# Inequalities and floors.

I've been presented with a question that I actually don't understand.

The question is:

Given $$\lfloor a\rfloor\leq a<\lfloor a\rfloor+1$$ Write an inequality for $\lfloor a\rfloor$

I'm fine with the floor function - that isn't unfamiliar. And while I can read and understand what the inequality is saying above, I don't know what it means to "write an inequality for $\lfloor a\rfloor$". We need to use the inequality we get to show a further expression which is in the form:

[something]$\leq$[something]<[something], so from this I know that the inequality signs don't change in the inequality I must write, but that's all I know!

Does it mean $\lfloor a\rfloor$ has to be in the middle?

## 2 Answers

Rearranging

$$a-1<\lfloor a\rfloor\leq a$$

• So when we are asked to write an inequality for $x$, $x$ must be bound by something potentially less than and something potentially greater than $x$ (i.e. $x$ is in the middle)? – Old mate Sep 22 '14 at 3:41
• Basically -- yes. The inequality I wrote puts lower and upper bounds on the floor in terms of its argument. – RRL Sep 22 '14 at 3:43
• floor(a) can be no bigger than a and never less than a-1 – RRL Sep 22 '14 at 3:43
• Awesome, thank you for clarifying! – Old mate Sep 22 '14 at 3:48
• @Eliot: You're welcome. – RRL Sep 22 '14 at 3:49

As @RRL pointed out, you could rearrange it that way. Depending on what you are doing, it might also be useful to note $$0 \leq a-\lfloor a \rfloor < 1$$