I am extremely poor at proofs and logical manipulation so I am stuck on a lot of these questions especially induction.
The question below I have been stuck at for a little over 1 hour and I can't get it, I have tried various substitutions and forms but I am stuck..
Using mathematical induction:
An integer is odd if it can be written as $n=2k+1$. Use induction to prove that the product of $m$ many odd integers is odd for every $m \geq 2$.
I have tried doing $(2k+1)(2(k+1)+1)$ and substituting $k=1$ for my basis step after that I am lost.
Best Answer
For the base case, we consider when $m=2$. For $a,b \in \mathbb{Z}$, that is, $(2a +1)(2b +1) = 4ab +2a+2b +1 = 2(2ab + a + b) +1$, where $2ab + a + b \in \mathbb{Z}$, so it is true for $m=2$.
By induction, we assume for arbitrary $k \in \mathbb{Z}$ that, $$\prod_{i=1}^{k}(2a_{i} +1)=2l+1$$ where $l \in \mathbb{Z}$. Since by our basis step we have that two odd numbers multiplied together gives two odd numbers, then for the $k+1^{th}$ odd integer $c$, we have that $$(2l+1)(2c+1)=2(2lc + l + c) +1$$ So, since $2lc + l + c \in \mathbb{Z}$, we have that $$\prod_{i=1}^{k+1}(2a_{i}+1)=2j+1$$ where $j \in \mathbb{Z}$. Therefore, by the Principle of Mathematical Induction, the product of $m$ odd integers is again odd.