Which fraction is bigger?

Thank you to Minhaj from St Ivo School for his very thorough solution to this problem.

Here is his solution to the warm up problem:

Minhaj thought of a few different ways to convince himself that $\frac{2}{5}$ is bigger than $\frac{1}{3}$

1. The lowest common denominator for $\frac{2}{5}$ and $\frac{1}{3}$ is $15$ and hence $\frac{2}{5}$ can be written as $\frac{6}{15}$ and $\frac{1}{3}$ can be written as $\frac{5}{15}$. It can now be seen that $\frac{2}{5} > \frac{1}{3}$ because when the denominator is $15$ for both fractions, the numerator is $1$ number larger for the fraction equivalent to $\frac{2}{5}$ which is $\frac{6}{15}$

2. If you were to simply do the division for both fractions, you would get $\frac{2}{5}= 0.4$ and $\frac{1}{3}= 0.3333”¦.$ Since the number in the tenths place for $\frac{2}{5}$ is $4$ and for $\frac{1}{3}$ it is $3$, this again shows that $\frac{2}{5}> \frac{1}{3}$ as $4$ is bigger than $3$

3. Imagine you got a string which has a length of 15cm. If you cut $\frac{2}{5}$ of this string, then you would a length of 6cm. If you cut $\frac{1}{3}$ of this string then you would get a length of 5cm. This also shows that $\frac{2}{5}> \frac{1}{3}$ as $\frac{2}{5}$ of $15$ is larger than $\frac{1}{3}$ of $15$. You can use this idea with any length or any other instance (e.g. time)

What if you had the fractions $\frac{2x}{5}$ and $\frac{x}{3}$?

The answer would only be the same if $x$ is some positive number which is not $0$. You can think of $\frac{2x}{5}$ as $x \times \frac{2}{5}$ and you can think of $\frac{x}{3}$ as $x \times \frac{1}{3}$. We have already established that $\frac{2}{5}>\frac{1}{3}$ and hence if $x$ is some positive constant which is not zero, the fractions are just getting multiplied by a fixed value and hence they are only getting scaled up or down at the same rate. If $x=0$, then $\frac{2x}{5}=\frac{x}{3}$ because you end up with a $0$ on the numerator for both fractions and $0$ divided $5$ or $3$ is just $0$. It gets interesting when $x$ is negative. If $x$ is negative, then $\frac{x}{3}> \frac{2x}{5}$. Because $\frac{2}{5}$ is bigger than $\frac{1}{3}$, this means if you multiply $\frac{2}{5}$ by a negative number, then the number is larger on the negative side (it is further down the number line on the negative side than $\frac{x}{3}$). What this consequently means is that $\frac{2x}{5}$ is smaller than $\frac{x}{3}$ as $\frac{x}{3}$ would be closer to $0$.

Here is Minhaj's solution to the main problem

$\frac{4x}{7}$ or $\frac{9}{14}$

$\frac{4x}{7}$ can be written as $\frac{8x}{14}$; this makes the denominators for both fractions the same and we only need to focus on the numerators. The only time these fractions are equal is when $8x=9$ and so when $x=\frac{9}{8}$ or $1.125$ then the fractions are equal. If $x$ is greater than $\frac{9}{8}$, then this means that $\frac{4x}{7}> \frac{9}{14}$. However if $x< 1.125$ then $\frac{9}{14}> \frac{4x}{7}$.

$\frac{5}{9}$ or $\frac{2x}{12}$

Again we should try to get the lowest common denominator. In this case it is $36$ and so $\frac{5}{9}= \frac{20}{36}$ and $\frac{2x}{12}= \frac{6x}{36}$. Same like last time, the only time the fractions are the same is if the numerators are equal. So when $6x=20$ and $x= \frac{10}{3}$, then $\frac{5}{9}= \frac{2x}{12}$. If $x$ is greater than $\frac{10}{3}$, then $\frac{2x}{12}> \frac{5}{9}$ but if $x$ is less than $\frac{10}{3}$ then $\frac{5}{9}> \frac{2x}{12}$.

$\frac{3x}{4}+1$ or $\frac{x}{4}+3$

$(\frac{3x}{4})+1=  (\frac{3x}{4})+(\frac{4}{4})= \frac{3x+4}{4}$,

$(\frac{x}{4})+3= (\frac{x}{4})+(\frac{12}{4})= \frac{x+12}{4}$. Both of the fractions have already got a common denominator and so when $3x+4=x+12$ i.e. $x=4$, then the fractions are the same as they got the same numerator (this is solving both linear expressions simultaneously). For $(\frac{3x}{4})+1$ to be greater than $(\frac{x}{4})+3$, then this means that $3x+4$ must be greater than $x+12$- in other words $x$ must be greater than $4$. Conversely, if $x<4$,  then $(\frac{x}{4})+3 > (\frac{3x}{4})+1$.

$\frac{8(1-x)}{5}$ or $\frac{x}{6}$

$\frac{8(1-x)}{5}= \frac{8-8x}{5}= \frac{48-48x}{30}$ and $\frac{x}{6}=\frac{5x}{30}$. The denominators are now the same so when the numerators are the same- when $48-48x=5x$ i.e. $x=\frac{48}{53}$, then the fractions are the same so $\frac{8(1-x)}{5}= \frac{x}{6}$. If $5x> 48-48x$ i.e. $x> \frac{48}{53}$, then $\frac{x}{6}> \frac{8(1-x)}{5}$. Conversely if $48-48x> 5x$ i.e. $x<\frac{48}{53}$, then $\frac{8(1-x)}{5}> \frac{x}{6}$.

$\frac{8}{2x}$ or $\frac{4x}{16}$

$\frac{8}{2x}= \frac{4}{x}$ and $\frac{4x}{16}=\frac{x}{4}$- so it can be seen that we are comparing the reciprocals of one another where $y=\frac{8}{2x}$ is a reciprocal graph and $y=\frac{4x}{16}$ is a linear graph with a gradient of $\frac{1}{4}$.  For the two fractions to be equal, then $\frac{4}{x}$ must be equal to $\frac{x}{4}$. This means $x^2=16$ and so $x=4$ or $x=-4$. This makes sense as $\frac{4}{-4}= \frac{-4}{4}=-1$ and $\frac{4}{4}=\frac{4}{4}=1$ (you get this by substituting $x=4$ or $-4$ into the functions).

Image

As can be seen by looking at the graph above, $\frac{4x}{16}>\frac{8}{2x}$ if $-4 < x < 0$ or $x > 4$, and $\frac{8}{2x}>\frac{4x}{16}$ when $0 < x < 4$ or $x < -4$

In the last case the fractions are incomparable when $x=0$ as $\frac{8}{2x}$ is not defined when $x=0$