Simple practice proof

Question

I am still trying to master the notion of proofs. It is quite a difficult transition, and I am not the best at it. Any suggestions or corrections on my logic would be extremely helpful. Thanks in advance.

Claim: If $n$ is an integer, then $n^2 \geq n$.

I rewrote this as: For all integers $n$, $n^2 \geq n$.

Proof: Suppose $n^2$ < $n$, for all integers $n$. Then, $n \cdot n < n \implies n < 1$. This is not true since $n \in \mathbb{Z}$ and $n$ can also be greater than or equal to $1$. Thus, contradicting our assumption.

I feel like I completely butchered this proof, but I am just starting to practice it more. Any help would be greatly appreciated.

Answer 1

The negation of "for all" is "there exists". If you want to proceed solving this question with a proof by contradiction, your starting assumption must be "Suppose $n^2<n$ for some integer $n\in \mathbb{Z}$.

Then, your need to be a little more careful in the writing of the proof. Indeed, you are trying to divide your inequality by $n$ to deduce a contradiction, but there are two things you need to be attentive to. First, $n$ must not be zero. The case $n=0$, which is trivial, must be mentioned separatedly. Second, if $n$ is negative, then the order of the inequality changes. So you need to treat the cases $n$ positive and $n$ negative separatedly (and conclude as you did).

Now, because of these cases that need to be distinguished, this kind of proof is not recommended.

You may show the result directly, by noticing that $n^2 \geq n$ if and only if $n(n-1)\geq 0$. A product of two quantities is non-negative if and only if both have the same sign. Thus, you are reduced to proving that for every $n \in \mathbb{Z}$, $n$ and $n-1$ have the same sign (with the meaning "both are non-negative or both are non-positive"). This is now rather straightforward.

Answer 2

Your proof is fine, after the modification in the comments. However, you can do a direct proof also:

Let $n$ be any integer. Thus $n \ge 1$. Multiplying this inequality* with $n \ge n$ we get $n^2 \ge n$.

*In the definition of $\ge$ it is said that if $a \ge b$ and $c \ge d$, and all quantities are positive, then $ac \ge bd$.

Answer 3

Claim: $n^2\geq n$ for all $n\in\mathbb{Z}$.

If there is even a single counterexample, then we know that the claim is false. So in fact, the true negation of the claim is: Suppose $n^2<n$ for some $n\in\mathbb{Z}$.

The rest of the proof follows like this:

If $n^2<n$ for some $n\in\mathbb{Z}$, then $n^2-n<0\implies n(n-1)<0$. Consider the following three cases:

$1)$ If $n=0$, then $n(n-1)=0\nless0$ so we arrive at a contradiction.

$2)$ If $n>0$, then $n-1\geq 0$ so $n(n-1)\nless 0$ and we arrive at a contradiction.

$3)$ If $n<0$, then $n-1<0$ so $n(n-1)>0$ and we arrive at a contradiction.

QED