How to prove that adding $n$ to the numerator and denominator will move the resultant fraction close to $1$?

Question

Given a fraction:

$$\frac{a}{b}$$

I now add a number $n$ to both numerator and denominator in the following fashion:

$$\frac{a+n}{b+n}$$

The basic property is that the second fraction is suppose to closer to $1$ than the first one. My question is how can we prove that?

What I have tried:

I know $\frac{n}{n} = 1$ so now adding numbers $a$ and $b$ to it would actually "move it away" from $1$. But I cannot understand why $\frac{a}{b}$ is actually farther away from $1$ than $\frac{a+n}{b+n}$.

Why is that? What does it mean to add a number to both the numerator and denominator?

Answer 1

Well, $\frac{a+n}{b+n} = \frac{\frac{a}{n}+1}{\frac{b}{n}+1}$. So if $n\rightarrow \infty$, 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.

Answer 2

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?

Answer 3

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)

Answer 4

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

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

Answer 5

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.

Answer 6

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

Answer 7

If $b$ and $d$ have the same sign, both $$ \frac ab-\frac{a+c}{b+d}=\frac1b\frac{ad-bc}{b+d}\tag1 $$ and $$ \frac{a+c}{b+d}-\frac cd=\frac1d\frac{ad-bc}{b+d}\tag2 $$ also have the same sign. Thus, $$ \frac{a+c}{b+d}\text{ is between }\frac ab\text{ and }\frac cd\tag3 $$ Therefore, if $bn\gt0$, $$ \frac{a+n}{b+n}\text{ is between }\frac ab\text{ and }\frac nn=1\tag4 $$

Answer 8

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

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

Answer 9

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

Perform the long division to get $$\frac{a+n}{b+n}=1+\frac{a-b}n\left(1-\frac{b}{n}+\frac{b^2}{n^2}-\frac{b^3}{n^3} +\cdots\right)=1+\frac{a-b}n\sum_{k=0}^\infty (-1)^k \left(\frac bn\right)^k$$

Answer 10

$$ f(x)=\frac{a+x}{b+x} $$ $ b > 0$

$$\lim_{x→ ∞} \frac{a+x}{b+x}=1$$ $$ f'(x)= \frac{b+x-a-x}{(b+x)^2}=\frac{b-a}{(b+x)^2}$$

We can conclude if $b > a$ the function is monotonically increasing to 1.

If $b < a $ function is monotonically decreasing to 1

If $b<0$ the conclusion doesnt follow because there exists a vertical asymptote at $x=-b$

Answer 11

Intuition?

For me the intuition is this: The absolute difference in size becomes less significant when we are comparing big things than when we are comparing small thing. e.g. If one person ways $100$ lbs more than another that is significant. If one elephant is $100$ lbs heavy then another that's noticeable if you look really close but not significant. If a building is $100$ lbs heavier than another it is ludicrous to even attempt to point that out (and darn near impossible to actually measure accurately). If a mastiff is $100$ lbs heavier than a rabbit... well, that shows they are entirely different things.

Adding a positive $n$ to both terms of a fraction "pushes" them both to a large frame of reference where the actual difference between them $(a-b)$ is less significant. $(a-b) = 2$ is a big part of $a = 3$ ($67\%$) and a big part of $b = 5$ ($40\%$) when it comes to comparing $a$ to $b$ the fact that they are not equal but apart by $2$ is going to make a big difference. But $(a-b) =2 $ not such a big deal when $a = 10$ (then $2$ is only $20$ percent) and $b = 12$ (then $2$ is only $17\%$) then the fact that they are not equal isn't that important because there are only $2$ which is a small proportion of either.

But that's just intuition. A proof needs to be done algebraicly and that's.... straightforward.

What does it mean to add a number to both the numerator and denominator?

Well, nothing mysterious. You are comparing the proportion of two numbers and adding $n$ to both means you are a different pair of numbers-- a pair where each term is $n$ more.

I guess a proof that is focused on this idea might be: if we assume $a - b =m$ ($m \ne 0$ but $m < 0$ is possible if $b < a$) then:

$\frac ab = \frac {b+m}b = 1 + \frac mb$. Whereas $\frac {a+n}{b+n} = \frac {b+m + n}{b+n} = 1 + \frac m{b+n}$

And $|\frac m{b+n}| < |\frac m{b+n}|$ so $1 + \frac m{b+n}$ is closer to $1$ than $1 + \frac m{b}$ is.

... or in other words...

If we notice that $\frac ab = 1 \pm \delta$ then $\delta = \frac {|numerator - denominator|}{denominator}$, then as the denominator becomes larger but the difference between the numerator and the denominator stay the same, $\delta$ becomes smaller and less significant.

.... or in my opinion best yet.....

Distance between $1$ and $\frac ab=|1 - \frac ab| = |\frac {b-a}b|$.

Distance between $1$ and $\frac {a+n}{b+n} =|1 - \frac {a+n}{b+n}| = |\frac {(b+n) - (a+n)}{b+n}| = |\frac {b-a}{b+n}|$.

An $|\frac {b-a}{b+n}| < |\frac{b-a}{b+n}|$.