Cfrgtkky

There's a very simple way to see this. Just take the difference between the two fractions and 1. You want to show that this is smaller in modulus for the second fraction.

You get $$ frac{a}{b} - 1 = frac{a-b}{b} $$ and
$$ frac{a+n}{b+n} -1 = frac{a-b}{b+n} $$

So the second is smaller in modulus (provided $b$ and $n$ are positive, although I supposed it also works if both are negative) because it has same numerator and larger (modulus) denominator, QED.

Visually: Consider the slope of the line segment from $(0, 0)$ to $(a+n, b+n$):

Mathematically (assuming $a, b, n > 0$): The distance
$$
left| frac {a+n}{b+n} - 1right| = frac{|a-b|}{b+n}
$$
is decreasing in $n$ (and approaches zero for $n to infty$).

You should start by thinking about particular cases. For instance, $dfrac{3+2}{7+2}=dfrac59$, which is indeed closer to $1$ than $dfrac37$.

Anyway, note that, if $a<b$ (and consequently, $a+n<b+n$, for which $frac ab<1$ and $frac{a+n}{b+n} < 1$), then$$frac{a+n}{b+n}-frac ab=frac{(a+n)b-a(b+n)}{(b+n)b}=frac{n(b-a)}{(b+n)b}>0$$
This shows $frac{a+n}{b+n}-frac ab>0$, and we already know both are $<1$, so:
$$frac ab<frac{a+n}{b+n}<1.$$So, yes, $dfrac{a+n}{b+n}$ is closer to $1$ than $dfrac ab$.

Can you deal with the case $a>b$ now?

If $b$ and $d$ have the same sign, both
$$
frac ab-frac{a+c}{b+d}=frac1bfrac{ad-bc}{b+d}tag1
$$
and
$$
frac{a+c}{b+d}-frac cd=frac1dfrac{ad-bc}{b+d}tag2
$$
also have the same sign. Thus,
$$
frac{a+c}{b+d}text{ is between }frac abtext{ and }frac cdtag3
$$
Therefore, if $bngt0$,
$$
frac{a+n}{b+n}text{ is between }frac abtext{ and }frac nn=1tag4
$$

Well, $frac{a+n}{b+n} = frac{frac{a}{n}+1}{frac{b}{n}+1}$. So if $nrightarrow infty$, then $frac{a}{n}rightarrow 0$ and $frac{b}{n}rightarrow 0$. Thus $frac{a+n}{b+n}rightarrow 1$.

As said in the comments, the answer is incorrect in that it does not address precisely what the OP asks, but gives some intuition as to why it is true.

Suppose $a,b,n in mathbb Q$, $0 < a < b$ and $n > 0$.

$$dfrac ab = dfrac{a(b+n)}{b(b+n)} = dfrac{ab+an}{b(b+n)}
<dfrac{ab+bn}{b(b+n)} = dfrac{b(a+n)}{b(b+n)} = dfrac{a+n}{b+n}
< dfrac{b+n}{b+n} = 1$$

Let $a=kb$. ($k$ doesnt necessarily have to be an integer). Then:

$$frac ab = k$$

$$frac{a+n}{b+n}=frac{k(b+n)-(k-1)n}{b+n}$$
$$=k-frac{kn-n}{b+n}$$

Can you show the extra term is positive when $k>1$, and negative when $k<1$? (Hint: let $k=1+t$ for first case and $k=1-t$ for the second)

You have to suppose $a,b >0$. Now, it is clear that, if $a<b,;$ i.e. $:smash{dfrac ab}<1$, $a+n<b+n$, hence $smash{dfrac{a+n}{b+n}}<1$, and similarly if $dfrac ab>1$.

• If $dfrac ab<1$, then $;dfrac ab<dfrac{a+n}{b+n}:(<1)$, which is equivalent to
  $$a(b+n)<b(a+n)iff an<bniff a<b.$$
  


• Similar proof that if $dfrac ab>1$, then $;dfrac ab>dfrac{a+n}{b+n}:(>1)$.

Just for the fun of it, since you already received very good answers.

Perform the long division to get
$$frac{a+n}{b+n}=1+frac{a-b}nleft(1-frac{b}{n}+frac{b^2}{n^2}-frac{b^3}{n^3} +cdotsright)=1+frac{a-b}nsum_{k=0}^infty (-1)^k left(frac bnright)^k$$