manipulation of subtraction

Question

I am trying to solve an induction problem and got stuck at this part.

$$ 1 - \frac{n+2}{(n+2)!} + \frac{n+1}{(n+2)!} = 1 - \frac{(n+2) - (n+1)}{(n+2)!} $$

Shouldn't it be

$$ 1 - \frac{n+2}{(n+2)!} + \frac{n+1}{(n+2)!} = 1 - \frac{(n+2) + (n+1)}{(n+2)!} $$

How do you get the left expression to the right expression?

Answer 1

Would you feel more comfortable with this?

$$1-\dfrac{n+2}{(n+2)!}+\dfrac{n+1}{(n+2)!}=1+\dfrac{(n+1)-(n+2)}{(n+2)!}$$

If you chose to, you could rewrite it from here to

$$1-[-\dfrac{(n+1)-(n+2)}{(n+2)!}]=1-\dfrac{(n+2)-(n+1)}{(n+2)!}$$

Answer 2

Note that $$-\frac{A+B}{C}=(-1)\times \left(\frac{A+B}{C}\right)=(-1)\times\left(\frac{A}{C}+\frac{B}{C}\right)=-\frac{A}{C}-\frac{B}{C}=\frac{-A-B}{C}.$$

The minus sign in front of the first fraction does influence the plus sign in front of $B$ of the numerator of the first fraction.