I think I got most of the proof, but feel free to critique anything you would like :).

First, notice that $aRa$ means $3a - 5a = a(3 - 5) = 2(-a)$. This is even, thus, $R$ is reflexive.

Second we show that $aRb \implies bRa$. Well, $aRb$ means that $3a - 5b$ is even. Then we have $3a - 5b = 2k$ for some integer k. Rearrange like so $$ \begin{equation} \begin{split} 3a - 5b &= 2k \\ -5b &= 2k - 3a \\ -5b &= 2(k - a) + (-a) \end{split} \end{equation} $$ Since, we stated that the LHS and RHS were both even in the beginning and still are (i think?), this forces $a$ and $b$ to be even. Thus, $3b - 5a$ is even because we are dealing with two even numbers. Therefore $bRa$ which means $R$ is symmetric.

Now for transitive. We have show that $(aRb \wedge bRc) \implies aRc$. Since $aRb \wedge bRc$, we have $3a - 5b$ even and $3b - 5c$ even. Note when we add, the sum is even. Observe $$ (3a - 5b) + (3b - 5c) \\ 3a - 2b - 5c \\ 2(-b) + (3a - 5c) $$ $2(-b)$ is even and this forces $3a - 5c$ to be even. Therefore $R$ is transitive. QED.

For the equivalences classes we know that if $a$ is even we have that $b$ must be even. Similarly, if $a$ is odd then $b$ is odd. Therefore, the equivalence classes are the set of odd integers and the set of even integers.

My question is on proving that $R$ is symmetric. Is it mathematically sound?

  • $\begingroup$ I see. Question @JMoravitz, how did you go from $3b - 8b + 8b - 5a +8a - 8a$ to $(5a - 3b) - 8a + 8b = 2(k- 4a + 4b)$? That is, it seems that you simply changed the sign of $3b$ and $5a$ to make it from $3b - 5a \to 5a - 3b $. $\endgroup$ – Owen Mar 16 at 20:22
  • $\begingroup$ We started with $3b-5a$ since this is what we want to show is even. We then "added zero" twice which is always allowed and doesn't change anything, zero here being $-8b+8b$ and $8a-8a$ since something minus itself is zero. We then used part of those expressions for zero to be grouped with the initial expression to make it look like $3a-5b$ since we know that is even and grouped the remaining portions of the expression, and factoring a $2$ out of everything at the end... showing that $3b-5a$ is also $2$ times an integer given that $3a-5b$ was as well. $\endgroup$ – JMoravitz Mar 16 at 20:24
  • $\begingroup$ and I seem to have a typo in the above, I meant of course $(3a-5b)-8a+8b$, not $(5a-3b)-8a+8b$. The point though still holds... you should be showing that $3b-5a$, the expression you are interested in, is equal to $2$ times an integer. $\endgroup$ – JMoravitz Mar 16 at 20:26
  • $\begingroup$ @JMoravitz, Ah, yes, that makes sense! That is where I was a little confused! Thank you for the explanation! $\endgroup$ – Owen Mar 16 at 20:27
  • $\begingroup$ It boils down to $3a - 5b = a -b + 2(a-2b)$ is even iff $a-b$ is even iff $a,b$ are both odd or both even. $\endgroup$ – lhf Mar 16 at 21:08

Suppose that $a\sim b$. That is to say, $3a-5b$ is even. That is to say, $3a-5b$ is equal to $2$ times some integer, we'll call it $k$ so $3a-5b=2k$

We ask whether or not this implies that $b\sim a$, that is if $3b-5a$ can be written as $2$ times an integer as well (note, not necessarily the same integer as before)

$$\begin{array}{l|l}~~~~3b-5a&\text{original}\\=3b+0-5a+0&\text{add zero twice}\\=3b+(-8b+8b)-5a+(8a-8a)&\text{replace zeroes}\\=(3b-8b)+8b+(-5a+8a)-8a&\text{adjust parentheses}\\=(3a-5b)-8a+8b&\text{simplify and rearrange}\\=2k-8a+8b&\text{use hypothesis}\\=2(k-4a+4b)&\text{factor out two}\end{array}$$

This shows that $3b-5a$ can also be written as $2$ times an integer and so is also even.

Similarly, for transitivity, we suppose $3a-5b$ is even and $3b-5c$ is even and we ask about $3a-5c$.

$3a-5c = 3a+0-5c=3a-5b+3b+2b-5c = (3a-5b)+(3b-5c)+2b$ is the sum of three even numbers and thus even as well.


3a-5b is even implie 3a=5b[2]

  • a$R$a implie $3a=5a[2]$$\Rightarrow $$-2a=0[2]$ so R is reflexive

    • a$R$b$\Rightarrow $$3a=5b[2]$$\Rightarrow $$ 9a=15b[2] $$\Rightarrow $$5a=3b[2]$$\Rightarrow $b$R$a

(because $9a=5a[2] $ and $15b=3b[2] $)

So R is symmetry

      • a$R$b$\Rightarrow $$3a=5b[2]$


b$R$c$\Rightarrow $$3b=5c[2]$ that implies $9b=15c[2]$$\Rightarrow $$5b=5c[2] $

So $3a=5c[2]$$\Rightarrow $ a$R$c

So a$R$b and b$R$c $\Rightarrow $a$R$c

So R is Transitive

Finally After (reflexivity, symmetry, transitivity) we Kan see R is equivalence relation


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