1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

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.