0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

Rearranging

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

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Old mate Sep 22 '14 at 3:41
  • 1
    $\begingroup$ Basically -- yes. The inequality I wrote puts lower and upper bounds on the floor in terms of its argument. $\endgroup$ – RRL Sep 22 '14 at 3:43
  • $\begingroup$ floor(a) can be no bigger than a and never less than a-1 $\endgroup$ – RRL Sep 22 '14 at 3:43
  • $\begingroup$ Awesome, thank you for clarifying! $\endgroup$ – Old mate Sep 22 '14 at 3:48
  • $\begingroup$ @Eliot: You're welcome. $\endgroup$ – RRL Sep 22 '14 at 3:49
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.