Fraction increases on adding 1?

If $a = b$ , then $\displaystyle \frac { a + 1 } { b + 1 } = \frac { a + 1 } { a + 1 } = 1$ because $a$ and $b$ are natural numbers.

And, $\displaystyle \frac { a + 1 } { a + 1 } = \frac { a } { a } = 1$ .

Hence, it is not true always.

$\frac {a + 1}{b + 1} > \frac {a}{b} \iff \frac {a + 1}{b + 1} - \frac {a}{b} > 0 \iff$ $\frac {ab + b - ab - a}{b(b + 1)} > 0 \iff$ $\frac {b - a}{b(b + 1)} > 0$ ,

and since b > 0 :

$\iff b - a > 0 \iff$ $\boxed {b > a}$ ,

which means that it is not true, if a ≥ b.

Therefore, our answer is $\boxed {No}$

Take for instance a = b = 1. Then the inequality will become,

$\dfrac{1+1}{1+1} > \dfrac{1}{1} \\ \implies 1 > 1$

Which is not true.

Let us consider $x = \dfrac {LHS}{RHS}$ . If $x > 1$ , $\implies LHS > RHS$ and if $x < 1$ , $\implies LHS < RHS$ .

$\begin{aligned} \frac {\frac {a+1}{b+1}}{\frac ab} & = \frac {b(a+1)}{a(b+1)} = \frac {ab+b}{ab+a} \implies \begin{cases} > 1 & \text{if } b > a \\ < 1 & \text{if } b < a \end{cases} \end{aligned}$

Therefore, $\dfrac {a+1}{b+1} > \dfrac ab$ is not always true. The answer is $\boxed{\text{no}}$ .