Prove $-0=0$ using the theorem of uniqueness of inverses.
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Theorem: Let $x$ be a real number. There exists one one real number $y$ such that $x+y=0$. Similarly, if $xneq 0$, there exists only one real number $y$ such that $xdot y$=1.
Using this theorem, prove $-0=0$.
I wrote down some steps of what is considered my scattered logic, but I am having a hard time forming a proof.
$0=-0$ (restating it)
$0=0+(-0)$ $rightarrow$ existence of additive inverse
$0=(-0)+0$ $rightarrow$ addition is commutative
$0=-0$ $rightarrow$ definition of additive identity
Is this correct? And if so, how can I form a proof using the theorem above?
real-analysis
asked Jan 15 at 16:10
RyanRyan
1448
$endgroup$
add a comment |
$begingroup$
Theorem: Let $x$ be a real number. There exists one one real number $y$ such that $x+y=0$. Similarly, if $xneq 0$, there exists only one real number $y$ such that $xdot y$=1.
Using this theorem, prove $-0=0$.
I wrote down some steps of what is considered my scattered logic, but I am having a hard time forming a proof.
$0=-0$ (restating it)
$0=0+(-0)$ $rightarrow$ existence of additive inverse
$0=(-0)+0$ $rightarrow$ addition is commutative
$0=-0$ $rightarrow$ definition of additive identity
Is this correct? And if so, how can I form a proof using the theorem above?
real-analysis
asked Jan 15 at 16:10
RyanRyan
1448
$endgroup$
1
$begingroup$
Apart from your "restating it", your proof is correct -- although I would call the last line "definition of additive identity", not "existence of additive identity", because you are using the equality $a = a + 0$, not the fact that an element satisfying that equality exists.
$endgroup$
– Mees de Vries
Jan 15 at 16:19
$begingroup$
Thanks! I knew I wasn't "restating" it. I am not sure why I included that. Its 6am here!
$endgroup$
– Ryan
Jan 15 at 16:29
add a comment |
$begingroup$
Theorem: Let $x$ be a real number. There exists one one real number $y$ such that $x+y=0$. Similarly, if $xneq 0$, there exists only one real number $y$ such that $xdot y$=1.
Using this theorem, prove $-0=0$.
I wrote down some steps of what is considered my scattered logic, but I am having a hard time forming a proof.
$0=-0$ (restating it)
$0=0+(-0)$ $rightarrow$ existence of additive inverse
$0=(-0)+0$ $rightarrow$ addition is commutative
$0=-0$ $rightarrow$ definition of additive identity
Is this correct? And if so, how can I form a proof using the theorem above?
real-analysis
asked Jan 15 at 16:10
RyanRyan
1448
$endgroup$
Theorem: Let $x$ be a real number. There exists one one real number $y$ such that $x+y=0$. Similarly, if $xneq 0$, there exists only one real number $y$ such that $xdot y$=1.
Using this theorem, prove $-0=0$.
I wrote down some steps of what is considered my scattered logic, but I am having a hard time forming a proof.
$0=-0$ (restating it)
$0=0+(-0)$ $rightarrow$ existence of additive inverse
$0=(-0)+0$ $rightarrow$ addition is commutative
$0=-0$ $rightarrow$ definition of additive identity
Is this correct? And if so, how can I form a proof using the theorem above?
real-analysis
real-analysis
asked Jan 15 at 16:10
RyanRyan
1448
asked Jan 15 at 16:10
RyanRyan
1448
edited Jan 15 at 16:34
Ryan
asked Jan 15 at 16:10
RyanRyan
1448
asked Jan 15 at 16:10
RyanRyan
1448
asked Jan 15 at 16:10
RyanRyan
1448
1448
1
$begingroup$
Apart from your "restating it", your proof is correct -- although I would call the last line "definition of additive identity", not "existence of additive identity", because you are using the equality $a = a + 0$, not the fact that an element satisfying that equality exists.
$endgroup$
– Mees de Vries
Jan 15 at 16:19
$begingroup$
Thanks! I knew I wasn't "restating" it. I am not sure why I included that. Its 6am here!
$endgroup$
– Ryan
Jan 15 at 16:29
add a comment |
1
$begingroup$
Apart from your "restating it", your proof is correct -- although I would call the last line "definition of additive identity", not "existence of additive identity", because you are using the equality $a = a + 0$, not the fact that an element satisfying that equality exists.
$endgroup$
– Mees de Vries
Jan 15 at 16:19
$begingroup$
Thanks! I knew I wasn't "restating" it. I am not sure why I included that. Its 6am here!
$endgroup$
– Ryan
Jan 15 at 16:29
1
1
$begingroup$
Apart from your "restating it", your proof is correct -- although I would call the last line "definition of additive identity", not "existence of additive identity", because you are using the equality $a = a + 0$, not the fact that an element satisfying that equality exists.
$endgroup$
– Mees de Vries
Jan 15 at 16:19
$begingroup$
Apart from your "restating it", your proof is correct -- although I would call the last line "definition of additive identity", not "existence of additive identity", because you are using the equality $a = a + 0$, not the fact that an element satisfying that equality exists.
$endgroup$
– Mees de Vries
Jan 15 at 16:19
$begingroup$
Thanks! I knew I wasn't "restating" it. I am not sure why I included that. Its 6am here!
$endgroup$
– Ryan
Jan 15 at 16:29
$begingroup$
Thanks! I knew I wasn't "restating" it. I am not sure why I included that. Its 6am here!
$endgroup$
– Ryan
Jan 15 at 16:29
add a comment |
2 Answers
2
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Note that $0+(-0)=0$ by the definition of inverse. On the other hand adding $0$ to a number doesn't change the number, hence $0+0=0$. That means $0,-0$ are both additive inverses of $0$. But the inverse is unique so they are equal.
answered Jan 15 at 16:17
MarkMark
6,913416
$endgroup$
add a comment |
$begingroup$
It's not clear how your proof is supposed to work: you state what you're trying to prove as the first line, making it appear that you're engaging in proof by assertion and/or circular reasoning.
You don't need the uniqueness of inverses (note that while the uniqueness of the identity is a theorem -- it can be derived from the axioms -- the existence of the identity is an axiom, not theorem).
$0+(-0)=0$ by definition of inverses, and $0+(-0)=-0$ by definition of identity. So $0$ and $-0$ are both equal to $0+(-0)$, and by transitivity of equality they are equal to each other.
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
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$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
Note that $0+(-0)=0$ by the definition of inverse. On the other hand adding $0$ to a number doesn't change the number, hence $0+0=0$. That means $0,-0$ are both additive inverses of $0$. But the inverse is unique so they are equal.
answered Jan 15 at 16:17
MarkMark
6,913416
$endgroup$
add a comment |
$begingroup$
Note that $0+(-0)=0$ by the definition of inverse. On the other hand adding $0$ to a number doesn't change the number, hence $0+0=0$. That means $0,-0$ are both additive inverses of $0$. But the inverse is unique so they are equal.
answered Jan 15 at 16:17
MarkMark
6,913416
$endgroup$
add a comment |
$begingroup$
Note that $0+(-0)=0$ by the definition of inverse. On the other hand adding $0$ to a number doesn't change the number, hence $0+0=0$. That means $0,-0$ are both additive inverses of $0$. But the inverse is unique so they are equal.
answered Jan 15 at 16:17
MarkMark
6,913416
$endgroup$
Note that $0+(-0)=0$ by the definition of inverse. On the other hand adding $0$ to a number doesn't change the number, hence $0+0=0$. That means $0,-0$ are both additive inverses of $0$. But the inverse is unique so they are equal.
answered Jan 15 at 16:17
MarkMark
6,913416
answered Jan 15 at 16:17
MarkMark
6,913416
answered Jan 15 at 16:17
MarkMark
6,913416
answered Jan 15 at 16:17
MarkMark
6,913416
6,913416
add a comment |
add a comment |
$begingroup$
It's not clear how your proof is supposed to work: you state what you're trying to prove as the first line, making it appear that you're engaging in proof by assertion and/or circular reasoning.
You don't need the uniqueness of inverses (note that while the uniqueness of the identity is a theorem -- it can be derived from the axioms -- the existence of the identity is an axiom, not theorem).
$0+(-0)=0$ by definition of inverses, and $0+(-0)=-0$ by definition of identity. So $0$ and $-0$ are both equal to $0+(-0)$, and by transitivity of equality they are equal to each other.
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
$endgroup$
$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
add a comment |
$begingroup$
It's not clear how your proof is supposed to work: you state what you're trying to prove as the first line, making it appear that you're engaging in proof by assertion and/or circular reasoning.
You don't need the uniqueness of inverses (note that while the uniqueness of the identity is a theorem -- it can be derived from the axioms -- the existence of the identity is an axiom, not theorem).
$0+(-0)=0$ by definition of inverses, and $0+(-0)=-0$ by definition of identity. So $0$ and $-0$ are both equal to $0+(-0)$, and by transitivity of equality they are equal to each other.
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
$endgroup$
$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
add a comment |
$begingroup$
It's not clear how your proof is supposed to work: you state what you're trying to prove as the first line, making it appear that you're engaging in proof by assertion and/or circular reasoning.
You don't need the uniqueness of inverses (note that while the uniqueness of the identity is a theorem -- it can be derived from the axioms -- the existence of the identity is an axiom, not theorem).
$0+(-0)=0$ by definition of inverses, and $0+(-0)=-0$ by definition of identity. So $0$ and $-0$ are both equal to $0+(-0)$, and by transitivity of equality they are equal to each other.
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
$endgroup$
It's not clear how your proof is supposed to work: you state what you're trying to prove as the first line, making it appear that you're engaging in proof by assertion and/or circular reasoning.
You don't need the uniqueness of inverses (note that while the uniqueness of the identity is a theorem -- it can be derived from the axioms -- the existence of the identity is an axiom, not theorem).
$0+(-0)=0$ by definition of inverses, and $0+(-0)=-0$ by definition of identity. So $0$ and $-0$ are both equal to $0+(-0)$, and by transitivity of equality they are equal to each other.
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
answered Jan 15 at 16:37
AcccumulationAcccumulation
6,9042618
6,9042618
$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
add a comment |
$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
$begingroup$
Duly noted -- thank you!
$endgroup$
– Ryan
Jan 15 at 16:41
add a comment |
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$begingroup$
Apart from your "restating it", your proof is correct -- although I would call the last line "definition of additive identity", not "existence of additive identity", because you are using the equality $a = a + 0$, not the fact that an element satisfying that equality exists.
$endgroup$
– Mees de Vries
Jan 15 at 16:19
$begingroup$
Thanks! I knew I wasn't "restating" it. I am not sure why I included that. Its 6am here!
$endgroup$
– Ryan
Jan 15 at 16:29