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2011-01-19, 07:37   #12
CRGreathouse

Aug 2006

2×3×977 Posts

Quote:
 Originally Posted by xilman I feel I ought to head off a possible misconception which, if you hold it, will cause confusion later. When CRGreathouse used the phrase "x is any random number" he clearly used "random" in the colloquial sense.
Correct, mirroring the colloquial sense sm used in posing the question. I actually agonized over that point, but decided that the technical meaning would not cause confusion here.

2011-01-19, 07:50   #13
xilman
Bamboozled!

May 2003
Down not across

1008210 Posts

Quote:
 Originally Posted by CRGreathouse Correct, mirroring the colloquial sense sm used in posing the question. I actually agonized over that point, but decided that the technical meaning would not cause confusion here.
Good. I thought that to be the case but wanted to head off any ambiguity and misunderstandings. Colloquial language is good and proper in its proper place but in the context of teaching mathematics I suggest that we should try to be more careful in the use of language.

If nothing else, sm is now aware of the distinction between colloquial and techincal language in this respect and has been given gentle advice to be careful in future.

Paul

Last fiddled with by xilman on 2011-01-19 at 07:53 Reason: Add final sentence

2011-01-19, 11:24   #14
Mr. P-1

Jun 2003

7·167 Posts

Quote:
 Originally Posted by science_man_88 I don't see why they use the word on to describe it yet but I might soon ( maybe it's so it doesn't get confused with $\in$ (meaning in ?). My next line of questioning is on the equation part, Does this mean for all $x \in S$ there's a $y \in S$ such that the relation on S always holds ? I'll be back around 8-9 am (UTC -4:00 time).
Science Man, I think it is great that you've accepted the criticism and the advice, and I don't mean this to be nasty or anything. In a math text "preliminaries" are the stuff you are already expected to know and be comfortable with. If you're struggling with it, then it's a clear indicator that the text is too advanced, and you need to back up.

Based upon the kinds of things you're not sure about (such as the meaning of $\in$), I suggest a introductory math course covering a typical high school curriculum.

2011-01-19, 12:11   #15
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by Mr. P-1 Science Man, I think it is great that you've accepted the criticism and the advice, and I don't mean this to be nasty or anything. In a math text "preliminaries" are the stuff you are already expected to know and be comfortable with. If you're struggling with it, then it's a clear indicator that the text is too advanced, and you need to back up. Based upon the kinds of things you're not sure about (such as the meaning of $\in$), I suggest a introductory math course covering a typical high school curriculum.
Yes, well maybe I was unsure because I've never used the symbol much before I know it has the Tex \in because I've been playing around with Tex but some other symbols use abbreviations so to assume outright meaning like in this set would be outrageous of me I think but i may be wrong as it's used a lot of the time where I'd imagine the word in to be.

2011-01-19, 12:29   #16
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by CRGreathouse No. For every x you find all y such that x is related to y (x ~ y) and call the set of all such y's "[x]". Suppose I define ~ over the real numbers as "x ~ y if and only if x - y is an integer". Then 0.2 ~ 1.2, 0.2 ~ 2.2, 0.2 ~ 3.2, etc., so that [0.2] = {..., 0.2, 1.2, 2.2, ...}. Is this ~ an equivalence relation? (Check if it has the three properties.)
This is going to kill me I know. by the looks of it it would fit x=x for the first and
x<y implies y>x which I believe fits the second , not so sure about the third I don't see a z. However since I see only 2 properties can I safely conclude no ? I'm slightly confused.

2011-01-19, 14:13   #17
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by Mr. P-1 Based upon the kinds of things you're not sure about (such as the meaning of $\in$), I suggest a introductory math course covering a typical high school curriculum.
Also from which country the math course I didn't have in high school were pre-calculus and calculus last I checked.

2011-01-19, 14:28   #18
Chris Card

Aug 2004

3×43 Posts

Quote:
 Originally Posted by science_man_88 This is going to kill me I know. by the looks of it it would fit x=x for the first and xx which I believe fits the second , not so sure about the third I don't see a z. However since I see only 2 properties can I safely conclude no ? I'm slightly confused.
Suppose x ~ y and y ~ z. Then by definition, x - y is an integer and y - z is an integer.
If we can prove that x - z is an integer then x ~ z which is the third property of an equivalence relation.
But x - z = x - y + y - z and since x - y and y - z are integers, their sum must also be an integer.

Chris

2011-01-19, 14:31   #19
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Chris Card Suppose x ~ y and y ~ z. Then by definition, x - y is an integer and y - z is an integer. If we can prove that x - z is an integer then x ~ z which is the third property of an equivalence relation. But x - z = x - y + y - z and since x - y and y - z are integers, their sum must also be an integer. Chris
He never gave a property of z so I can't assume anything about it. Since there is no z the third property listed in the text can't happen. Which leads me to think the answer is no, because not all the properties have been shown.

Last fiddled with by science_man_88 on 2011-01-19 at 14:32

2011-01-19, 15:01   #20
xilman
Bamboozled!

May 2003
Down not across

2×712 Posts

Quote:
 Originally Posted by science_man_88 He never gave a property of z so I can't assume anything about it. Since there is no z the third property listed in the text can't happen. Which leads me to think the answer is no, because not all the properties have been shown.
I'm not sure who is the "he" to whom you refer but in your earlier post you have

Quote:
 A binary relation ∼ on a set S is called an equivalence relation if for all x, y, z ∈ S, we have • x ∼ x (reﬂexive property), • x ∼ y implies y ∼ x (symmetric property), and • x ∼ y and y ∼ z implies x ∼ z (transitive property).
The first portion says for all x, y, z and the last gives two properties of z. Once again, please try to read carefully.

Paul

2011-01-19, 15:06   #21
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by xilman I'm not sure who is the "he" to whom you refer but in your earlier post you have The first portion says for all x, y, z and the last gives two properties of z. Once again, please try to read carefully. Paul
I know, But if you look at CRG's example which is what I said no to it has no properties of z but Chris seems to be arguing back, that properties of a supposedly existent z not mentioned in CRG's post that I must be flawed ?

2011-01-19, 15:20   #22
Chris Card

Aug 2004

8116 Posts

Quote:
 Originally Posted by science_man_88 I know, But if you look at CRG's example which is what I said no to it has no properties of z but Chris seems to be arguing back, that properties of a supposedly existent z not mentioned in CRG's post that I must be flawed ?
CRG's example gives a definition for a relation ~ on the set of real numbers. Since the the relation is binary, you only need two symbols to define it. CRG used x and y but that was just arbitrary.

Chris

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