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#1
Old 03-28-2003, 03:35 AM
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a=b, b=c, therefore a=c -- what theorem is this?

thanks
#2
Old 03-28-2003, 03:38 AM
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Transitivity.
#3
Old 03-28-2003, 03:39 AM
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Oh, and it's not a theorem; it's more of a postulate. You can't exactly prove it.
#4
Old 03-28-2003, 03:57 AM
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How about the C theroem?

int main
{
int a;
int b;
int c;

a=b=c=0;
}
#5
Old 03-28-2003, 04:00 AM
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Yeah, it's not a theorem. In math, you start with axioms or postulates. These are assumed to be true without proof. Transitivity is assumed to hold for equality. In general an equivalence relation is one that has these properties:[list=1][*]Reflexive -- a=a[*]Symmetric -- If a=b, then b=a[*]Transitive -- If a=b and b=c, then a=c[/list=1]
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#6
Old 03-28-2003, 04:09 AM
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So for instance, for people, "lives in the same country as" would be an equivalence relation.

"Is at least as tall as" would not be an equivalence relation because it's not symmetric.

"Has been in the same room as" would not be an equivalence relation because it is not transitive.
#7
Old 03-28-2003, 09:37 AM
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I think that this is one of Euclid's "Common notions" or postulates.

"Things that are equal to the same thing are equal to each other."
#8
Old 03-28-2003, 09:47 AM
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Actually, the transitive property of equality *is* a theorem. If you have reflexivity (for all a, a = a) and the principle of substitution (if t = s and P(t) is a statement about t, then P(s) has the same truth value as P(t)), you can derive symmetry (for all a and b, if a = b then b = a) and transitivity (as in the OP).

But this is an advanced matter. 99% of the mathematicians out there regard it as a postulate.
#9
Old 03-28-2003, 10:50 AM
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The Principle of Substitution doesn't look like math. It looks like logic.
#10
Old 03-28-2003, 10:52 AM
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Quote:
Originally posted by Urban Ranger
The Principle of Substitution doesn't look like math. It looks like logic.
You think there's a difference?
#11
Old 03-28-2003, 11:02 AM
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Worse, ultrafilter's definition of the principle of substitution uses the equality of t=s, which as DrMatrix mentioned is usually defined by using the idea of transitivity. So, it'd be tough to use that to derive transitivity.
#12
Old 03-28-2003, 11:14 AM
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Quote:
Originally posted by RM Mentock
Worse, ultrafilter's definition of the principle of substitution uses the equality of t=s, which as DrMatrix mentioned is usually defined by using the idea of transitivity. So, it'd be tough to use that to derive transitivity.
Actually, '=' is an undefined symbol in that definition, which just happens to satisfy that property. I'll give you a more complete answer later.
#13
Old 03-28-2003, 11:47 AM
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Quote:
Originally posted by ultrafilter
I'll give you a more complete answer later.
Not necessary. It's fairly easy to take a group of axioms and their derived theorems, and change the designation--make a theorem an axiom and derive a former axiom from it. But that doesn't make everything a theorem, and nothing an axiom. Clearly, you have to start somewhere sometime. Context is key.
#14
Old 03-28-2003, 12:01 PM
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David Simmons: yes. It's an algebraic equivalent of the first General Axiom from Euclid's 'The Elements'. To Euclid, it was one of a number of self-evident notions on which further his further reasoning was based.
#15
Old 03-28-2003, 12:02 PM
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Actually, isn't the transitive property just an application of substitution?

since a=b we can replace any occurance of b with a and still have a true statement.
#16
Old 03-28-2003, 12:11 PM
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Quote:
Originally posted by Hobie the One
Actually, isn't the transitive property just an application of substitution?

since a=b we can replace any occurance of b with a and still have a true statement.
I think that's what ultrafilter is saying, leaving equality undefined. This could get messy.
#17
Old 03-28-2003, 02:13 PM
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Messy? You don't know the half of it.

What I should've written is this:

For any binary relation R, if R(a, a) for all a, and R(t, s) implies that P(t) has the same truth value as P(s) for any statement P, then R(a, b) implies R(b, a) for all a, b, and R(a, b) and R(b, c) implies R(a, c) for all a, b, c.

In such case, R has all the same properties as '=', so that's how we usually denote it.

We do use '=' to represent weaker relations--for instance, most equivalence relations do not have the principle of substitution. That's why this is not just a rearrangement of axioms into theorems and vice versa. You can have reflexivity, transitivity, and symmetry without being able to substitute.

Note that for any particular area, the concept of 'a = b' is not undefined, but the symbols are undefined until we're talking about a particular area.

Is this a little clearer?
#18
Old 03-28-2003, 07:08 PM
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Quote:
Originally posted by ultrafilter
Is this a little clearer?
Seems to be just DrMatrix's definition, in a different form, as opposed to using substitution to define equivalence.
#19
Old 03-28-2003, 08:05 PM
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Quote:
Originally posted by RM Mentock
Seems to be just DrMatrix's definition, in a different form, as opposed to using substitution to define equivalence.
No, because DrMatrix's definition is not sufficient to imply substitution.
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