Question

$Z$ is the set of integers and $N$ is the set of natural numbers. If $b\in Z,c\in Z$ and $n\in N,$ if $b$ and $c$ are said to be congruent with respect to modulo $n,$ then symbolically it is written as $b\equiv c\left(modn\right)$.

Let $a\equiv b\left(modn\right),a^{\prime }\equiv b^{\prime }\left(modn\right)$ and $d,m\in N-\left[1\right]$, then which of the following need not be true?

• A. $a+a^{\prime }\equiv b+b^{\prime }\left(modn\right)$
• B. $a{a}^{\prime }\equiv b{b}^{\prime }\left(modn\right)$
• C. $a^m\equiv b^m\left(modn\right)$
• D. $\frac{a}{d}\equiv \frac{b}{d}\left(modn\right)$

Answer 1

The correct option is D $\frac{a}{d}\equiv \frac{b}{d}\left(modn\right)$

$Z=$ set of integers

$N=$ set of natural numbers$.$

$b\in Zc\in Zn\in N,n>0$

We say $a$ is congruent to be modulo $n.$

$T1:$

$a=b$$($modulo $n$$)$ of $n$ divides $b-a$

$T2:$

$a=b($modulo $n)$ and $c=d($modulo $n)$

$a+c=b+d($modulo $n)$

$ac=bd($modulo $n)$

$\because$ It is given that

$a=b($ modulo $n)$

$a^{\prime }=b^{\prime }($ modulo $n)$

$\Rightarrow a+a^{\prime }=b+b^{\prime }($modulo $n)$ $(T$-$2)$

also $a{a}^{\prime }=b{b}^{\prime }($ modulo $n)$$(T$-$2)$

$\because a=b($ modulo $n)$ $(T$-$2)$

$\Rightarrow a^m=b^m($ modulo $n)$ $(T$-$2)$

but $\frac{a}{m}-\frac{a}{d}\ne \frac{b}{d}$ modulo $n.$

as $n$ divides $b-a$

but $n$ does not divide $\left(\frac{b-a}{d}\right).$