# Define a relation $R$ on $\mathbb{Z}$ by $aRb$ iff $3a - 5b$ is even. Prove $R$ is an equivalence relation and describe equivalence classes.

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 $$$$\begin{split} 3a - 5b &= 2k \\ -5b &= 2k - 3a \\ -5b &= 2(k - a) + (-a) \end{split}$$$$ Since, we stated that the LHS and RHS were both even in the beginning and still are (i think?), this forces $$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?

• 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 $3b - 5a \to 5a - 3b$. – Owen Mar 16 at 20:22
• 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. – JMoravitz Mar 16 at 20:24
• 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. – JMoravitz Mar 16 at 20:26
• @JMoravitz, Ah, yes, that makes sense! That is where I was a little confused! Thank you for the explanation! – Owen Mar 16 at 20:27
• 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. – 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$$\Rightarrow 3a=5b[2]$$

and

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