An algebra problem by Chung Kevin

Let's convert all of the fractions into decimals.

$\frac12 = 0.5$
$\frac{1+3}{2+4} = \frac46 = \frac23 = 0.666666\ldots$
$\frac34 = 0.75$

Thus, we conclude that $\frac12 < \frac{1+3}{2+4} < \frac34$

Now, lets look at other fun ways that I came up with to solve this problem. (And that's the real reason why I wrote this solution)

2nd solution We can solve this by applying the Farey Sequence .

$F_4 = \left \{ \frac01 , \frac14,\frac13, \frac12,\frac23 ,\frac34, \frac11 \right \}$

So, $\frac12 < \frac23 < \frac34$ .

With $\frac23 = \frac46 = \frac{1+3}{2+4}$ . We have the desired inequality: $\frac12 < \frac{1+3}{2+4} < \frac34$ .

3rd solution Alternatively (again), we can prove that for all positive intgers $a,b$ , the inequality $\frac ab < \frac{a+1}{b+1}$ is true for $a< b$ .

Proof by contradiction: Suppose otherwise, that $\frac ab \geq \frac{a+1}{b+1}$ then $a(b+1) \geq b(a+1) \Rightarrow a \geq b$ which contradicts the criteria.

So $\frac ab < \frac{a+1}{b+1}$ is indeed true. Thus $\frac12 < \frac{1+1}{2+1} < \frac{1+1+1}{2+1+1}$ or $\frac12 < \frac{1+3}{2+4} < \frac34$ .

4th solution Alternatively (yet again!), let $A,B,C$ denote the values of these three fractions respectively. Then $1 - A = \frac12 , 1-B = \frac14, 1-C = \frac13$ .

Because $2 < 3 < 4$ , then $\frac14 < \frac13 < \frac12$ or $1-B < 1-C < 1-A$ . Equivalently, $-B<-C<-A$ or $A<C<B$ .

Thus $\frac12 < \frac{1+3}{2+4} < \frac24$ .

5th solution Alternatively (not again!), we set the denominators of all of these fractions to be equal. With $\frac{1+3}{2+4}$ simplifies to $\frac23$ . then the denominators of all three fractions are $2,3,4$ respectively. And the lowest common multiple of these three numbers is $2^2 \times 3 = 12$ .

So $\frac12 = \frac6{12}$ , $\frac{1+3}{2+4} = \frac23 = \frac8{12}$ , $\frac34 = \frac9{12}$ .

With $\frac6{12} < \frac8{12} < \frac9{12}$ or $\frac12 < \frac{1+3}{2+4} < \frac34$ .

6th solution Alternatively (please stop!), consider the reciprocal of these fractions. That is $2, \frac43, \frac32$ .

Because $1 < 2 < 3$ , then $\frac13 < \frac12 < 1$ or $1 + \frac13 < 1 + \frac12 < 2$ . Which is equivalent to $\frac43 < \frac32 < \frac21$ .

Reciprocate back gives $\frac12 < \frac23 < \frac34$ . With $\frac23 = \frac{1+3}{2+4}$ . So the numbers arranged in ascending order is $\frac12 < \frac{1+3}{2+4} < \frac34$ .

Python 3.3:

1
2
3
4

from fractions import Fraction as frac
nums = [frac(1,2), frac(3, 4), frac(1+3,2+4)]
nums.sort()
print("{0} < {1} < {2}".format(nums[0], nums[1], nums[2]))

$\implies \frac{1}{2} < \frac{2}{3} < \frac{3}{4} \implies \boxed{\frac{1}{2} < \frac{1+3}{2+4} < \frac{3}{4}}$