equivalence relation vs equality
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Apparently the equivalence class made with equality can only hold one element. The way that I see it, this means that 5-3 = 6-4 is true only because 6-4 is an element of an equivalence class that represents 1, like 7-6 and 111-110 is, and 5-3 is of the same equivalence class. That way we can say that X^R (Where X^R is the equivalence class of x) is equal to Y^R, when XRY. Because now the equivalence class containing X^R only holds a single element. Am I correct in assuming this? I think I'm being stupid...
Never thought I'd be befuddled by subtraction...
logic
asked Nov 22 at 3:43
pdf1234
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Apparently the equivalence class made with equality can only hold one element. The way that I see it, this means that 5-3 = 6-4 is true only because 6-4 is an element of an equivalence class that represents 1, like 7-6 and 111-110 is, and 5-3 is of the same equivalence class. That way we can say that X^R (Where X^R is the equivalence class of x) is equal to Y^R, when XRY. Because now the equivalence class containing X^R only holds a single element. Am I correct in assuming this? I think I'm being stupid...
Never thought I'd be befuddled by subtraction...
logic
asked Nov 22 at 3:43
pdf1234
6
6-4 represents 1? Don't you mean 2?
– coffeemath
Nov 22 at 3:52
...oops. lol...
– pdf1234
Nov 22 at 4:11
add a comment |
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0
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favorite
up vote
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down vote
favorite
Apparently the equivalence class made with equality can only hold one element. The way that I see it, this means that 5-3 = 6-4 is true only because 6-4 is an element of an equivalence class that represents 1, like 7-6 and 111-110 is, and 5-3 is of the same equivalence class. That way we can say that X^R (Where X^R is the equivalence class of x) is equal to Y^R, when XRY. Because now the equivalence class containing X^R only holds a single element. Am I correct in assuming this? I think I'm being stupid...
Never thought I'd be befuddled by subtraction...
logic
asked Nov 22 at 3:43
pdf1234
6
Apparently the equivalence class made with equality can only hold one element. The way that I see it, this means that 5-3 = 6-4 is true only because 6-4 is an element of an equivalence class that represents 1, like 7-6 and 111-110 is, and 5-3 is of the same equivalence class. That way we can say that X^R (Where X^R is the equivalence class of x) is equal to Y^R, when XRY. Because now the equivalence class containing X^R only holds a single element. Am I correct in assuming this? I think I'm being stupid...
Never thought I'd be befuddled by subtraction...
logic
logic
asked Nov 22 at 3:43
pdf1234
6
asked Nov 22 at 3:43
pdf1234
6
asked Nov 22 at 3:43
pdf1234
6
asked Nov 22 at 3:43
pdf1234
6
asked Nov 22 at 3:43
pdf1234
6
6
6-4 represents 1? Don't you mean 2?
– coffeemath
Nov 22 at 3:52
...oops. lol...
– pdf1234
Nov 22 at 4:11
add a comment |
6-4 represents 1? Don't you mean 2?
– coffeemath
Nov 22 at 3:52
...oops. lol...
– pdf1234
Nov 22 at 4:11
6-4 represents 1? Don't you mean 2?
– coffeemath
Nov 22 at 3:52
6-4 represents 1? Don't you mean 2?
– coffeemath
Nov 22 at 3:52
...oops. lol...
– pdf1234
Nov 22 at 4:11
...oops. lol...
– pdf1234
Nov 22 at 4:11
add a comment |
1 Answer
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The equivalence relation $R$, with the equivalence sets $(a,b)^R$ defined as ${(c,d)inBbb N^{+2}:a-b=c-d}$ is not an equality relation because it is not the case that every equivalence class contains a single element.
As you noticed, there are many pairs that are R-equivalence to any pair.$$(5,3)^R={(2,0),(3,1),(4,2),(5,3),(6,4),(7,5),ldots}$$
You may be confused by what they mean by "made with equality".
They mean that the equivalence class $X^=$ is defined as ${Y: Y=X}$ . That's not equating some function of the terms, but the terms themselves.
$$begin{split}(a,b)^= &={(c,d)inBbb N^{+2}: (c,d)=(a,b)}\&={(c,d)inBbb N^{+2}: c=a land d=b}\&={(a,b)}end{split}$$
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
add a comment |
1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The equivalence relation $R$, with the equivalence sets $(a,b)^R$ defined as ${(c,d)inBbb N^{+2}:a-b=c-d}$ is not an equality relation because it is not the case that every equivalence class contains a single element.
As you noticed, there are many pairs that are R-equivalence to any pair.$$(5,3)^R={(2,0),(3,1),(4,2),(5,3),(6,4),(7,5),ldots}$$
You may be confused by what they mean by "made with equality".
They mean that the equivalence class $X^=$ is defined as ${Y: Y=X}$ . That's not equating some function of the terms, but the terms themselves.
$$begin{split}(a,b)^= &={(c,d)inBbb N^{+2}: (c,d)=(a,b)}\&={(c,d)inBbb N^{+2}: c=a land d=b}\&={(a,b)}end{split}$$
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
add a comment |
up vote
0
down vote
The equivalence relation $R$, with the equivalence sets $(a,b)^R$ defined as ${(c,d)inBbb N^{+2}:a-b=c-d}$ is not an equality relation because it is not the case that every equivalence class contains a single element.
As you noticed, there are many pairs that are R-equivalence to any pair.$$(5,3)^R={(2,0),(3,1),(4,2),(5,3),(6,4),(7,5),ldots}$$
You may be confused by what they mean by "made with equality".
They mean that the equivalence class $X^=$ is defined as ${Y: Y=X}$ . That's not equating some function of the terms, but the terms themselves.
$$begin{split}(a,b)^= &={(c,d)inBbb N^{+2}: (c,d)=(a,b)}\&={(c,d)inBbb N^{+2}: c=a land d=b}\&={(a,b)}end{split}$$
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
add a comment |
up vote
0
down vote
up vote
0
down vote
The equivalence relation $R$, with the equivalence sets $(a,b)^R$ defined as ${(c,d)inBbb N^{+2}:a-b=c-d}$ is not an equality relation because it is not the case that every equivalence class contains a single element.
As you noticed, there are many pairs that are R-equivalence to any pair.$$(5,3)^R={(2,0),(3,1),(4,2),(5,3),(6,4),(7,5),ldots}$$
You may be confused by what they mean by "made with equality".
They mean that the equivalence class $X^=$ is defined as ${Y: Y=X}$ . That's not equating some function of the terms, but the terms themselves.
$$begin{split}(a,b)^= &={(c,d)inBbb N^{+2}: (c,d)=(a,b)}\&={(c,d)inBbb N^{+2}: c=a land d=b}\&={(a,b)}end{split}$$
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
The equivalence relation $R$, with the equivalence sets $(a,b)^R$ defined as ${(c,d)inBbb N^{+2}:a-b=c-d}$ is not an equality relation because it is not the case that every equivalence class contains a single element.
As you noticed, there are many pairs that are R-equivalence to any pair.$$(5,3)^R={(2,0),(3,1),(4,2),(5,3),(6,4),(7,5),ldots}$$
You may be confused by what they mean by "made with equality".
They mean that the equivalence class $X^=$ is defined as ${Y: Y=X}$ . That's not equating some function of the terms, but the terms themselves.
$$begin{split}(a,b)^= &={(c,d)inBbb N^{+2}: (c,d)=(a,b)}\&={(c,d)inBbb N^{+2}: c=a land d=b}\&={(a,b)}end{split}$$
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
edited Nov 22 at 4:16
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
answered Nov 22 at 4:07
Graham Kemp
84.6k43378
84.6k43378
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
add a comment |
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
Does that mean then that 5-3 can not equal 2? I think that I am not understanding what "Equal" means. How can you define equal? If I say that a Hippo is equal to a Croc and that Hippo^= contains two element, doesn't that make the equality invalid?
– pdf1234
Nov 22 at 4:37
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
@pdf1234 We have a choice of viewing $5-3$ and $6-4$ as two non-identical terms or as the same number, 2. The terms are in the same equivalence class of the equivalence relation "$t_1sim t_2$ iff $t_1$ and $t_2$ evaluate to the same number." When we decide we don't care about the terms themselves, only what they evaluate to, we quotient by this equivalence relation, effectively treating $sim$ as equality. This is the beauty of equivalence relations: they allow us to abstract away distinctions that don't matter and focus on the ones that do.
– spaceisdarkgreen
Nov 22 at 5:29
add a comment |
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6-4 represents 1? Don't you mean 2?
– coffeemath
Nov 22 at 3:52
...oops. lol...
– pdf1234
Nov 22 at 4:11