Alternative route:
Write $x=n+r$ and $y=m+s$ where $n$ and $m$ are integers and $r,s\in\left[0,1\right)$.
Then $\lfloor2x\rfloor+\lfloor2s\rfloor=2n+2m+\lfloor2r\rfloor+\lfloor2s\rfloor$
and $\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor=2n+2m+\lfloor r+s\rfloor$.
This shows that it is enough to prove $\lfloor2r\rfloor+\lfloor2s\rfloor\geq\lfloor r+s\rfloor$
For this discern the cases $r,s\in\left[0,0.5\right)$ and $r\in\left[0.5,1\right)\vee s\in\left[0.5,1\right)$.
You are correct that mathematical induction consists of three steps:
- Prove the base case.
- Assume that the claim is true for some $n\in\Bbb N$ (possibly including $0$ if needed).
- Use the above assumption to prove the claim for $n+1$.
Let us see this in practice with standard example:
$$1+2+3+\ldots+n = \frac{n(n+1)}{2}\tag{1}$$
Base case: $$1=\frac{1(1+1)}{2}\quad\small\text{(obviously true)}$$
Assume that formula $(1)$ is true for some $n$. Now we need to prove that $$\underbrace{1+2+3+\ldots +n}_{\text{use the assumption on this}}+(n+1) = \frac{(n+1)(n+2)}{2}$$ and we can use the assumption to get $$1+2+3+\ldots +n+(n+1) = \frac{n(n+1)}{2} +(n+1) = \frac{(n+1)(n+2)}{2}$$ so we are done.
Now, in your example, what is used is version of induction called strong induction. The difference is that in step 2, we assume something stronger, but it turns out this is equivalent to "weak" induction (just happens to be more convenient in some cases).
- Prove the base case.
- Fix some $n\in\Bbb N$ and assume that the claim is true for all natural numbers $k$ when $k< n$.
- Use the above assumption to prove the claim for $n$.
I will now try to clarify the steps in your proof. Let us define $m=a+b\in\Bbb N$. We will do induction on $m$.
Base case is $m = 2$. Since $a+b = 2$ and $a,b>0$, we must have $a=b=1$, so $$\gcd(na,nb) = \gcd(n,n) = |n| = |n|\gcd(a,b).$$
Now let $m\in\Bbb N$ and assume that $$\gcd(na',nb') = |n|\gcd(a',b'),\quad(\forall a',b'\in\Bbb N)\ a'+b' < m.\tag{2}$$ We want to prove that the claim is true for $a,b$ (remember that $a+b=m$).
Using Euclid's division, we have $a = qb + r$, $0\leq |r|<b$, and thus $$\gcd(a,b) = \gcd(b,r).$$ But if we multiply everything by $n$, we get $na = qnb + nr$, $0\leq |nr| < nb$, so we have that $$\gcd(na,nb) = \gcd(nb,nr).$$
Notice that if we can prove that $$\gcd(nb,nr) = |n|\gcd(b,r),\tag{3}$$ we will have $$\gcd(na,nb) = |n|\gcd(a,b)$$ as well. So let us see if we can employ our assumption $(2)$ - for it to work, we must check that $b+r<m$, and you can check with your proof how it is done. Hence, we can apply induction assumption to conclude that $(3)$ is true.
Also, notice that we indeed needed strong induction, since we don't know that $b + r + 1 = m$, which we would want in ordinary induction.
Best Answer
Let $a,b>0$, $d=(a,b)$. Let $a'=ad^{-1},b'=bd^{-1}$. Suppose that $e\mid a',b'$. Then $de\mid a,b$, so $de\mid (a,b)=d$. This means $e\mid 1$. It follows that $(a',b')=1$.