Proof for $\frac{a}{b} < \frac{a+1}{b+1}$

Question

Supposing $b > 0$ and $a < b$, how could I prove:

$$ \frac{a}{b} < \frac{a+1}{b+1} $$

Related: How to prove that adding $n$ to the numerator and denominator will move the resultant fraction close to $1$? — Martin R, Sep 28 '19 at 21:47

score 5 · Answer 1 · answered Sep 28 '19 at 20:54

5

It's equivalent to $$ab + {\color{red}a} < ab +{\color{red}b}$$

answered Sep 28 '19 at 20:54

Bman72

score 2 · Answer 2 · edited Sep 28 '19 at 20:58

2

Hint:

$\dfrac{a}{b} = \dfrac{a(b+1)}{b(b+1)} = \dfrac{ab+a}{b(b+1)}$. However, $\dfrac{a+1}{b+1} = \dfrac{(a+1)(b)}{(b)(b+1)} = \dfrac{ab+b}{b(b+1)}$.

If $a<b$ then what about $ab+a$ vs $ab+b$?

edited Sep 28 '19 at 20:58

Sebastiano

answered Sep 28 '19 at 20:55

Mike

2 Answers2