Comparing two fractions

Question

I saw this problem from an elementary textbook:

Let

$$ A = \frac{2014}{2015} + \frac{2015}{2016} $$

and

$$ B = \frac{2014 + 2015}{2015 + 2016} $$

Compare $A$ and $B$.

I know the answer is $A \gt B$ because $$\frac{A}{B} \gt 1.$$ But I cannot find a precise way to explain it. Can anyone please help? Thank you.

Answer 1

Aditya's comment gives the simplest way to see that $A>B$.

Another approach is to note that if $a, b, c,$ and $d$ are positive and

$$\frac{a}{b}<\frac{c}{d}$$

then

$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$

This is a very useful fact about ratios. When you add the numerators and denominators, you get a fraction in between the two you started with.

In your case, if we say $A=X+Y$, with $X=2014/2015$ and $Y=2015/2016$, then it is obvious that $X<Y$, and the above fact implies we have $X<B<Y$. Therefore $B<Y<Y+X=A$.

In other words, we must have

$$B<\frac{2015}{2016}$$

so of course $B<A$.

Answer 2

We are comparing $\frac{a}{b}+\frac{b}{c}$ with $\frac{a+b}{b+c}$

$\frac{a}{b}+\frac{b}{c}=\frac{a(\frac{b+c}{b})+b(\frac{b+c}{c})}{b+c}$ so we can compare the numerators.

$a(\frac{b+c}{b})=a(1+\frac{c}{b})\gt a$

Likewise,

$b(\frac{b+c}{c})=b(1+\frac{b}{c})\gt b$

Therefore, $\frac{a}{b}+\frac{b}{c}=\frac{a(\frac{b+c}{b})+b(\frac{b+c}{c})}{b+c}\gt \frac{a+b}{b+c}$ so $A\gt B$

Answer 3

Beside the general proofs given in answers$$A = \frac{2014}{2015} + \frac{2015}{2016} = \frac{2014+1-1}{2015} + \frac{2015+1-1}{2016}=1-\frac{1}{2015}+1-\frac{1}{2016}$$ $$A=2-\frac{1}{2015}-\frac{1}{2016}=2-small_1$$ $$B = \frac{2014 + 2015}{2015 + 2016}=\frac{2014+1-1 + 2015+1-1}{2015 + 2016}=\frac{2015-1 + 2016-1}{2015 + 2016}$$ $$B=1-\frac 2{2015+2016}=1-small_2$$