can i use division when im inducting a proof? So I recently started learning Math by myself and I have a question about induction.
So, I was trying to prove $$n^2>n+1$$
the induction step:
$$(n+1)^2>(n+1)+1$$
$$(n+1)(n+1)>(n+1)+1$$
so, I thought that I could divide each side by $(n+1)$
then
$$1+n> 1+ 1/(n+1)$$
Concluding that $1+n$ is bigger cause 1 plus an integer is bigger than 1 plus $1/(n+1)$
however, I'm not sure if it is valid to use division.
 A: I Don's see a problem with dividing, however, you need to show that $(n+1)^2 > (n+1)+1$ based on the fact that you have assumed that $n^2 > n+1$
I would work one side of $(n+1)^2 > (n+1)+1$ until you showed that the relation holds.
$(n+1)^2 = n^2 + 2n + 1 > (n+1) + 2n+1 = 3n+2 > n+2$ when $n>1 $
A: As $n\geq 2$, you can indeed divide by $n+1$ there, that's not really a problem. All that remains is to prove the base case, i.e. when $n=2$.
In case you're interested, an alternative argument for the entire problem without induction would be to notice that, as $n\geq 2$,
$$n^2\geq 2n=n+n> n+1.$$
A: Based on the comments, It seems you want to prove this statement: $n^2 > n+1$ for all integers $n \ge 2$. To do this with induction, you would first show this statement is true for $n=2$, (simple enough, 2^2 > 2+1), then show if the statement is true for an integer $n \ge 2$, then it is true for $n+1$.
In the induction step, we assume that $n^2 > n+1$ is true, and want to show that $(n+1)^2 > (n+1)+1$ is true. In other words, we don't know whether $(n+1)^2 > (n+1)+1$ is true, that is what we need to verify.
So, you must work starting from something you know is true (such as $n \ge \frac{1}{n+1}$ for all $n \ge 2$) and deduce that $(n+1)^2 > (n+1)+1$ with the assumption that $n^2 > n+1$.
So in your proof, you're working a bit backwards by starting at $(n+1)^2 > (n+1)+1$, which is a normal way to figure out the solution. But your actual solution should be going the other way. So the heart of you question is: Is division logically reversible, and the answer is yes. More precisely, for $c > 0$, $a > b$ if and only if $\frac{a}{c} > \frac{b}{c}$.
Now in fact, you didn't even use induction to prove this, since your argument doesn't use the assumption $n^2 > n+1$. You've written a direct proof that $(n+1)^2 > (n+1)+1$. An inductive proof would look like the following:
Base Case: $2^2 > 2 + 1$ is true
Inductive step: Let $n \ge 2$ and assume that $n^2 > n+1$. Then,
$$ (n+1)^2 = n^2 + 2n + 1 > (n+1) + 2n + 1 = 3n+2 > (n+1)+1$$
