Question

Is the following relation an equivalence relation on $\mathbf{ℝ}\mathbf{:}\mathit{a}\mathbf{~}\mathit{b}$

if and only if there exists $\mathit{k}\mathbf{\in }\mathbf{\mathbb{Z}}$such that $\mathit{a}\mathbf{=}{\mathbf{10}}^{\mathbf{k}}\mathit{b}$.

Short answer

We proved that $~$is an equivalence relation.

Step-by-step solution

To mention given data

Let be defined on $\mathrm{ℝ}:a~b$by:

$a~b$if and only if $a={10}^{k}b$for $k\in \mathrm{\mathbb{Z}}$.

We have to prove that $~$is an equivalence relation.

In order to prove that $~$is an equivalence relation, we need to prove the following conditions:

1. ‘$~$’ is reflexive.
2. ‘$~$’ is symmetric.
3. ‘$~$’ is transitive.

To prove that ~is reflexive

Let $a\in \mathrm{ℝ}$.

Then,

$$\begin{gathered} a~a \\ \iff a=1\times a={10}^0\times a \\ \iff a={10}^0\times a,\forall a\in \mathrm{ℝ},0\in \mathrm{\mathbb{Z}} \end{gathered}$$

Thus, there exists $0\in \mathrm{\mathbb{Z}}$such that $a={10}^{0}a$if and only $a~a$.

Hence, $~$is reflexive.

To prove that ~ is symmetric

Let$a,b\in \mathrm{ℝ}$.

Then suppose that

$$\begin{gathered} a~b \\ \iff a={10}^{k}b,(k\in \mathrm{\mathbb{Z}}) \\ \iff {10}^{-k}a={10}^{-k}{10}^{k}b,(-k\in \mathrm{\mathbb{Z}}) \\ \iff {10}^{-k}a={10}^{-k+k}b \\ \iff {10}^{-k}a=b\cdots \cdots (\because {10}^{-k+k}={10}^0=1) \\ \iff b={10}^{n}a\cdots \cdots (\because -k=n\in Z) \end{gathered}$$
$\iff b~a$
Hence,$~$is symmetric.

To prove that ~is transitive

Let$a,b,c\in R^{*}$.

Then suppose that:

$$\begin{gathered} a~b,b~c \\ \iff a={10}^{k}b,b={10}^{k\text{'}}c,k,k^{\text{'}}\in \mathrm{\mathbb{Z}} \\ \iff a={10}^k\left({10}^{k^{\text{'}}}c\right) \\ \iff a={10}^{k+k^{\text{'}}}c \\ \iff a={10}^{n}c\cdots \cdots (\because k+k^{\text{'}}=n\in \mathrm{\mathbb{Z}}) \end{gathered}$$

$\iff a~c,\forall a,b,c\in \mathrm{ℝ},k,k^{\text{'}}\in \mathrm{\mathbb{Z}}$

Hence, $~$is transitive.

Hence, $~$is an equivalence relation since all the conditions are satisfied.