Prove that $\frac{n-a}{n} < \frac{n+1-a}{n+1}$?

Question

I have this math question that I'm kind of stuck on.

Prove that $\frac{n-a}{n} < \frac{n+1-a}{n+1}$

So far I have that: $\frac{n-a}{n} < \frac{n+1-a}{n+1} = n-a < \frac{n(n+1-a)}{n(n+1)} = n-\frac{n(n+1-a)}{n(n+1)} < a$

However, I'm not sure where to go from here. Thanks

Answer 1

$$\frac {n-a}n = 1 - \frac a n < 1-\frac a{n+1}=\frac{n+1-a}{n+1}$$