# How do you insert the filled square "End of proof" unicode character [U+220E] in a socratic.org answer?

## The empty square can be inserted like this $\square$, however I cannot find the filled square QED character Thanks :-)

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

3
mason m Share
Jun 15, 2016

You can also add, within the hashtags:

hashtag " "" "" "" "" "" "square hashtag

Basically, it's a string of " " units next to one another, which creates space, and then square, which when between hashtags makes a $\square$.

Within the context of a problem, this could look something like:

Hence $T - {r}_{i} = \left\{m + k + 1 - {r}_{i} , \ldots , m + r + {r}_{i}\right\}$ is a segment of length at most $r - k < r$ of ${B}_{0}$. Since $m + k + 1 - {r}_{i} < m$, this contradicts the definition of $S \text{. "" "" "" "" } \square$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Dec 7, 2015

Here's what you can do.

#### Explanation:

To add the q.e.d. symbol to the Socratic edit, write it in Microsoft Word using Unicode, then copy/paste it in the answer in between hashtags.

So, I wrote the q.e.d. symbol in Word and got this

If I use hashtags

hashtag ▄ hashtag

I will get

▄

Now just use it anytime you want. To take an example you used in one of your answers, you will have

$\csc \left(x\right) - \sin \left(x\right) = \cos \left(x\right) \cot \left(x\right) \Rightarrow$

$\frac{1}{\sin} \left(x\right) - \sin \left(x\right) = \cos \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right) \Rightarrow$

$\left[\frac{1}{\sin} \left(x\right) - \sin \left(x\right) = \cos \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right)\right] \cdot \sin \left(x\right) \Rightarrow$

$\sin \frac{x}{\sin} \left(x\right) - {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right) \cdot \sin \frac{x}{\sin} \left(x\right) \Rightarrow$

$1 - {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right) \Rightarrow$

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ ▄

The alignment is not perfect, but I think it gets the job done.

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