Is this direct proof of an inequality wrong?

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."

Solution 1:

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

etc.

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.

Solution 2:

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.