Question

pls answer these Questions too ​

Answer 1

3.  Assume that $\sf{(a,\ a)\in R}$ such that, by definition,

$\longrightarrow\sf{a\leq a^3}$

$\longrightarrow\sf{a^3-a\geq 0}$

$\longrightarrow\sf{a(a-1)(a+1)\geq 0}$

Solution to this inequality is,

$\longrightarrow\sf{a\in[-1,\ 0]\cup[1,\ \infty)}$

Since $\sf{R}$ is defined on set of all real numbers $\mathbb{R,}$

$\Longrightarrow\sf{(a,\ a)\notin R\quad\!\forall a\in\mathbb{R}}$

Therefore, R is not reflexive.

We see $\sf{(-2,\ -2)\notin R}$ hence R is not reflexive.

We see $\sf{(-1,\ 2)\in R}$ but $\sf{(2,\ -1)\notin R.}$ Hence R is not symmetric.

We see $\sf{(9,\ 3)\in R}$ and $\sf{(3,\ 2)\in R}$ but $\sf{(9,\ 2)\notin R.}$ Hence R is not transitive.

4. (a)  Since $\sf{a-a=0}$ is divisible by every non - zero real numbers, 0 is also divisible by 2, hence,

$\longrightarrow\sf{(a,\ a)\in R\quad\!\forall a\in\mathbb{Z}}$

since R is defined on set of all integers $\mathbb{Z.}$

Therefore, R is reflexive.

By definition, let,

$\longrightarrow\sf{a-b=2m,\quad\!m\in\mathbb{Z}}$

We see that,

$\longrightarrow\sf{b-a=-(a-b)}$

$\longrightarrow\sf{b-a=-2m}$

$\Longrightarrow\sf{2\ divides\ (b-a)}$

$\Longrightarrow\sf{(b,\ a)\in R}$

Therefore, R is symmetric.

Assume $\sf{(b,\ c)\in R}$ such that 2 divides (b - c).

Since 2 divides both (a - b) and (b - c), 2 can divide their sum too.

$\longrightarrow\sf{2\ divides\ (a-b)+(b-c)}$

$\longrightarrow\sf{2\ divides\ (a-c)}$

Therefore, R is transitive.

Hence R is an equivalence relation.

4. (b)  In an equivalence relation $\sf{R,}$ equivalence class of $\sf{a}$ denoted by $\sf{[a]}$ is the set of all possible values of $\sf{b}$ for which $\sf{(a,\ b)\in R.}$

According to the question let $\sf{a=0.}$ Then by definition,

$\longrightarrow\sf{2\ divides\ (0-b)}$

$\longrightarrow\sf{2\ divides\ (-b)}$

This implies $\sf{b}$ must be an even number. Hence, since R is defined on $\mathbb{Z,}$

$\longrightarrow\sf{b\in\{2n:n\in\mathbb{Z}\}}$

This set is the same as the equivalence class [0].

$\longrightarrow\underline{\underline{\sf{[0]=\{2n:n\in\mathbb{Z}\}}}}$

or,

$\longrightarrow\underline{\underline{\sf{[0]=\{0,\ \pm2,\ \pm4,\ \pm6,\,\dots\}}}}$

Similarly, take $\sf{a=1.}$ Then by definition,

$\longrightarrow\sf{2\ divides\ (1-b)}$

This implies $\sf{b}$ must be odd since 1 is an odd number. Hence,

$\longrightarrow\sf{b\in\{2n+1:n\in\mathbb{Z}\}}$

Therefore,

$\longrightarrow\underline{\underline{\sf{[1]=\{2n+1:n\in\mathbb{Z}\}}}}$

or,

$\longrightarrow\underline{\underline{\sf{[1]=\{\pm1,\ \pm3,\ \pm5,\,\dots\}}}}$

4. (c)  We see that,

$\longrightarrow\sf{[0]\cap[1]=\phi}$

and,

$\longrightarrow\sf{[0]\cup[1]=\{0,\ \pm1,\ \pm2,\ \pm3,\ \pm4,\,\dots\}}$

$\longrightarrow\sf{[0]\cup[1]=\mathbb{Z}}$

Hence we can say $\sf{[0]}$ and $\sf{[1]}$ form a partition of $\mathbb{Z.}$

5. (a)  Here R is reflexive since $\sf{x-x=0}$ is an integer, i.e., $\sf{(x,\ x)\in R\quad\!\forall x\in\mathbb{Z}.}$

5. (b)  Let $\sf{x-y=m}$ be an integer.

We see $\sf{y-x=-m}$ is also an integer. Hence $\sf{(y,\ x)\in R.}$

Therefore, R is symmetric.

Assume there exists $\sf{(y,\ z)\in R}$ such that $\sf{y-z=n}$ is an integer.

Since sum of two integers is always an integer, we see,

$\longrightarrow\sf{(x-y)+(y-z)=m+n}$

$\longrightarrow\sf{x-z=m+n}$

Here m and n are integers, so are $\sf{m+n}$ and $\sf{x-z.}$

Hence, $\sf{(x,\ z)\in R.}$

Therefore, R is transitive.

Hence, R is an equivalence relation.