Ask Uncle Colin: A Fractional Kerfuffle

Dear Uncle Colin,

I was trying to work out $\frac{\frac{3}{7+h}-\frac{3}{7}}{h}$, and I got it down to $\frac{\frac{3}{h}}{h}$ - but that's not the answer in the book! What have I done wrong?

- Likely I've Mistreated It Terribly

Hi, LIMIT, and thank you for your message! I'm afraid you're right, you have mistreated it.

Your problem is that $\frac{3}{7+h} - \frac{3}{7}$ is definitely not $\frac{3}{h}$ - try it with any $h$ you like and you'll see it doesn't work out. For $h=7$, for example, you'd have $\frac{3}{14} - \frac{3}{7} = \frac{3}{14}-\frac{6}{14} = -\frac{3}{14}$, which definitely isn't $\frac{3}{h}$.

Be careful when subtracting fractions

Before you subtract fractions, you need to make sure they have the same denominator (as I did above). It's a bit more complicated when there's algebra involved, but not that much.

Here, we have $\frac{3}{7+h} - \frac{3}{7}$ on top - I haven't forgotten the $h$ on the bottom, I'm just dealing with the top on its own first.

We need a common denominator, and $7(7+h)$ will do the trick. We'd multiply the first fraction by $\frac{7}{7}$ to get $\frac{21}{7(7+h)}$ and the second by $\frac{7+h}{7+h}$ to get $\frac{21 + 3h}{7(7+h)}$.

Subtracting those, you get $\frac{21 - (21 + 3h)}{7(7+h)}$, or $\frac{-3h}{7(7+h)}$ if you're careful with the minus sign.

Now bring the $h$ back in

Of course, we have to divide all of that by $h$, which leaves us with $\frac{-3}{7(7+h)}$.

Presumably you would then say "as $h \to 0$, this evaluates to $-\frac{3}{49}$."

Hope that helps!

- Uncle Colin

* Edited 2018-04-25 to correct my misreading of the question. Thanks to Adam Atkinson for pointing out my error. Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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