Is this correct approach to proving $xy=0$? [duplicate]
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This question already has an answer here:
Need help with a proof that if $xy=0$ then $x=0$ or $y=0$
3 answers
Is this a sufficient way of proving that for for all real numbers $x$ and $y$, if $xy=0$ then $x=0$ or $y=0$?
My attempt:
$$xy=0$$
$$y=frac{0}{x}$$
$$x=frac{0}{y}$$
Is this a correct approach using axioms of real numbers?
proof-verification proof-writing proof-explanation
edited Jan 16 at 1:23
David G. Stork
11.6k41533
asked Jan 16 at 1:19
ForextraderForextrader
988
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marked as duplicate by Key Flex, Chris Custer, Cesareo, A. Pongrácz, ancientmathematician Jan 16 at 7:48
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.
add a comment |
$begingroup$
This question already has an answer here:
Need help with a proof that if $xy=0$ then $x=0$ or $y=0$
3 answers
Is this a sufficient way of proving that for for all real numbers $x$ and $y$, if $xy=0$ then $x=0$ or $y=0$?
My attempt:
$$xy=0$$
$$y=frac{0}{x}$$
$$x=frac{0}{y}$$
Is this a correct approach using axioms of real numbers?
proof-verification proof-writing proof-explanation
edited Jan 16 at 1:23
David G. Stork
11.6k41533
asked Jan 16 at 1:19
ForextraderForextrader
988
$endgroup$
marked as duplicate by Key Flex, Chris Custer, Cesareo, A. Pongrácz, ancientmathematician Jan 16 at 7:48
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.
2
$begingroup$
A mathematical proof consists of logical statements, not just symbols.
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– user587192
Jan 16 at 1:22
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Perhaps you want to prove that if $xnot=0$, then $y$ must be zero.
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– Michael Burr
Jan 16 at 1:23
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Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said.
$endgroup$
– timtfj
Jan 16 at 1:32
add a comment |
$begingroup$
This question already has an answer here:
Need help with a proof that if $xy=0$ then $x=0$ or $y=0$
3 answers
Is this a sufficient way of proving that for for all real numbers $x$ and $y$, if $xy=0$ then $x=0$ or $y=0$?
My attempt:
$$xy=0$$
$$y=frac{0}{x}$$
$$x=frac{0}{y}$$
Is this a correct approach using axioms of real numbers?
proof-verification proof-writing proof-explanation
edited Jan 16 at 1:23
David G. Stork
11.6k41533
asked Jan 16 at 1:19
ForextraderForextrader
988
$endgroup$
This question already has an answer here:
Need help with a proof that if $xy=0$ then $x=0$ or $y=0$
3 answers
Is this a sufficient way of proving that for for all real numbers $x$ and $y$, if $xy=0$ then $x=0$ or $y=0$?
My attempt:
$$xy=0$$
$$y=frac{0}{x}$$
$$x=frac{0}{y}$$
Is this a correct approach using axioms of real numbers?
This question already has an answer here:
Need help with a proof that if $xy=0$ then $x=0$ or $y=0$
3 answers
proof-verification proof-writing proof-explanation
proof-verification proof-writing proof-explanation
edited Jan 16 at 1:23
David G. Stork
11.6k41533
asked Jan 16 at 1:19
ForextraderForextrader
988
edited Jan 16 at 1:23
David G. Stork
11.6k41533
asked Jan 16 at 1:19
ForextraderForextrader
988
edited Jan 16 at 1:23
David G. Stork
11.6k41533
edited Jan 16 at 1:23
David G. Stork
11.6k41533
edited Jan 16 at 1:23
David G. Stork
11.6k41533
11.6k41533
asked Jan 16 at 1:19
ForextraderForextrader
988
asked Jan 16 at 1:19
ForextraderForextrader
988
asked Jan 16 at 1:19
ForextraderForextrader
988
988
marked as duplicate by Key Flex, Chris Custer, Cesareo, A. Pongrácz, ancientmathematician Jan 16 at 7:48
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 Key Flex, Chris Custer, Cesareo, A. Pongrácz, ancientmathematician Jan 16 at 7:48
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.
2
$begingroup$
A mathematical proof consists of logical statements, not just symbols.
$endgroup$
– user587192
Jan 16 at 1:22
$begingroup$
Perhaps you want to prove that if $xnot=0$, then $y$ must be zero.
$endgroup$
– Michael Burr
Jan 16 at 1:23
$begingroup$
Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said.
$endgroup$
– timtfj
Jan 16 at 1:32
add a comment |
2
$begingroup$
A mathematical proof consists of logical statements, not just symbols.
$endgroup$
– user587192
Jan 16 at 1:22
$begingroup$
Perhaps you want to prove that if $xnot=0$, then $y$ must be zero.
$endgroup$
– Michael Burr
Jan 16 at 1:23
$begingroup$
Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said.
$endgroup$
– timtfj
Jan 16 at 1:32
2
2
$begingroup$
A mathematical proof consists of logical statements, not just symbols.
$endgroup$
– user587192
Jan 16 at 1:22
$begingroup$
A mathematical proof consists of logical statements, not just symbols.
$endgroup$
– user587192
Jan 16 at 1:22
$begingroup$
Perhaps you want to prove that if $xnot=0$, then $y$ must be zero.
$endgroup$
– Michael Burr
Jan 16 at 1:23
$begingroup$
Perhaps you want to prove that if $xnot=0$, then $y$ must be zero.
$endgroup$
– Michael Burr
Jan 16 at 1:23
$begingroup$
Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said.
$endgroup$
– timtfj
Jan 16 at 1:32
$begingroup$
Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said.
$endgroup$
– timtfj
Jan 16 at 1:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $x$ and $y$ be two real numbers and suppose $xy=0$. There are two cases:
$x=0$;
$xne 0$.
If $x=0$, then "$x=0$ or $y=0$" is true.
If $xneq 0$, then $x^{-1}$ exists and multiplying this number on both sides of $xy=0$ gives $y=0$, which implies that "$x=0$ or $y=0$" is true.
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x$ and $y$ be two real numbers and suppose $xy=0$. There are two cases:
$x=0$;
$xne 0$.
If $x=0$, then "$x=0$ or $y=0$" is true.
If $xneq 0$, then $x^{-1}$ exists and multiplying this number on both sides of $xy=0$ gives $y=0$, which implies that "$x=0$ or $y=0$" is true.
$endgroup$
add a comment |
$begingroup$
Let $x$ and $y$ be two real numbers and suppose $xy=0$. There are two cases:
$x=0$;
$xne 0$.
If $x=0$, then "$x=0$ or $y=0$" is true.
If $xneq 0$, then $x^{-1}$ exists and multiplying this number on both sides of $xy=0$ gives $y=0$, which implies that "$x=0$ or $y=0$" is true.
$endgroup$
add a comment |
$begingroup$
Let $x$ and $y$ be two real numbers and suppose $xy=0$. There are two cases:
$x=0$;
$xne 0$.
If $x=0$, then "$x=0$ or $y=0$" is true.
If $xneq 0$, then $x^{-1}$ exists and multiplying this number on both sides of $xy=0$ gives $y=0$, which implies that "$x=0$ or $y=0$" is true.
$endgroup$
Let $x$ and $y$ be two real numbers and suppose $xy=0$. There are two cases:
$x=0$;
$xne 0$.
If $x=0$, then "$x=0$ or $y=0$" is true.
If $xneq 0$, then $x^{-1}$ exists and multiplying this number on both sides of $xy=0$ gives $y=0$, which implies that "$x=0$ or $y=0$" is true.
answered Jan 16 at 1:30
user587192
add a comment |
add a comment |
2
$begingroup$
A mathematical proof consists of logical statements, not just symbols.
$endgroup$
– user587192
Jan 16 at 1:22
$begingroup$
Perhaps you want to prove that if $xnot=0$, then $y$ must be zero.
$endgroup$
– Michael Burr
Jan 16 at 1:23
$begingroup$
Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said.
$endgroup$
– timtfj
Jan 16 at 1:32