with an unjustified assumption, which makes the whole 'proof' void, while FoobazJohn does not. Instead, FoobazJohn shows, step by step, an equivalence of the given expression (inequality) to the one you can easily see is true, then he can say all previous expressions, back to the original inequality, are true, too. $\endgroup$ – CiaPan Apr 13 '16 at 12:56
  • 5
    $\begingroup$ @ThalesPereira: even when speaking, the teacher said "Therefore" when what was necessary for the proof was "implied by", and said "assume that" when assuming it doesn't contribute to any valid proof. There's no rule of inference that says "if assuming X allows us to prove a true statement, then X is true", but that's the incorrect structure of the teacher's argument. There is a rule of inference that says, "if X is equivalent to a true statement then it's true", and that's the structure of FooBazJohn's correct proof. Even in speech there's a difference. $\endgroup$ – Steve Jessop Apr 13 '16 at 20:39
  • 3
    $\begingroup$ @ThalesPereira The notation is easily expressed verbally by saying something of the form, "<statement> is equivalent to <statement>." Math is difficult enough when you are precise; imprecision only makes it worse. (Of course, if the teacher misspoke, this is okay since everyone misspeaks at times, but correcting and clarifying when it happens is vital. So the student is absolutely right to ask the question and seek calrification.) $\endgroup$ – jpmc26 Apr 14 '16 at 3:37
  • 12
    $\begingroup$ It's also possible that the teacher presented exactly FoobazJohn's proof and the questioner recorded it incorrectly. It's easy to think that because someone writes something on a blackboard they must be assuming it, whereas this proof merely manipulates it using rules known to generate equivalent inequalities. (In fact a proof that the rules work could go by assuming the left-hand-side to deduce the right, and separately assuming the right to deduce the left. But when we apply the rule, those assumptions are already spent). $\endgroup$ – Steve Jessop Apr 14 '16 at 10:37
33
$\begingroup$

Your teacher is incorrect. In order to prove something is true, you do not assume it first. Here is a correct, logical way.

Let $a$ and $b$ be real numbers. Then, $(a-b)^{2}\geq 0$. Hence, $a^{2}-2ab+b^{2}\geq 0$, and thus, $a^{2}+b^{2}\geq 2ab$.

$\endgroup$
  • 2
    $\begingroup$ Why did it work then? $\endgroup$ – The Cryptic Cat Apr 12 '16 at 17:09
  • 41
    $\begingroup$ @TheCrypticCat: Because all of the steps were reversible. If you tried to prove $1=2$ by multiplying both sides by $0$ to get $0=0$ (a true statement), the proof would not work since you can't "undo" a multiplication by $0$. $\endgroup$ – Deusovi Apr 12 '16 at 17:12
  • 5
    $\begingroup$ Suppose we have two logical statements $A$ and $B$. If you want to prove that $A$ implies $B$, you can not start by assuming $B$ first. You must assume $A$ and deduce $B$. It just so happens that in this case the statements $(a-b)^{2}\geq 0$ and $a^{2}+b^{2}\geq 2ab$ are equivalent to each other (each one implies the other), but logically your teacher did not prove what he/she intended to $\endgroup$ – ervx Apr 12 '16 at 17:13
  • 8
    $\begingroup$ @TheCrypticCat: it did work, because the statement is true. That does not make the proof correct. $\endgroup$ – ipsec Apr 12 '16 at 18:18
  • 2
    $\begingroup$ @user1717828 All steps were reversible in the case of the OP. What Deusovi is illustrating is that the reasoning is flawed. As ipsec pointed, the statement is true, only the proof is not correct. $\endgroup$ – filipos Apr 13 '16 at 12:51
22
$\begingroup$

If the teacher worded it exactly as you have, then he is absolutely wrong.

However, there's a proof technique that follows a similar structure and is correct, which might have been what the teacher was going for. I'm talking about having the inference direction opposite to the writing direction.

Usually a proof starts with something that is true, makes a deduction from it, and continues until you reach your result: $a \Rightarrow b \Rightarrow c \Rightarrow d$.

Instead, you can start with what you want to prove, but instead of showing what follows from it, show what it follows from: $d \Leftarrow c \Leftarrow b \Leftarrow a$. If you end up with something true, then your original claim is true as well.

This is what your teacher was trying to do: $a^2+b^2\ge 2ab \Leftarrow a^2+b^2-2ab \ge 0 \Leftarrow (a-b)^2 \ge 0$. The last statement is true, so this completes the proof.

Some of the confusion might arise from the fact that in this case, all these inferences are reversible - it's true that $a^2+b^2\ge 2ab \Leftarrow a^2+b^2-2ab \ge 0$, but it's also true that $a^2+b^2\ge 2ab \Rightarrow a^2+b^2-2ab \ge 0$. But it's only the first of these inferences that is relevant for the proof. And of course, in other proofs you might use inferences which are not reversible.

This technique is logically the same as the usual forward method, but useful when you don't know what premise to start from; instead, you take what you want to prove, and figure out what you need in order for it to be true.

$\endgroup$
  • 3
    $\begingroup$ Another way to say this is, when is $a^2+b^2\ge 2ab$? When $a^2+b^2-2ab \ge 0$. And when is $a^2+b^2-2ab \ge 0$? When $(a-b)^2 \ge 0$. And when is $(a-b)^2 \ge 0$? Always. $\endgroup$ – David K Apr 13 '16 at 14:52
18
$\begingroup$

My rep is too low to comment, but others are correct in saying this works because each statement is an equivalence relation. If your teacher wants to prove something by starting with an assumption, it's better to assume the opposite: $a^2+b^2<2ab$, then work through the same steps to arrive at $(a-b)^2<0$ which is a contradiction if $a$ and $b$ are real. That proves $a^2+b^2 \geq 2ab$ by contradiction.

$\endgroup$
  • 1
    $\begingroup$ This makes for a better answer than a comment. $\endgroup$ – user1717828 Apr 13 '16 at 9:53
6
$\begingroup$

You can not actually. You should rather start with $(a-b)^2 \geq 0$ and the show that $a^2+b^2 \geq 2ab$, not the other way round.

$\endgroup$
5
$\begingroup$

You can't.

By assuming first what he needed to prove, your teacher has proven $True \implies True$ which is quite useless.

$\endgroup$
4
$\begingroup$

The way it's written is wrong, and you're right, we can't assume something and prove that it is right.

However, instead of one way implication at each step (like $a^2+b^2\geq 2ab$ therefore $a^2+b^2 -2ab\geq 0$), we in fact have an if and only if condition. Since each statement then becomes equivalent, if you can prove any one of them you are done. And that is the case since $(a-b)^2$ is indeed greater than equal to $0$.

$\endgroup$
  • $\begingroup$ What is the meaning of if and only if statement? $\endgroup$ – The Cryptic Cat Apr 12 '16 at 17:16
  • 2
    $\begingroup$ $A$ if and only if $B$ means that if $A$ is true, then so is $B$, and if $B$ is true then so is $A$. $\endgroup$ – Seven Apr 12 '16 at 17:18
3
$\begingroup$

Contrary to what other people are saying, your teacher's proof was mostly right - it was just missing one last sentence which he might have left out: "Every step done so far is also valid in reverse". Since that is true, you can read the steps backwards and get to the premise given the conclusion. This sort of reasoning is quite common in math since the most obvious way to prove is not necessarily the way the proof will actually read.

Reading backwards gives:

We know that $(a-b)^2 \geq 0$.

$\implies a^2+b^2-2ab \geq 0$

$\implies a^2+b^2 \geq 2ab$

So it can be converted to the more "traditional" way of writing the proof.

$\endgroup$
  • 4
    $\begingroup$ Not. At. All. ! The 'proof' is not 'mostly right', because it starts with a phrase 'First assume that it is in fact, true that $a^2 +b^2 \ge 2ab$', which makes all the rest void. Once you assume something is true, you no longer can, and no longer need to, prove it. $\endgroup$ – CiaPan Apr 14 '16 at 20:37
2
$\begingroup$

For real numbers the square of a number is non negative. $a-b$ is a real number because $a$ and $b$ are real. Therefore $(a-b)^2$ is non negative.

The inequality $a^2+b^2\geq 2ab$ can be transformed to

$a^2-2ab+b^2\geq 0$ by substracting $2ab$ on both sides.

This is equal to $(a-b)^2$. Just multiply out the brackets.

$\endgroup$
  • $\begingroup$ For real numbers, the square of a number is positive or zero. $\endgroup$ – Mark Pattison Apr 12 '16 at 20:22
  • $\begingroup$ @MarkPattison Thanks: I corrected it. $\endgroup$ – callculus Apr 12 '16 at 20:24
  • $\begingroup$ @MarkPattison *shrug* There are plenty enough people who use the terms "positive" and "strictly positive" where you would use "nonnegative" and "positive". Arguing about it is like arguing about whether zero is a natural number. $\endgroup$ – David Richerby Apr 14 '16 at 4:29
  • $\begingroup$ @DavidRicherby With respect, I disagree, at least in the context of mathematics. Whether zero is a natural number is a matter of arbitrary definition. Mathematicians can afford to be precise about whether, for example, the inequality above is ">" or ">=". $\endgroup$ – Mark Pattison Apr 14 '16 at 6:32
  • $\begingroup$ @MarkPattison It's also a matter of arbitrary definition whether "$x$ is positive" means "$x>0$" or "$x\geq 0$". People who use the first definition use the term "nonnegative" to mean "$\geq 0$"; people who use the second definition use the term "strictly positive" to mean "$>0$". $\endgroup$ – David Richerby Apr 14 '16 at 6:44
1
$\begingroup$

Okay, so first take the simple case where (a == b).

Imagine two large blocks, each arranged with unit squares. On the left is A, a square with (a) rows of (a) columns. On the right is B, a square with (b) rows of (b) columns. The two squares are the same, and the total is a^2 + b^2, which is also 2ab.

Now, let's tackle the unequal case: increase (b) by exactly one. Imagine that both A and B turn into rectangles with (a) columns and (b) rows. That gives us a total of (2ab) unit squares -- but that's not what we want. We want to take the top row from block A, rotate it 90 degrees, and slap it on as the rightmost column of block B. Draw this out to see what I'm describing. Notice that now, A is again a square, with total area a^2. Block B is incomplete however: it is exactly one less than b^2.

In short, if b==(a+1), then 2ab + 1 == (a^2 + b^2). This problem gets worse as the difference between (a) and (b) increases. It follows that (a^2 + b^2) >= 2ab.

On a sidenote, if you continue to chew on this relationship, you'll discover why factoring the difference between squares as (a+b)(a-b) works, and if you play further, you can discover why F.O.I.L. works.

$\endgroup$
  • 1
    $\begingroup$ This doesn't answer the question. The question asks whether a specific proof is valid, not whether it's possible to prove the given statement by some other method. $\endgroup$ – David Richerby Apr 14 '16 at 4:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.