0

This assingment is preparation for exam.

I need to prove with $(a_1, b_1)\sim(a_2, b_2)\Leftrightarrow\ a_1=a_2\land b_1=b_2$ that $\sim$ is equivalence relatio.

Can you tell me how to do this.

Thanks!!!

frabala
  • 3,732
depecheSoul
  • 913
  • 7
  • 13

1 Answers1

1

Simple. There are three properties an equivalence relation above the set $X$ must fulfill:

  • For any $x\in X$, you must have $x\sim x$ (reflexive)
  • For any $x,y\in X$ for which $x\sim y$, you must have $y\sim x$ (symmetric)
  • For any $x,y,z\in X$, for which $x\sim y$ and $y\sim z$, you must have $x\sim z$ (transitive)

Try proving these properties one by one for your relation. Post results, then we will help you if something goes wrong.

Note: i intentionaly wrote $x$ instead of $(a,b)$. In your case, the set $X$ contains ordered pairs, which may be confuzing but does not change a thing.

5xum
  • 123,496
  • 6
  • 128
  • 204
  • For reflexiv part I put $(a,b)\sim(a,b)$, and that is obiviusly true, to this is refleksiv. Or it may be that I am wrong, with staring condition, I mean $(a,b)\sim(a,b)$ :( – depecheSoul Feb 12 '14 at 12:35
  • 1
    You are correct. You must take an arbitrary element of the set, in your case an arbitrary pair $(a,b)$, and prove that $(a,b)\sim (a,b)$. But don't just say obvious. Write down that it is true because $a=a$ and $b=b$ and both these statements hold by definition. My professor always said you don't get to use the word 'obvious' until you get a PhD. – 5xum Feb 12 '14 at 12:38
  • 1
    Reflexive means $(a,b) \sim (a,b)$, as you says; this is what you have to prove... Apply the condition : obviously, $a = a$ and $b = b$, so that you have the reflexivity. Sorry for the "obviously" ... – Mauro ALLEGRANZA Feb 12 '14 at 12:39
  • 1
    ... then go on with symmetry; it is also easy. After that, transitivity will be quite easy. – Mauro ALLEGRANZA Feb 12 '14 at 12:41
  • For symmetric part, I have $(a_x, b_x)\sim(a_y, b_y) \Rightarrow (a_y, b_y)\sim (a_x, b_x)$. Because of symetry of $=$ sign $(x=y)\Rightarrow (y=x)$ , I can tell that this is symetric. And Thanks for helping me. :) – depecheSoul Feb 12 '14 at 12:43
  • 1
    Correct! Notice how all you really need to use is the fact that "$=$" represents an equivalence relation. – 5xum Feb 12 '14 at 12:44
  • For transitive part, I have that $(a_x, b_x)\sim(a_y, b_y) \land (a_y, b_y)\sim(a_z, b_z)\Rightarrow (a_x, b_x)\sim (a_z, b_z)$. Here I will presume that, $(a_x, b_x)\sim(a_y, b_y)$ and $(a_y, b_y)\sim(a_z, b_z)$, are true, so that gives me that $(a_x, b_x)\sim (a_z, b_z)$ is also true. ??? – depecheSoul Feb 12 '14 at 12:52
  • 1
    How does that give you that? I mean, it does, but you wrote nothing to prove that it does. – 5xum Feb 12 '14 at 12:56
  • This is a shorthand, of what I wrote, I wrote more in my book. – depecheSoul Feb 12 '14 at 12:59
  • Well, if you wrote that $a_x = a_y$ and $a_y=a_z$ implies $a_x=a_z$, that should be sufficient. – 5xum Feb 12 '14 at 13:00
  • I wrote a little more, but that is the end result. $\lnot ((a_x, a_y)\land(a_y, a_z))\lor (a_x, a_y)$. And That gives me that $\lnot(T\land T) \lor T; \lnot(T)\lor T; F\lor T; T$ – depecheSoul Feb 12 '14 at 13:03
  • 1
    Then you are done! – 5xum Feb 12 '14 at 13:04
  • I wrote I the commet above how I did it, can you please check it out. Thanks. – depecheSoul Feb 12 '14 at 13:08
  • That makes no sense whatsoever. Nowhere have you defined what $(a,b)\wedge (c,d)$ even means. Ordered pairs are not logical statements, you cannot build conjunctive statements with them! $(a,b)\sim (c,d)$ is a statement. $(a,b)$ is just an element. You cannot replace $(a_x, a_y)$ with $T$. – 5xum Feb 12 '14 at 13:18
  • Second solution: $(a_x, b_x)\sim(a_y, b_y)\land (a_y, b_y)\sim(a_z, b_z)\Rightarrow (a_x, b_x)\sim(a_z, b_z)$. Now I had presume something like this $if\ (a_x, b_x)\sim(a_y, b_y)\land (a_y, b_y)\sim(a_z, b_z)\text{is true, them } (a_x, b_x)\sim(a_z, b_z)\text{ is also true}$ and that gives me that $(a_x, b_x)\sim (a_z, b_z)$. This if..., then... I pulled out form implies. Any good? – depecheSoul Feb 12 '14 at 13:31
  • let us continue this discussion in chat – 5xum Feb 12 '14 at 13:59
  • 1
    You said: if$(a_x, b_x)\sim(a_y, b_y)\land (a_y, b_y)\sim(a_z, b_z)\text{is true, them } (a_x, b_x)\sim(a_z, b_z)\text{ is also true}$. You cannot just presume this statement is true! This is what you are trying to prove! – 5xum Feb 12 '14 at 14:03
  • Let e try like this: – depecheSoul Feb 12 '14 at 14:39
  • Ok, I found the solution here http://answers.yahoo.com/question/index?qid=20101024094452AAGUoO6, and I understand it, but there is still one thing that confusses me. Why is $(a,b)=(c,d)$ equal to $ad=bc$. It makes more sence like this $ac=bd$. Thanks – depecheSoul Feb 12 '14 at 15:06
  • 1
    The solution you posted is not the solution to your task. The relation there is different. – 5xum Feb 12 '14 at 15:12
  • I really dont understand how to this. Can you give me some advice!! Please – depecheSoul Feb 12 '14 at 15:20
  • So I did it like this (I have x,y,z elements) $x=y \land y=z$. Bassicly I solved this like a equation. I know that $y=z$ so I just put it inside $x=y$ so that I got $x=z$, snd so I got that is transsitive. Any better??? – depecheSoul Feb 12 '14 at 15:55
  • It's not really that hard. You can assume this: $(a_x, b_x)\sim (a_y, b_y)\wedge (a_y,b_y)\sim (a_z,b_z)$. From the first equivalence, you know that $a_x=a_y$ and $b_x=b_y$. From the second, you know that $a_y = a_z$ and $b_y = b_z$. Now, from $a_x=a_y$ and $a_y = a_z$, it follows that $a_x=a_z$. In the same way, you can prove that $b_x=b_z$. From $a_x=a_z$ and $b_x=b_y$, you can by definition claim that $(a_x,b_x)=(a_z,b_z)$. – 5xum Feb 12 '14 at 16:54