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Is this proof of $ab = 0$ correct?
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I'm trying to prove this following theorem:
If $x, y in mathbbZ$, use the cancellation law for $mathbbZ$ to demonstrate that $xy = 0 implies$ $x = 0$ or $y = 0$
The proof I came up with doesn't quite seem definitive enough. I know how to prove this without using the cancellation law, but this requirement seems to make things much more difficult.
So, clearly $0 = 0 cdot x = 0 cdot y$ $forall x, y in mathbbZ$. So, we can clearly say that this holds for $x neq 0$ and $y neq 0$.
So, first we can write that $xy = 0 cdot x$ for $x neq 0$. So, by the cancellation law, $y = 0$. Similarily, we can write that $xy =0 cdot y$ for $y neq 0$, so $x = 0$ by the cancellation law.
It seems to me that we can write $xy = 0 cdot a$ for any integer, so this doesn't quite "prove" anything, though in such a case we wouldn't be able to use the cancellation law, so that wouldn't at all be a pertinent fact.
How does this sound?
proof-verification
asked Jul 25 at 17:45
Matt.P
671313
marked as duplicate by Holo, Community♦ Jul 25 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Is this proof of $ab = 0$ correct?
4 answers
I'm trying to prove this following theorem:
If $x, y in mathbbZ$, use the cancellation law for $mathbbZ$ to demonstrate that $xy = 0 implies$ $x = 0$ or $y = 0$
The proof I came up with doesn't quite seem definitive enough. I know how to prove this without using the cancellation law, but this requirement seems to make things much more difficult.
So, clearly $0 = 0 cdot x = 0 cdot y$ $forall x, y in mathbbZ$. So, we can clearly say that this holds for $x neq 0$ and $y neq 0$.
So, first we can write that $xy = 0 cdot x$ for $x neq 0$. So, by the cancellation law, $y = 0$. Similarily, we can write that $xy =0 cdot y$ for $y neq 0$, so $x = 0$ by the cancellation law.
It seems to me that we can write $xy = 0 cdot a$ for any integer, so this doesn't quite "prove" anything, though in such a case we wouldn't be able to use the cancellation law, so that wouldn't at all be a pertinent fact.
How does this sound?
proof-verification
asked Jul 25 at 17:45
Matt.P
671313
marked as duplicate by Holo, Community♦ Jul 25 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Which is the "cancellation law" (for $Bbb Z$) ?
– dan_fulea
Jul 25 at 17:53
I believe it is that, for $x, y, z in mathbbZ$, $x cdot y = y cdot z wedge z neq 0 implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it.
– Matt.P
Jul 25 at 17:55
Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$.
– dan_fulea
Jul 25 at 17:57
>"So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law" But wait, if $xne 0$ then $x=0$, no need the second part.
– Holo
Jul 25 at 18:03
"It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0*a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody born in Maine so it doesn't prove anything" um.... why not?
– fleablood
Jul 25 at 18:23
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up vote
1
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up vote
1
down vote
favorite
This question already has an answer here:
Is this proof of $ab = 0$ correct?
4 answers
I'm trying to prove this following theorem:
If $x, y in mathbbZ$, use the cancellation law for $mathbbZ$ to demonstrate that $xy = 0 implies$ $x = 0$ or $y = 0$
The proof I came up with doesn't quite seem definitive enough. I know how to prove this without using the cancellation law, but this requirement seems to make things much more difficult.
So, clearly $0 = 0 cdot x = 0 cdot y$ $forall x, y in mathbbZ$. So, we can clearly say that this holds for $x neq 0$ and $y neq 0$.
So, first we can write that $xy = 0 cdot x$ for $x neq 0$. So, by the cancellation law, $y = 0$. Similarily, we can write that $xy =0 cdot y$ for $y neq 0$, so $x = 0$ by the cancellation law.
It seems to me that we can write $xy = 0 cdot a$ for any integer, so this doesn't quite "prove" anything, though in such a case we wouldn't be able to use the cancellation law, so that wouldn't at all be a pertinent fact.
How does this sound?
proof-verification
asked Jul 25 at 17:45
Matt.P
671313
This question already has an answer here:
Is this proof of $ab = 0$ correct?
4 answers
I'm trying to prove this following theorem:
If $x, y in mathbbZ$, use the cancellation law for $mathbbZ$ to demonstrate that $xy = 0 implies$ $x = 0$ or $y = 0$
The proof I came up with doesn't quite seem definitive enough. I know how to prove this without using the cancellation law, but this requirement seems to make things much more difficult.
So, clearly $0 = 0 cdot x = 0 cdot y$ $forall x, y in mathbbZ$. So, we can clearly say that this holds for $x neq 0$ and $y neq 0$.
So, first we can write that $xy = 0 cdot x$ for $x neq 0$. So, by the cancellation law, $y = 0$. Similarily, we can write that $xy =0 cdot y$ for $y neq 0$, so $x = 0$ by the cancellation law.
It seems to me that we can write $xy = 0 cdot a$ for any integer, so this doesn't quite "prove" anything, though in such a case we wouldn't be able to use the cancellation law, so that wouldn't at all be a pertinent fact.
How does this sound?
This question already has an answer here:
Is this proof of $ab = 0$ correct?
4 answers
proof-verification
asked Jul 25 at 17:45
Matt.P
671313
asked Jul 25 at 17:45
Matt.P
671313
asked Jul 25 at 17:45
Matt.P
671313
asked Jul 25 at 17:45
Matt.P
671313
671313
marked as duplicate by Holo, Community♦ Jul 25 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Holo, Community♦ Jul 25 at 18:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Which is the "cancellation law" (for $Bbb Z$) ?
– dan_fulea
Jul 25 at 17:53
I believe it is that, for $x, y, z in mathbbZ$, $x cdot y = y cdot z wedge z neq 0 implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it.
– Matt.P
Jul 25 at 17:55
Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$.
– dan_fulea
Jul 25 at 17:57
>"So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law" But wait, if $xne 0$ then $x=0$, no need the second part.
– Holo
Jul 25 at 18:03
"It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0*a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody born in Maine so it doesn't prove anything" um.... why not?
– fleablood
Jul 25 at 18:23
 |Â
show 1 more comment
Which is the "cancellation law" (for $Bbb Z$) ?
– dan_fulea
Jul 25 at 17:53
I believe it is that, for $x, y, z in mathbbZ$, $x cdot y = y cdot z wedge z neq 0 implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it.
– Matt.P
Jul 25 at 17:55
Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$.
– dan_fulea
Jul 25 at 17:57
>"So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law" But wait, if $xne 0$ then $x=0$, no need the second part.
– Holo
Jul 25 at 18:03
"It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0*a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody born in Maine so it doesn't prove anything" um.... why not?
– fleablood
Jul 25 at 18:23
Which is the "cancellation law" (for $Bbb Z$) ?
– dan_fulea
Jul 25 at 17:53
Which is the "cancellation law" (for $Bbb Z$) ?
– dan_fulea
Jul 25 at 17:53
I believe it is that, for $x, y, z in mathbbZ$, $x cdot y = y cdot z wedge z neq 0 implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it.
– Matt.P
Jul 25 at 17:55
I believe it is that, for $x, y, z in mathbbZ$, $x cdot y = y cdot z wedge z neq 0 implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it.
– Matt.P
Jul 25 at 17:55
Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$.
– dan_fulea
Jul 25 at 17:57
Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$.
– dan_fulea
Jul 25 at 17:57
>"So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law" But wait, if $xne 0$ then $x=0$, no need the second part.
– Holo
Jul 25 at 18:03
>"So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law" But wait, if $xne 0$ then $x=0$, no need the second part.
– Holo
Jul 25 at 18:03
"It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0*a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody born in Maine so it doesn't prove anything" um.... why not?
– fleablood
Jul 25 at 18:23
"It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0*a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody born in Maine so it doesn't prove anything" um.... why not?
– fleablood
Jul 25 at 18:23
 |Â
show 1 more comment
2 Answers
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The cancellation law says that if $cne 0$ and $ac = bc$ then $a=b$.
If $xy=0$ and $xne 0$, then since $xy = x0$, $y=0$.
answered Jul 25 at 17:57
John Douma
4,62011217
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0
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It seems to me that we can write $xy=0⋅a$ for any integer, so this doesn't quite "prove" anything,
And if $yne 0$ you can write $x = x*y*y^-1 = 0*a*y^-1 = 0*(a*y^-1) = 0$ for any integer. So that proves everything.
answered Jul 25 at 18:27
fleablood
60.3k22575
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The cancellation law says that if $cne 0$ and $ac = bc$ then $a=b$.
If $xy=0$ and $xne 0$, then since $xy = x0$, $y=0$.
answered Jul 25 at 17:57
John Douma
4,62011217
add a comment |Â
up vote
0
down vote
The cancellation law says that if $cne 0$ and $ac = bc$ then $a=b$.
If $xy=0$ and $xne 0$, then since $xy = x0$, $y=0$.
answered Jul 25 at 17:57
John Douma
4,62011217
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The cancellation law says that if $cne 0$ and $ac = bc$ then $a=b$.
If $xy=0$ and $xne 0$, then since $xy = x0$, $y=0$.
answered Jul 25 at 17:57
John Douma
4,62011217
The cancellation law says that if $cne 0$ and $ac = bc$ then $a=b$.
If $xy=0$ and $xne 0$, then since $xy = x0$, $y=0$.
answered Jul 25 at 17:57
John Douma
4,62011217
answered Jul 25 at 17:57
John Douma
4,62011217
answered Jul 25 at 17:57
John Douma
4,62011217
answered Jul 25 at 17:57
John Douma
4,62011217
4,62011217
add a comment |Â
add a comment |Â
up vote
0
down vote
It seems to me that we can write $xy=0⋅a$ for any integer, so this doesn't quite "prove" anything,
And if $yne 0$ you can write $x = x*y*y^-1 = 0*a*y^-1 = 0*(a*y^-1) = 0$ for any integer. So that proves everything.
answered Jul 25 at 18:27
fleablood
60.3k22575
add a comment |Â
up vote
0
down vote
It seems to me that we can write $xy=0⋅a$ for any integer, so this doesn't quite "prove" anything,
And if $yne 0$ you can write $x = x*y*y^-1 = 0*a*y^-1 = 0*(a*y^-1) = 0$ for any integer. So that proves everything.
answered Jul 25 at 18:27
fleablood
60.3k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It seems to me that we can write $xy=0⋅a$ for any integer, so this doesn't quite "prove" anything,
And if $yne 0$ you can write $x = x*y*y^-1 = 0*a*y^-1 = 0*(a*y^-1) = 0$ for any integer. So that proves everything.
answered Jul 25 at 18:27
fleablood
60.3k22575
It seems to me that we can write $xy=0⋅a$ for any integer, so this doesn't quite "prove" anything,
And if $yne 0$ you can write $x = x*y*y^-1 = 0*a*y^-1 = 0*(a*y^-1) = 0$ for any integer. So that proves everything.
answered Jul 25 at 18:27
fleablood
60.3k22575
answered Jul 25 at 18:27
fleablood
60.3k22575
answered Jul 25 at 18:27
fleablood
60.3k22575
answered Jul 25 at 18:27
fleablood
60.3k22575
60.3k22575
add a comment |Â
add a comment |Â
Which is the "cancellation law" (for $Bbb Z$) ?
– dan_fulea
Jul 25 at 17:53
I believe it is that, for $x, y, z in mathbbZ$, $x cdot y = y cdot z wedge z neq 0 implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it.
– Matt.P
Jul 25 at 17:55
Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$.
– dan_fulea
Jul 25 at 17:57
>"So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law" But wait, if $xne 0$ then $x=0$, no need the second part.
– Holo
Jul 25 at 18:03
"It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0*a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody born in Maine so it doesn't prove anything" um.... why not?
– fleablood
Jul 25 at 18:23