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

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?

• @SikFengCheong I dont understand. What does it mean? – ng.newbie Feb 6 '19 at 8:36
• If $a \lt b$, I just think of the two fractions as batting averages. If you start with a batting average of $a/b$ and get $n$ hits in your next $n$ at bats, your batting average will go up. – Robert Shore Feb 6 '19 at 9:01
• Adding one to the numerator and denominator of $\frac{-2}{-3}$ increases the distance from $1$. – Martin R Feb 6 '19 at 9:17
• Compute the limit as $n \rightarrow \infty$. – chepner Feb 6 '19 at 21:00
• It's not true in general, but it is true for $n, a, b > 0$. – chepner Feb 6 '19 at 21:06

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 $$\left| \frac {a+n}{b+n} - 1\right| = \frac{|a-b|}{b+n}$$ is decreasing in $$n$$ (and approaches zero for $$n \to \infty$$).

• [+1] The value of a fraction as a slope is too rarely found. More generally, there are not enough answers that use graphical representations permiting to have a different, complementary, mental representation. It is what I try to do as well in many of my answers. – Jean Marie Feb 7 '19 at 6:50

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 $$a (and consequently, $$a+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?

• I dont understand how $$(a+n)b-a(b+n) = n(b-a)$$ – ng.newbie Feb 6 '19 at 8:46
• Because $(a+n)b-a(b+n)=ab+nb-ab-an=n(b-a)$. – José Carlos Santos Feb 6 '19 at 8:47
• You don't understand that $a+n<b+n\iff\frac{a+n}{b+n}<1$? – José Carlos Santos Feb 6 '19 at 9:05
• No! I proved that dividing both sides of the inequality $a+n<b+n$ by $b+n$. – José Carlos Santos Feb 6 '19 at 9:06
• Yes, that is what I proved, except that I did not mention the “decimal value of that fraction”, whatever that is. – José Carlos Santos Feb 6 '19 at 9:14

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 $$bn\gt0$$, $$\frac{a+n}{b+n}\text{ is between }\frac ab\text{ and }\frac nn=1\tag4$$

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.

• Comments are not for extended discussion; this conversation has been moved to chat. – Pedro Tamaroff Feb 7 '19 at 14:58
• 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, so please modify it or otherwise delete it, since the vote tally 17/12 which is now at 5 might mislead other users into thinking it is correct. I can also convert it into a comment. Thank you, – Pedro Tamaroff Feb 7 '19 at 14:58
• @PedroTamaroff: I'm not sure why Xander Henderson's precise objection has been moved to chat, since it is obviously not an extended discussion. In any case, it is false that this answer gives any intuition whatsoever. Surely you won't say that $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ for positive integer $n$ intuitively 'based' on the fact that $\sqrt[n]{n} → 0$ as $n → ∞$? I've just given an explicit example to show that the notion in this post is useless. – user21820 Feb 8 '19 at 8:09
• Contrary to what a mod (!) pretends above, this answer does not give any intuition as to why the fraction (a+n)/(b+n) "is suppose[d] to [be] closer to 1 than" the fraction a/b, which is what the OP is asking. (For the record, I also share @user21820's puzzlement about the reason why a precise mathematical objection formulated in the comments by another user to this effect, has been moved to chat.) – Did Feb 8 '19 at 8:11
• @user170039 Why these comments? Again, here is the situation: 1. The result is true. 2. Its proof is simple. 3. Such a proof is not in the answer above. 4. The answer above does not "give [any] intuition as to why it is true". 5. A mod deleted some comments explaining why. 6. And asserted wrongly that the answer above "gives some intuition as to why it is true". 7. Some users (but perhaps not all of them) find that mathematically wrong answers should be signalled as such. 8. Some users (but perhaps not all of them) find annoying that a mod endorses some wrong mathematical content ... – Did Feb 8 '19 at 14:12

You have to suppose $$a,b >0$$. Now, it is clear that, if $$a i.e. $$\:\smash{\dfrac ab}<1$$, $$a+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 $$a(b+n)
• Similar proof that if $$\dfrac ab>1$$, then $$\;\dfrac ab>\dfrac{a+n}{b+n}\:(>1)$$.

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

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

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

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

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 $$x=-b$$

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

• +1 because it is correct unlike the heavily upvoted wrong answer. – user21820 Feb 20 '19 at 6:50

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)

• Can you please how $a+b = k(b+n)-(k-1)n$ ?? I don't understand how you got that. – ng.newbie Feb 6 '19 at 11:21
• Thats not what it is. Its $(a+n)=(bk+n)$, and then we use $n=kn-(k-1)n$ – Rhys Hughes Feb 6 '19 at 11:49

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