Associative or Commutative of Binary Operation
$begingroup$
I need to figure out whether these binary operations are commutative or associative. And then whether a unity exists (but I don't know what that means).
- M=$mathbb{Z}$; a*b=a-b
- M=$mathbb{Q}$; a*b=$frac{1}{2}$ab
I understand commutative property but not associative for these.
binary-operations
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
$endgroup$
add a comment |
$begingroup$
I need to figure out whether these binary operations are commutative or associative. And then whether a unity exists (but I don't know what that means).
- M=$mathbb{Z}$; a*b=a-b
- M=$mathbb{Q}$; a*b=$frac{1}{2}$ab
I understand commutative property but not associative for these.
binary-operations
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
$endgroup$
add a comment |
$begingroup$
I need to figure out whether these binary operations are commutative or associative. And then whether a unity exists (but I don't know what that means).
- M=$mathbb{Z}$; a*b=a-b
- M=$mathbb{Q}$; a*b=$frac{1}{2}$ab
I understand commutative property but not associative for these.
binary-operations
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
$endgroup$
I need to figure out whether these binary operations are commutative or associative. And then whether a unity exists (but I don't know what that means).
- M=$mathbb{Z}$; a*b=a-b
- M=$mathbb{Q}$; a*b=$frac{1}{2}$ab
I understand commutative property but not associative for these.
binary-operations
binary-operations
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
asked Oct 7 '15 at 1:34
ematth7ematth7
3891513
3891513
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
Commutative means
$a*b = b*a$.
Associative means
$a*(b*c) = (a*b)*c$.
A unity element $e$ is one such that
$a*e = a$ for all $a$
(this is a right unity -
a left unity is defined similarly).
So,
put these definitions into
your operator definitions
and see which holds and which do not.
For the unity,
see what properties the
unity element must have.
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Commutative means
$a*b = b*a$.
Associative means
$a*(b*c) = (a*b)*c$.
A unity element $e$ is one such that
$a*e = a$ for all $a$
(this is a right unity -
a left unity is defined similarly).
So,
put these definitions into
your operator definitions
and see which holds and which do not.
For the unity,
see what properties the
unity element must have.
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
add a comment |
$begingroup$
Commutative means
$a*b = b*a$.
Associative means
$a*(b*c) = (a*b)*c$.
A unity element $e$ is one such that
$a*e = a$ for all $a$
(this is a right unity -
a left unity is defined similarly).
So,
put these definitions into
your operator definitions
and see which holds and which do not.
For the unity,
see what properties the
unity element must have.
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
add a comment |
$begingroup$
Commutative means
$a*b = b*a$.
Associative means
$a*(b*c) = (a*b)*c$.
A unity element $e$ is one such that
$a*e = a$ for all $a$
(this is a right unity -
a left unity is defined similarly).
So,
put these definitions into
your operator definitions
and see which holds and which do not.
For the unity,
see what properties the
unity element must have.
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
$endgroup$
Commutative means
$a*b = b*a$.
Associative means
$a*(b*c) = (a*b)*c$.
A unity element $e$ is one such that
$a*e = a$ for all $a$
(this is a right unity -
a left unity is defined similarly).
So,
put these definitions into
your operator definitions
and see which holds and which do not.
For the unity,
see what properties the
unity element must have.
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
answered Oct 7 '15 at 2:25
marty cohenmarty cohen
73.3k549128
73.3k549128
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
add a comment |
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
So for the associative property for ab=a-b, I would have (ab)*c=(a-b)*c=(a-b)-c=a-b-c a*(bc)=a*(b-c)=a-(b-c)=a-b+c therefore (ab)*c does not equal a*(b-c)
$endgroup$
– ematth7
Oct 8 '15 at 20:53
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
$begingroup$
Correct. Now do the same for everything else.
$endgroup$
– marty cohen
Oct 8 '15 at 21:22
add a comment |
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