Question

Verify associativity for the given three mappings: $f:N\to Z_0$ (the set of non-zero integers), $g:Z_0\to Q$ and $h:Q\to R$, given by $f\left(x\right)=2x,g\left(x\right)=\frac{1}{x}$ and $h\left(x\right)=e^x$.

Answer 1

$f\circ g$ means $g\left(x\right)$ function is in $f\left(x\right)$ function.

$g\circ f$ means $f\left(x\right)$ function is in $g\left(x\right)$ function.

$f:N\to Z_0,g:Z_0\to Q$ and $h:Q\to R$...... [ Given ]

$g\circ f:N\to Q$ and $h\circ g:Z_0\to R.$

$\Rightarrow$ $h\circ (g\circ f):N\to R$ and $(h\circ g)\circ f:N\to R$

So, both have the same domains.

$\Rightarrow$ $(g\circ f)\left(x\right)=g\left[f\right(x\left)\right]$

$=g\left(2x\right)$ ......... [ Since, $f\left(x\right)=2x$ ]

$=\frac{1}{2x}$ ----- ( 1 )

$\Rightarrow$ $(h\circ g)\left(x\right)=h\left[g\right(x\left)\right]$

$=h\left(\frac{1}{x}\right)$

$=e^{\frac{1}{x}}$ ---- ( 2 )

Now,

$\Rightarrow$ $[h\circ (g\circ f\left)\right]\left(x\right)=h\left[\right(g\circ f\left)\right(x\left)\right]$

$=h\left(\frac{1}{2x}\right)$......... [ From ( 1 ) ]

$=e^{\frac{1}{2x}}$......... [ From ( 2 ) ]

$\Rightarrow$ $\left[\right(h\circ g)\circ f]\left(x\right)=(h\circ g)\left[f\right(x\left)\right]$

$=(h\circ g)\left(2x\right)$......... [ Since, $f\left(x\right)=2x$ ]

$=e^{\frac{1}{2x}}$ ......... [ From ( 2 ) ]

$\Rightarrow$ $[h\circ (g\circ f\left)\right]\left(x\right)=\left[\right(h\circ g)\circ f]\left(x\right)$, $\forall x\in N$

So, $h\circ (g\circ f)=(h\circ g)\circ f$

Hence, the associate property has been verified.