Cancellation law on a commutative and associative binary operation on a set $S$
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I need to Show: Let$*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$.(This z may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and ac=bc, then $a=b$.
contest-math
edited Dec 24 '18 at 7:25
bof
52.4k558121
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
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add a comment |
$begingroup$
I need to Show: Let$*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$.(This z may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and ac=bc, then $a=b$.
contest-math
edited Dec 24 '18 at 7:25
bof
52.4k558121
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
There's an archive of Putnam problems with solutions: kskedlaya.org/putnam-archive
$endgroup$
– the_fox
Nov 12 '18 at 10:27
add a comment |
$begingroup$
I need to Show: Let$*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$.(This z may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and ac=bc, then $a=b$.
contest-math
edited Dec 24 '18 at 7:25
bof
52.4k558121
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
$endgroup$
I need to Show: Let$*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$.(This z may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and ac=bc, then $a=b$.
contest-math
contest-math
edited Dec 24 '18 at 7:25
bof
52.4k558121
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
edited Dec 24 '18 at 7:25
bof
52.4k558121
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
edited Dec 24 '18 at 7:25
bof
52.4k558121
edited Dec 24 '18 at 7:25
bof
52.4k558121
edited Dec 24 '18 at 7:25
bof
52.4k558121
52.4k558121
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
asked Nov 12 '18 at 10:15
nafhgoodnafhgood
1,803422
1,803422
$begingroup$
There's an archive of Putnam problems with solutions: kskedlaya.org/putnam-archive
$endgroup$
– the_fox
Nov 12 '18 at 10:27
add a comment |
$begingroup$
There's an archive of Putnam problems with solutions: kskedlaya.org/putnam-archive
$endgroup$
– the_fox
Nov 12 '18 at 10:27
$begingroup$
There's an archive of Putnam problems with solutions: kskedlaya.org/putnam-archive
$endgroup$
– the_fox
Nov 12 '18 at 10:27
$begingroup$
There's an archive of Putnam problems with solutions: kskedlaya.org/putnam-archive
$endgroup$
– the_fox
Nov 12 '18 at 10:27
add a comment |
1 Answer
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Suppose ac=bc. Find $x,y$ so that $acx=a$ and $ay=b$. Then
$$a=acx=bcx=aycx=acxy=ay=b.$$
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose ac=bc. Find $x,y$ so that $acx=a$ and $ay=b$. Then
$$a=acx=bcx=aycx=acxy=ay=b.$$
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
$endgroup$
add a comment |
$begingroup$
Suppose ac=bc. Find $x,y$ so that $acx=a$ and $ay=b$. Then
$$a=acx=bcx=aycx=acxy=ay=b.$$
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
$endgroup$
add a comment |
$begingroup$
Suppose ac=bc. Find $x,y$ so that $acx=a$ and $ay=b$. Then
$$a=acx=bcx=aycx=acxy=ay=b.$$
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
$endgroup$
Suppose ac=bc. Find $x,y$ so that $acx=a$ and $ay=b$. Then
$$a=acx=bcx=aycx=acxy=ay=b.$$
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
answered Nov 12 '18 at 12:33
bofbof
52.4k558121
52.4k558121
add a comment |
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$begingroup$
There's an archive of Putnam problems with solutions: kskedlaya.org/putnam-archive
$endgroup$
– the_fox
Nov 12 '18 at 10:27