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I just want to know what this combination of symbols means:

$\exists !$

I know ∃ means 'there exists', but what does it mean when it is paired with a '!'? I have written down 'there exists unique" but I am not 100% sure this is correct?

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  • $\begingroup$ $\exists!$ can mean there exists unique or there exists only one. $\endgroup$
    – Mr Pie
    May 12, 2018 at 12:50

3 Answers 3

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“There exist unique” $$\exists!m\in\mathbb R:\forall a\in\mathbb R:a\cdot m=a$$

In this example $\exists!$ means that $m$ is unique, that this number exist ($m=1$) but no other does have this property.

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  • $\begingroup$ This is false when a is zero. You need to swap your quantifiers. $\endgroup$
    – H.Durham
    Dec 15, 2013 at 13:56
  • $\begingroup$ You're right @user10193, corrected. $\endgroup$
    – Carlos Eugenio Thompson Pinzón
    Dec 15, 2013 at 13:58
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$\exists ! x:$ "There exists a unique x..."

To use Carlos' example: $$\exists ! m \in \mathbb R,\,\forall a \in \mathbb R( a\cdot m = a)$$

This is equivalent to the statement:

$$\exists m \in \mathbb R, \forall a \in \mathbb R\Big(a\cdot m = a \land \forall y \in \mathbb R((a \cdot y = a) \implies y = m)\Big)$$

As you can see, the uniqueness quantifier greatly simplifies the logical statement.

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    $\begingroup$ This is the only answer that gets to the point, in my opinion. $\endgroup$
    – Git Gud
    Dec 15, 2013 at 14:09
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    $\begingroup$ The final formula is not quite right, it should be $(\exists m \in \mathbb{R})[(\forall a \in \mathbb{R})[a\cdot m = a] \land (\forall y \in \mathbb{R})[(\forall a \in \mathbb{R})(a\cdot y = a) \to y = m]]$. In general $(\exists! x)\phi(x)$ means $(\exists x)[\phi(x) \land (\forall y)[\phi(y) \to y=x]]$ $\endgroup$
    – Carl Mummert
    Dec 15, 2013 at 15:52
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The symbol $\exists$ is the existential quantifier and with a $!$ in front of it, it means that there exists at most one element...

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    $\begingroup$ I'd replace "at most" with "exactly", because this implies there may be less than one, which is not true because of the ∃ symbol. $\endgroup$
    – Darrel Hoffman
    Dec 15, 2013 at 15:46

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