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I want to prove for any $x \in R$, then $2[x] ≤ [2x] ≤ 2[x] + 1$ where $[x]$ is the largest integer smaller than $x$.

Proof: By definition, $[x] ≤ x ≤ [x] + 1$. Multiplying everything by 2, we see that $2[x] ≤ 2x ≤ 2[x] + 2$, so $[2[x]] = 2[x] ≤ [2x]$. At this point, I'm not sure how to prove the other inequality.

I know that $[2x] ≤ 2x ≤ 2[x] + 2$ and similarly, $[2x] ≤ 2x ≤ [2x] + 1$. But how do I show $[2x] ≤ 2[x] + 1$?

Thank you!

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    @dxiv did you end up deleting your answer? – 2017-01-25
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    @Cici No, that was me. – 2017-01-25
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    @Cici Wasn't *my* answer. That said, I elaborated on my previous (now deleted) comment and posted it as an alternative answer. – 2017-01-25
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    Thank you to both of you! I understand this now. – 2017-01-25

3 Answers 3

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Let $x=[x]+\alpha$ where $\alpha$ satisfies $0 \le \alpha < 1$, Note that if $$[2x]=[2[x]+2\alpha] > 2[x]+1 $$ Then we have $$ 2[x]+2>2[x]+2\alpha \ge[2[x]+2\alpha]> 2[x]+1 $$ So $[2x]$ becomes a integer that is between two consecutive integers, $2[x]+1$ and $2[x]+2$. Contradiction.

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    Thank you for this alternative answer too! – 2017-01-25
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    +1 for a proof by contradiction, which is definitely one of my favorite proof techniques. – 2017-01-26
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By definition, $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$.

The second inequality is in fact strict: $\;\lfloor x \rfloor \le x \lt \lfloor x \rfloor + 1\,$.

I know that $[2x] ≤ 2x ≤ 2[x] + 2$

Reworking the same steps with the strict inequality on the RHS results in $\;\lfloor 2x \rfloor \le 2x \lt 2\lfloor x \rfloor + 2\,$, which gives $\;\lfloor 2x \rfloor \lt 2\lfloor x \rfloor + 2\,$. Note that in this last inequality both sides are integers, and for integers $m \lt n \iff m \le n-1$, so the inequality can be rewritten as $\;\lfloor 2x \rfloor \le 2\lfloor x \rfloor + 1\,$.

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We can also do a proof by cases.

If $x$ is a real number, let $x = [x]+\epsilon$ where $0 \leq \epsilon < 1$ is the fractional part of $x$. Then multiplying by $2$ gives

$$2x = 2[x]+2\epsilon \, \, \, \textrm{where} \, \, \, 0 \leq 2 \epsilon < 2$$

If we take the floor of both sides, we have

$$[2x] = [2[x] + 2\epsilon] $$

Since $2[x]$ is an integer and $2\epsilon$ is a real number, we can "distribute" the floor symbol across the binomial

$$[2x] = 2[x] + [2\epsilon]$$

Suppose that $0 \leq 2\epsilon < 1$. Then $[2\epsilon] = 0$, so that

$$[2x] = 2[x]$$

Therefore

$$2[x] = [2x] < 2[x] + 1$$

Finally, if $1 \leq 2\epsilon < 2$, we have $[2\epsilon] = 1$, so

$$[2x] = 2[x] + 1$$

We then get

$$2[x] < [2x] = 2[x]+1$$

In both cases we have

$$2[x] \leq [2x] \leq 2[x]+1,$$

but $[2x]$ either equals $2[x]$ or $2[x]+1$ depending on the fractional part $\epsilon$.