# A student asks: upper bounds

A student asks:

When you’ve got a value to the nearest whole number, why is the upper bound something $.5$ rather than $.4$? Doesn’t $.5$ round up?

So I don’t have to keep writing something$.5$, let’s pick a number, and say we’ve got 12 to the nearest whole number.

$12.5$ does indeed round up (at least in the GCSE maths convention that you break a tie by going up; in some sciences, the convention is that you round to the nearest even number, so you don’t introduce an upward bias in your data), but $12.4$ certainly isn’t the upper bound - for example, $12.49$ would still round down. So would $12.499$. And $12.4999$. And, for that matter, $12.4999999999$.

In fact, you can carry this on forever and say the upper bound has to be $12.4\dot9$ - which is technically a correct answer. However, we already have a name for $12.4\dot9$ - it’s the same as $12.5$.

(Aside: don’t believe me? If $x = 12.4\dot 9$, then $10x = 124.\dot9$. Take them away and you get $9x = 112.5$. Divide by 9… $x = 12.5$.)

You should get the mark if you write $12.4\dot 9$, but why risk it? Saying something is 12 to the nearest whole number is the same as saying $11.5 \le x \lt 12.5$ - the upper bound (the *supremum*, if you want the technical term) is 12.5.