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I have been stuck on this problem for a couple of days now. I know that I need to prove that this is reflexive, symmetrical and transitive, but I have zero clue where to start..

Let U be the set of all linear functions mx + b where m ∈ Z and b ∈ Z. The relation ≈ is defined by: $m_1x + b_1 ≈ m_2x + b_2$ if $b_1 = b_2$ Prove that $≈$ is an equivalence relation.

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    do you mean $m_1 x+b_1 \sim m_2x+b_2$ otherwise it doesn't make sense to me – 2017-02-23
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    Hint. In words this algebra says two lines are related if they have the same $y$-intercept. Can you show that relation is an equivalence relation? – 2017-02-23
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    @Sentinel135 I have no clue I asked if it was about 4 hours ago never got answer, but if it doesn't makes I'll assume it is. – 2017-02-23
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    So a couple of hints: 1) lets denote a linear function say $a: \frac{x}{3}+5$ is that related to itself? 2) lets denote another linear function say $b: 6x+5$ is that related to $a$ as we defined it? 3) if we have linear functions $c$, $d$, and $e$ such that $c$ is related to $d$ and $d$ is related to $e$ does this imply that $c$ is related to $e$? – 2017-02-23
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    1) yes thats reflexive, 2) I would think so given that it follows $mx+b$ 3) yes that's transitive. – 2017-02-23

1 Answers 1

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I'm going to try to show you how the proof is written without actually proving it.

Part I) show the relation is reflexive. in the hints I gave you a specific case you will need to show this is true in general.

Part II) this is the tricky part. like in example 2) the key here is that if the $b$'s equal then they have an relation. So $a$ is related to $b$ but if we defined another one say $k: 6x+1$ it has no relation to $a$ as $5\neq1$. Now what's important here is showing that if $a$ is related to $b$ then $b$ is related to $a$. again needs to be in general not specific cases.

Part III) Show transitivity. This seems easy since you pointed it out fairly quickly.

If you have any questions please leave a comment below.

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    So when showing the symmetry is it looking at $mx + b$ as a whole or just the $b$ portion? ex because $b_1 = b_k$... – 2017-02-23
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    let's look at the example we have. Defining $a: \frac{x}{3}+5$ and $b:6x+5$. You've already stated that $a \sim b$ and you can show this by comparison of $b_a=5$ and $b_b=5$. Now to show that it is symetric in this instance you would need to show that $b \sim a$, So, lets compare. $b_b=5=b_a$. Retorical question: is $b \sim a$? – 2017-02-23