Let us consider $S=\mathbb R\times\mathbb C$.

We write $a$ and $b$ two elements of $S$: $a=(x_a,z_a)$, and $b=(x_b,z_b)$.

We define the binary operation $∘$ as:

$a∘b=(x_a+x_b,x_a+ix_b+z_a+z_b)$, with $i$ the imaginary unit $i^2=−1$.

We say that $a\sim b$ if and only if $a∘b=b∘a$.

We write $[a]$ the equivalence class of $a$. Only one of the following is correct.

a. For $a=(x_a,z_a)$ one has $[a]=\{(x,z_a);∀x∈\mathbb R\}$

b. For $a=(x_a,z_a)$ one has $[a]=\{(x_a,z);∀z∈\mathbb C\}$

c. For $a=(x_a,z_a)$ one has $[a]=\{(x_a+R(z_a),z);∀z∈\mathbb C\}$

d. For $a=(x_a,z_a)$ one has $[a]=\{(x_a+I(z_a),z);∀z∈\mathbb C\}$

e. For $a=(x_a,z_a)$ one has $[a]=\{(x_a+|z_a|^2,z);∀z∈\mathbb C\}$

I am working on equivalence class questions but I'm so stuck on this one. It's already given on the question that $a\sim b$ so I don't have to show the reflexive, symmetric and transitive.

I'm not sure how to find out the equivalence class of complex number, but I think I can get rid of option a. as it says $∀x∈\mathbb R$ which is incorrect. I'm not sure about the remaining 4.

Any help will be appreciated, thanks.

My working:

Working from $x_a+ix_b=x_b+ix_a$, we have $x_a=x_b$, $ix_b=ix_a$ and $z_a+z_b=z_b+z_a$. $z_a+z_b=z_b+z_a$ is just $z$ , $ix_b=ix_a$ where $ix_b,ix_a$ are imagery numbers. Since $S=R×C, a∈S, x_a∈R$ and $z_a∈C,[a]=${$(x_a+I(z_a),z);∀z∈C$}.

Am I right?


The condition $a\circ b = b\circ a$ means that $$(x_a+x_b, x_a + ix_b + z_a+z_b) = (x_b+x_a,x_b+ix_a+z_b+z_a),$$ i.e., $x_a+x_b = x_b+x_a$ and $$x_a+ix_b + z_a+z_b = x_b+ix_a+z_b+z_a.$$ The latter means that $x_a+ix_b = x_b+ix_a$. Since $x_a,x_b$ are real numbers, it follows that $x_a=x_b$. Could you complete it from here?

  • $\begingroup$ I'm not sure if understand it correctly. So working from $x_a+ix_b=x_b+ix_a$, we have $x_a=x_b$, $ix_b=ix_a$ and $z_a+z_b=z_b+z_a$. $z_a+z_b=z_b+z_a$ is just $z$ , $ix_b=ix_a$ where $ix_b$,$ix_a$ are imagery numbers. Since $S=R \times C$, $a∈S$, $x_a∈R$ and $z_a ∈ C$,$ [a]=${$(x_a+I(z_a),z);∀z∈C$}. So d. is the answer, am I right? $\endgroup$ – BlackSky Nov 3 '18 at 17:06
  • 1
    $\begingroup$ To the proposer: $a\sim b \iff x_a=x_b.$ So $[a]=\{b: x_b=x_a\}=\{x_a\}\times \Bbb C.$ $\endgroup$ – DanielWainfleet Nov 3 '18 at 21:32
  • $\begingroup$ I am confused about the imagery part. If $[a]=${$b:x_b=x_a$}$=${$x_a$}$×C$ which I believe it is same as $[a]={(x_a,z);∀z∈C}$ but since $z_a+z_b=z_b+z_a$ $z_a$ and $z_b$ should be real numbers? If $[a]={(x_a,z);∀z∈C}$ is true, where does the real number $z$ goes? $\endgroup$ – BlackSky Nov 4 '18 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.