An equation is represented externally using the binary operator =. An expression which is an equation has two operands, the left-hand side and the right-hand side.  The names = and equation are known to the type function.

In 2-D math, when you type <>, <=, or >=, it is displayed as $\ne$, $\le$, or $\ge$, respectively.

There are three internal data types for inequalities, corresponding to the operators <>, <, and <=. Inequalities involving the operators > and >= are converted to the latter two cases for purposes of representation. An inequality has two operands, the left-hand side and the right-hand side. The names <>, <, <= are known to the type function.

Comparisons of numeric values are carried out in the corresponding numeric computation environment. For example, the test 3.141 < 3.142 is evaluated by subtraction in the floating-point environment determined by Digits. Hence, if Digits > 3, this returns true.  If Digits <= 3, this test returns false.

These operators are viewed as relational operators in a Boolean context or by the evalb function.  For more information, see boolean.

The equation and inequality operators are thread safe as of Maple 15.

The Equations and Inequalities, =, <>, <, <=, >, >= command is thread-safe as of Maple 15.

For more information on thread safety, see index/threadsafe.

$e≔a=b$

$e≔a=b$

$\mathrm{type}\left(e\&comma;'\mathrm{equation}'\right)$

$\mathrm{true}$

$e≔f\left(x\right)<g\left(x\right)$

$e≔f\left(x\right)<g\left(x\right)$

$\mathrm{type}\left(e\&comma;'\mathrm{equation}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(e\&comma;\mathrm{`<`}\right)$

$\mathrm{true}$

$\mathrm{lhs}\left(e\right)$

$f\left(x\right)$

$\mathrm{rhs}\left(e\right)$

$g\left(x\right)$

$\mathrm{eqs}≔\left\{\mathrm{a1}=\mathrm{b1}\&comma;\mathrm{a2}=\mathrm{b2}\&comma;\mathrm{a3}=\mathrm{b3}\right\}$

$\mathrm{eqs}≔\left\{\mathrm{a1}=\mathrm{b1}\&comma;\mathrm{a2}=\mathrm{b2}\&comma;\mathrm{a3}=\mathrm{b3}\right\}$

$\mathrm{map}\left(\mathrm{lhs}\&comma;\mathrm{eqs}\right)$

$\left\{\mathrm{a1}\&comma;\mathrm{a2}\&comma;\mathrm{a3}\right\}$

$a=2$

$a=2$

$a$

$a$

$\mathrm{assign}\left(\right)$

$a$

$2$