Mathematical induction can I assume this? let $a,b\in\mathbb R$ so that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$ prove that $a^n+b^n\in\mathbb Q$.
So I tried to prove using mathematical induction and got to:
$$(a^k+b^k)(a+b)-ab(a^{k-1}+b^{k-1})$$
Now we know that $(a^k+b^k)(a+b)\in\mathbb Q$ from the Induction assumption.
My question is that if I assume something is true for all $n=k$ does it  mean it's also true for $k-1$? 
and why?
 A: You can write a perfectly valid proof by induction that assumes the statement is true for $k$ and $k-1$ and deduces that it is true for $k+1$. However, in order to do this you need to check two base cases (i.e. that it works for $n=1$ and $n=2$) rather than one. This is because the first case that isn't a base case has to depend on two previous cases, which both have to be base cases.
This works because once you know it's true for $1$ and $2$ you can show it's true for $3$; you already know $2$ so it's true for $2$ and $3$, which allows you to show it for $4$; and so on...
This is something you often see in e.g. proving things about the Fibonacci numbers, where each one is defined in terms of the previous two.
A: Yes, you can assume that.
Suppose you have your predicate $P(n)$ to prove and consider the predicate $Q(n): \forall j\le n: P(j)$. Of course we see that $Q(n)\Rightarrow P(n)$ so if we prove that $Q(n)$ is true for all natural number then so is $P(n)$
Now if you can prove that $Q(n)\Rightarrow P(n+1)$ you use the fact that $Q(n)\land P(n+1)\Rightarrow Q(n+1)$ So that means that we know that $Q(n)\Rightarrow Q(n+1)$.
So given that $Q(0)$ is true and $Q(n)\Rightarrow P(n+1)$ then we have that $Q(n)$ is true for all natural numbers and therefore also $P(n)$.
Note that your assumption is slightly different however. If you have the predicate $R(n): P(n) \land P(n-1)$ that doesn't make sense if $n<1$, but otherwise we have that $Q(n)\Rightarrow R(n)$. You would to use this need to dafine $R(0)$ specially or prove $P(1)$ explicitely.
