My professor graded my proof as a zero, and I'm having a hard time seeing why it would be graded as such. Either he made a mistake while grading or I'm lacking in my understanding. Hopefully someone can help sort it out. The proof is as follows:

Goal: If $n$ is a positive integer, then $\frac{n}{n+1} > \frac{n}{n+2}$.

Proof: Assume $n$ is a positive integer.

Observe, $\frac{n}{n+1} > \frac{n}{n+2}$

$\frac{n(n+1)}{n+1} > \frac{n(n+1)}{n+2}$

$n > \frac{n^2+n}{n+2}$

$n(n+2) > \frac{(n^2 + n)(n+2)}{n+2}$

$n^2 + 2n > n^2 + n$

$n^2 - n^2 + 2n > n^2 - n^2 + n$

$2n > n$

Since $2n > n$ for all positive integers, then $\frac{n}{n+1} > \frac{n}{n+2}$ for all positive integers.

Therefore, if $n$ is a positive integer, then $\frac{n}{n+1} > \frac{n}{n+2}$. Q.E.D.

Here are the notes on the problem by the professor:

"You assumed Q! You cannot assume your conclusion!"

Shows that $2n > n$ reduces down to $n > 0$ and points an arrow to 'Assume n is a positive integer' "Circular logic."

"By the way... reducing to falsehood is a valid truth technique(proof by contradiction) but reduction to truth tells you nothing."

  • $\begingroup$ As other's have pointed out..your poof boils down to: If X then X or $a \implies a$.... $\endgroup$ – user237392 Apr 1 '16 at 22:30
  • $\begingroup$ I'm not proving that n is a positive integer, I'm proving that the inequality holds true when n is a positive integer. $\endgroup$ – Ravarion Apr 1 '16 at 22:38
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    $\begingroup$ What you are doing is trying to do a proof by assuming true and getting a different true result and concluding your premise was true. That's faulty logic. Although a false conclusion means a false premise. A true conclusion doesn't mean anything at all. $\endgroup$ – fleablood Apr 1 '16 at 22:44
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    $\begingroup$ Your proof would have worked if you stated each line was an "if and only if" statement. But you have to state that and the have to be if and only if statements. $\endgroup$ – fleablood Apr 1 '16 at 22:55

The problem is that you did not state that those are equivalences (usually denoted by $\iff$) between your lines. And you do not even need the equivalences, you only need the implicatiosn from the bottom to the top, so you should perhaps write your proof "upside down".

The way your proof is presented right now makes it look like the top implies the bottom. (Which it does) but that doesn't mean that the bottom implies the top. So I'd write down:

For any positiv integer $n$ the following inequality obviously holds:

$$2n > n$$

This implies $$n^2-n^2+2n > n^2-n^2+n$$


Notice that this is also the way you'd want to read your proof so that people understand it. And it is important to realize that this is not necessarly the way you found the proof.

Whenever you find a proof somewhere you can be quite sure that the way it is presented to you has nothing to do with how the one who proved it found the proof. It is really just written down nicely in order for the reader to be able to nicely follow the chain of arguments.

But don't worry, writing "nice" proofs does take a while at the beginning of your math career=)

Regarding the comment: You could prove the inequality by contradicion, e.g. suppose that the inequality does not hold, and then find a contradiction, but this is not necessary here.

EDIT: It seems to me that you are unfamiliar with the concept of logical implications. If two mathematical statements $A,B$ (e.g. equations, inequalities etc.) mean the same thing, they are called equivalent, denoted by the bidirectional double arrow. $$A \iff B$$

Example: Let $x$ be a real number. Then following equivalence holds: $$x -5 = 0 \iff x = 5$$

But if $A$ only hold if $B$ holds, then we say $A$ implies $B$ or alternatively, "if $A$ holds then $B$ must hold too." This is denoted by a simple doublearrow: $$A \implies B$$

Example: Let $x$ be a real number. Then following implication holds:

$$x = 5 \implies x^2 = 25$$ But "the other way around" is not necessarily true, as the right statement is also true for $x=-5$ but not the left one.

  • $\begingroup$ So you're saying that reducing n/(n+1) > n/(n+2) does not suffice as proof, but expanding 2n > n, does? The question isn't so much about how to make my proof better, but instead, does it prove? It's important because it was worth half of my exam and I got zero for this problem. $\endgroup$ – Ravarion Apr 1 '16 at 22:36
  • $\begingroup$ Exactly. You need the direction $2n > n \implies \frac{n}{n+1} > \frac{n}{n+2}$. But proving $\frac{n}{n+1} > \frac{n}{n+2} \implies 2n > n $ does not help anything with your task. $\endgroup$ – flawr Apr 1 '16 at 22:38
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    $\begingroup$ Well to make it a proof you should have denoted that each line holds if and only if ("$\iff$") the next line holds. If you do not write anything (which is not a good practise), this might suggest to a reader that you write down implications. I'd accept your proof as correct (but not really nicely written) if you'd have written "$\iff$" between all those lines. $\endgroup$ – flawr Apr 1 '16 at 23:11
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    $\begingroup$ I see, then it would have been a rigorous proof. I didn't know about this bi directional implication. I made an assumption that it was obvious to the reader that the implication was intended to be seen as reversible. Thank you for the explanation, that helped me to understand why the logic I saw wasn't shared. $\endgroup$ – Ravarion Apr 1 '16 at 23:19
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    $\begingroup$ Note that you can present the proof in the order it was found, and one does not need bi-implications $\iff$ to do so. But you really need to use the right connectives. Starting to "observe" the conclusion is a no-no (except if it is to admire its aesthetic qualities) and the professor rightly scorned you for it. Instead say "We want to prove" and it is OK. Then for each next line say (something like) "for this it suffices to have" (which corresponds to the logical connective $\Leftarrow$) and at the end "which is true" and you've got a proper (though somewhat unusual) proof. $\endgroup$ – Marc van Leeuwen Apr 2 '16 at 12:15

Your proof doesn't work because concluding a true statement is irrelevant. A proof by contradiction works because a true premise only yields true results, therefore if you get a false result your premise had to be false.

It doesn't work the other way. Both true and false premises can yield true results so getting a true result yields nothing.

Consider this proof that 5=-5. Asume 5 = -5 then $5^2=(-5)^2 \implies 25=25\implies 0=0$ As 0=0, this is true.

Well, so what? That doesn't mean our first comment was true.

Your professor made 3 good comments. Heed them.

  • $\begingroup$ The problem with your false proof is that it doesn't reverse, which is a different case than mine. 25 = 25 does not imply 5^2 = (-5)^2 $\endgroup$ – Ravarion Apr 1 '16 at 22:49
  • $\begingroup$ @Ravarion Thats why the implication ("$\implies$") was written only in one direction here, and not as an equivalence ("$\iff$"). $\endgroup$ – flawr Apr 1 '16 at 22:59
  • $\begingroup$ But my implications go both ways, so I'm not sure your example is the same case as my proof. $\endgroup$ – Ravarion Apr 1 '16 at 23:04
  • $\begingroup$ Thanks for the help. I can see that your answer is also correct. $\endgroup$ – Ravarion Apr 1 '16 at 23:25
  • $\begingroup$ I didn't think your implications went both ways. Why would I if they weren't stated? I thought you were assuming what you were trying to prove which you can't do. Then when you concluded 2n > n I didn't see any significance to that at all. And as you assumed it was positive I thought that was circular. It wasn't at all clear that not only were your steps were reversable but that was the intent. Also you need to show that some are reversable. I'd go back and point out that each step was reversable and explain you thought you intent was clear and hope for 1/4 or 1/3 credit. $\endgroup$ – fleablood Apr 1 '16 at 23:33

The many answers here correctly point out your error - you are working backwards. You seem to have trouble understanding those answers. Take comfort in the fact that you're not alone. Many students struggle with this problem.

Here's a critique of your argument that may help you (it's essentially a rewording of @flawr 's accepted answer).

The first line of your proof is

Observe, $\frac{n}{n+1} > \frac{n}{n+2}$

That asks your reader to


Well if it were easy to observe then you wouldn't have to prove it. You need an argument that starts from something easy to observe and proceeds logically to the conclusion you want. So begin

Observe that $2n > n$ (since $n$ is a positive integer)

and then use your algebraic skills to get to what you want to convince your reader must be true.

PS In third grade kids learn that it's easy to compare fractions with the same numerator: the smaller the denominator the larger the fraction since the pie is cut into fewer hence bigger pieces. But your instructor would probably not accept that as a proof ... The question would have been more interesting if it asked you to compare $\frac{n}{n+1}$ to $\frac{n+1}{n+2}$.


You could have succeeded in this method if you did a proof by contradiction.

In particular, you could have supposed that $\frac{n}{n+1} \leq \frac{n}{n+2}$ and then deduced that this implies $n \geq 2n$ which is absurd (for positive integer $n$).

  • $\begingroup$ I agree that contradiction would have worked here, but I'm not sure that my proof does not also succeed. $\endgroup$ – Ravarion Apr 1 '16 at 22:30
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    $\begingroup$ @Ravarion The fundamental reason is that it is possible for a false statement to imply a true statement. On the other hand, a true statement can imply only a true statement (which is why starting with $2n>n$ would have been better). If you wanted to do a proof by contradiction instead, as I suggest in my answer, you're essentially asserting that a particular statement ($\frac{n}{n+1} \leq \frac{n}{n+2}$) implies a false statement, which implies the original statement is false. $\endgroup$ – MathematicsStudent1122 Apr 1 '16 at 22:36
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    $\begingroup$ @MathematicsStudent1122 Saying it is possible that a false statement to imply a true statement is a big understatement: a false statement implies any other statement (either true or false)! ;-) $\endgroup$ – egreg Apr 1 '16 at 22:47

The problem is that you assume the thing you're trying to prove and then derive some other true fact from your assumption. But that doesn't prove anything: perhaps your assumption is false and the fact is true for some other reason.

For example, consider the following "proof".

Suppose $\pi=3$. Then $\pi\cdot3^2=9\pi=27 > 16=4^2$, which is true because a square of side $4$ fits inside a circle of radius $3$. Therefore, $\pi=3$.

Well, $9\pi$ is bigger than $16$ but that doesn't mean that $\pi=3$: indeed, $9\pi$ would be bigger than $16$ if $\pi$ was $2$ or $3.14159\ldots$ or $4$ or a million.

Other answers already include the relationship between this and proof by contradiction so I'll not repeat that.

You've shown that the inequality might be true, in that it seems to imply reasonable things. That would be a useful initial step if you didn't know whether the inequality was true or not. But it's not a proof. This is how a lot of research starts: you come up with a hypothesis and see if its consequences seem sensible. If the consequences seem reasonable then you try to come up with a proof.


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