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Is this proof of $ab = 0$ correct?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have to prove the following theorem(Apostol's Calculus I, exercise 1 page 19):
If $ab = 0$ then either $a = 0$ or $b = 0$.
My attempt to solve it was: $ab = 0$ can be rewritten as $ab = a0$, because $a0 = 0$(already been proved).
So we now can cut a on both sides(already been proved too), so we have $b = 0$.
Also, we could rewrite the original equation as $ab = b0$ and b on both sides of equation and turn it into $a = 0$.
I think my proof covers the basic steps but I don't think it's asserting anything.
proof-verification
asked Jul 23 at 1:07
Victor Feitosa
445
add a comment |Â
up vote
0
down vote
favorite
I have to prove the following theorem(Apostol's Calculus I, exercise 1 page 19):
If $ab = 0$ then either $a = 0$ or $b = 0$.
My attempt to solve it was: $ab = 0$ can be rewritten as $ab = a0$, because $a0 = 0$(already been proved).
So we now can cut a on both sides(already been proved too), so we have $b = 0$.
Also, we could rewrite the original equation as $ab = b0$ and b on both sides of equation and turn it into $a = 0$.
I think my proof covers the basic steps but I don't think it's asserting anything.
proof-verification
asked Jul 23 at 1:07
Victor Feitosa
445
No, this isn't a proof of $ab=0$, remotely.
– Kenny Lau
Jul 23 at 1:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to prove the following theorem(Apostol's Calculus I, exercise 1 page 19):
If $ab = 0$ then either $a = 0$ or $b = 0$.
My attempt to solve it was: $ab = 0$ can be rewritten as $ab = a0$, because $a0 = 0$(already been proved).
So we now can cut a on both sides(already been proved too), so we have $b = 0$.
Also, we could rewrite the original equation as $ab = b0$ and b on both sides of equation and turn it into $a = 0$.
I think my proof covers the basic steps but I don't think it's asserting anything.
proof-verification
asked Jul 23 at 1:07
Victor Feitosa
445
I have to prove the following theorem(Apostol's Calculus I, exercise 1 page 19):
If $ab = 0$ then either $a = 0$ or $b = 0$.
My attempt to solve it was: $ab = 0$ can be rewritten as $ab = a0$, because $a0 = 0$(already been proved).
So we now can cut a on both sides(already been proved too), so we have $b = 0$.
Also, we could rewrite the original equation as $ab = b0$ and b on both sides of equation and turn it into $a = 0$.
I think my proof covers the basic steps but I don't think it's asserting anything.
proof-verification
asked Jul 23 at 1:07
Victor Feitosa
445
asked Jul 23 at 1:07
Victor Feitosa
445
asked Jul 23 at 1:07
Victor Feitosa
445
asked Jul 23 at 1:07
Victor Feitosa
445
445
No, this isn't a proof of $ab=0$, remotely.
– Kenny Lau
Jul 23 at 1:59
add a comment |Â
No, this isn't a proof of $ab=0$, remotely.
– Kenny Lau
Jul 23 at 1:59
No, this isn't a proof of $ab=0$, remotely.
– Kenny Lau
Jul 23 at 1:59
No, this isn't a proof of $ab=0$, remotely.
– Kenny Lau
Jul 23 at 1:59
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
This is almost perfect but you missing a bit:
My attempt to solve it was: $ab=0$ can be rewritten as $ab=a0$, because $a0=0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b=0$.
Here you need to add "assuming that $ane 0$" to be able to justify the last part.
Same thing with:
Also, we could rewrite the original equation as $ab=b0$ and b on both sides of equation and turn it into $a=0$.
You need to first assume that $bne 0$.
Now you left with the case that $a=b=0$, in which case it is trivial.
As @Theo point out we can do it using only $2$ cases:
Either $ane 0$ hence we can do the cancellation: $ab=a0implies b=0$
Or $a=0$, in this case we are done.
answered Jul 23 at 1:15
Holo
4,1962528
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
add a comment |Â
up vote
2
down vote
Proof: Let $a,b in mathbbR$ with $ab = 0$.
Then, if $a neq 0$, we know there exists $a^-1 in mathbbR$ such that $acdot a^-1= 1$. Thus,
$$ab = 0 implies a^-1 (ab) = a^-1 0 = 0$$But,$$beginalign* a^-1 (ab) = 0 &implies (a^-1a)b = 0 \ &implies 1b = 0\&implies b = 0. endalign*$$
answered Jul 23 at 1:36
Key Flex
4,278423
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
add a comment |Â
up vote
0
down vote
Consider the contrapositive of the statement:
if $aneq 0$ and $bneq0$, then $abneq0$. Which is trivial.
answered Jul 23 at 2:09
abc...
1,190524
2
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Proof by contradiction
Assume:
$ab= 0$ implies $a not =0$ and $b not = 0$.
If $a not =0$, and $b not =0$, $a, b$ have inverses $a^-1,b^-1$.
$(b^-1a^-1)(ab)=b^-1(a^-1a)b=$
$ b^-1(1b)= b^-1b=1;$
By assumption we have $ab=0$, hence
$(b^-1a^-1)(ab)= (b^-1a^-1)0=0.$
Hence $1=0$, a contradiction.
answered Jul 23 at 7:46
Peter Szilas
7,9252617
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is almost perfect but you missing a bit:
My attempt to solve it was: $ab=0$ can be rewritten as $ab=a0$, because $a0=0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b=0$.
Here you need to add "assuming that $ane 0$" to be able to justify the last part.
Same thing with:
Also, we could rewrite the original equation as $ab=b0$ and b on both sides of equation and turn it into $a=0$.
You need to first assume that $bne 0$.
Now you left with the case that $a=b=0$, in which case it is trivial.
As @Theo point out we can do it using only $2$ cases:
Either $ane 0$ hence we can do the cancellation: $ab=a0implies b=0$
Or $a=0$, in this case we are done.
answered Jul 23 at 1:15
Holo
4,1962528
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
add a comment |Â
up vote
3
down vote
accepted
This is almost perfect but you missing a bit:
My attempt to solve it was: $ab=0$ can be rewritten as $ab=a0$, because $a0=0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b=0$.
Here you need to add "assuming that $ane 0$" to be able to justify the last part.
Same thing with:
Also, we could rewrite the original equation as $ab=b0$ and b on both sides of equation and turn it into $a=0$.
You need to first assume that $bne 0$.
Now you left with the case that $a=b=0$, in which case it is trivial.
As @Theo point out we can do it using only $2$ cases:
Either $ane 0$ hence we can do the cancellation: $ab=a0implies b=0$
Or $a=0$, in this case we are done.
answered Jul 23 at 1:15
Holo
4,1962528
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is almost perfect but you missing a bit:
My attempt to solve it was: $ab=0$ can be rewritten as $ab=a0$, because $a0=0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b=0$.
Here you need to add "assuming that $ane 0$" to be able to justify the last part.
Same thing with:
Also, we could rewrite the original equation as $ab=b0$ and b on both sides of equation and turn it into $a=0$.
You need to first assume that $bne 0$.
Now you left with the case that $a=b=0$, in which case it is trivial.
As @Theo point out we can do it using only $2$ cases:
Either $ane 0$ hence we can do the cancellation: $ab=a0implies b=0$
Or $a=0$, in this case we are done.
answered Jul 23 at 1:15
Holo
4,1962528
This is almost perfect but you missing a bit:
My attempt to solve it was: $ab=0$ can be rewritten as $ab=a0$, because $a0=0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b=0$.
Here you need to add "assuming that $ane 0$" to be able to justify the last part.
Same thing with:
Also, we could rewrite the original equation as $ab=b0$ and b on both sides of equation and turn it into $a=0$.
You need to first assume that $bne 0$.
Now you left with the case that $a=b=0$, in which case it is trivial.
As @Theo point out we can do it using only $2$ cases:
Either $ane 0$ hence we can do the cancellation: $ab=a0implies b=0$
Or $a=0$, in this case we are done.
answered Jul 23 at 1:15
Holo
4,1962528
edited Jul 23 at 1:29
answered Jul 23 at 1:15
Holo
4,1962528
answered Jul 23 at 1:15
Holo
4,1962528
answered Jul 23 at 1:15
Holo
4,1962528
4,1962528
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
add a comment |Â
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
This can be done in two cases: if $a = 0$, you're done. Otherwise, division by $a$ is possible, so $b = 0$.
– Theo Bendit
Jul 23 at 1:26
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
@TheoBendit Correct, I didn't thought about this, I'll add this to the answer
– Holo
Jul 23 at 1:27
add a comment |Â
up vote
2
down vote
Proof: Let $a,b in mathbbR$ with $ab = 0$.
Then, if $a neq 0$, we know there exists $a^-1 in mathbbR$ such that $acdot a^-1= 1$. Thus,
$$ab = 0 implies a^-1 (ab) = a^-1 0 = 0$$But,$$beginalign* a^-1 (ab) = 0 &implies (a^-1a)b = 0 \ &implies 1b = 0\&implies b = 0. endalign*$$
answered Jul 23 at 1:36
Key Flex
4,278423
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
add a comment |Â
up vote
2
down vote
Proof: Let $a,b in mathbbR$ with $ab = 0$.
Then, if $a neq 0$, we know there exists $a^-1 in mathbbR$ such that $acdot a^-1= 1$. Thus,
$$ab = 0 implies a^-1 (ab) = a^-1 0 = 0$$But,$$beginalign* a^-1 (ab) = 0 &implies (a^-1a)b = 0 \ &implies 1b = 0\&implies b = 0. endalign*$$
answered Jul 23 at 1:36
Key Flex
4,278423
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Proof: Let $a,b in mathbbR$ with $ab = 0$.
Then, if $a neq 0$, we know there exists $a^-1 in mathbbR$ such that $acdot a^-1= 1$. Thus,
$$ab = 0 implies a^-1 (ab) = a^-1 0 = 0$$But,$$beginalign* a^-1 (ab) = 0 &implies (a^-1a)b = 0 \ &implies 1b = 0\&implies b = 0. endalign*$$
answered Jul 23 at 1:36
Key Flex
4,278423
Proof: Let $a,b in mathbbR$ with $ab = 0$.
Then, if $a neq 0$, we know there exists $a^-1 in mathbbR$ such that $acdot a^-1= 1$. Thus,
$$ab = 0 implies a^-1 (ab) = a^-1 0 = 0$$But,$$beginalign* a^-1 (ab) = 0 &implies (a^-1a)b = 0 \ &implies 1b = 0\&implies b = 0. endalign*$$
answered Jul 23 at 1:36
Key Flex
4,278423
answered Jul 23 at 1:36
Key Flex
4,278423
answered Jul 23 at 1:36
Key Flex
4,278423
answered Jul 23 at 1:36
Key Flex
4,278423
4,278423
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
add a comment |Â
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
Good alternative proof, very elegant
– Victor Feitosa
Jul 23 at 1:44
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
@VictorFeitosa Thanks!
– Key Flex
Jul 23 at 1:45
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
You could also start with "If $a=0$, then we are done. If $a ne 0$ then... so $b=0"
– steven gregory
Jul 23 at 1:47
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
@stevengregory Yes, that would be a short and easy way.
– Key Flex
Jul 23 at 1:54
add a comment |Â
up vote
0
down vote
Consider the contrapositive of the statement:
if $aneq 0$ and $bneq0$, then $abneq0$. Which is trivial.
answered Jul 23 at 2:09
abc...
1,190524
2
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
add a comment |Â
up vote
0
down vote
Consider the contrapositive of the statement:
if $aneq 0$ and $bneq0$, then $abneq0$. Which is trivial.
answered Jul 23 at 2:09
abc...
1,190524
2
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider the contrapositive of the statement:
if $aneq 0$ and $bneq0$, then $abneq0$. Which is trivial.
answered Jul 23 at 2:09
abc...
1,190524
Consider the contrapositive of the statement:
if $aneq 0$ and $bneq0$, then $abneq0$. Which is trivial.
answered Jul 23 at 2:09
abc...
1,190524
answered Jul 23 at 2:09
abc...
1,190524
answered Jul 23 at 2:09
abc...
1,190524
answered Jul 23 at 2:09
abc...
1,190524
1,190524
2
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
add a comment |Â
2
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
2
2
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
Why is it trivial? Easy? Sure. Fundamental? Certainly. But, there's still something to show here.
– Theo Bendit
Jul 23 at 3:42
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Proof by contradiction
Assume:
$ab= 0$ implies $a not =0$ and $b not = 0$.
If $a not =0$, and $b not =0$, $a, b$ have inverses $a^-1,b^-1$.
$(b^-1a^-1)(ab)=b^-1(a^-1a)b=$
$ b^-1(1b)= b^-1b=1;$
By assumption we have $ab=0$, hence
$(b^-1a^-1)(ab)= (b^-1a^-1)0=0.$
Hence $1=0$, a contradiction.
answered Jul 23 at 7:46
Peter Szilas
7,9252617
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Proof by contradiction
Assume:
$ab= 0$ implies $a not =0$ and $b not = 0$.
If $a not =0$, and $b not =0$, $a, b$ have inverses $a^-1,b^-1$.
$(b^-1a^-1)(ab)=b^-1(a^-1a)b=$
$ b^-1(1b)= b^-1b=1;$
By assumption we have $ab=0$, hence
$(b^-1a^-1)(ab)= (b^-1a^-1)0=0.$
Hence $1=0$, a contradiction.
answered Jul 23 at 7:46
Peter Szilas
7,9252617
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Correct me if wrong.
Proof by contradiction
Assume:
$ab= 0$ implies $a not =0$ and $b not = 0$.
If $a not =0$, and $b not =0$, $a, b$ have inverses $a^-1,b^-1$.
$(b^-1a^-1)(ab)=b^-1(a^-1a)b=$
$ b^-1(1b)= b^-1b=1;$
By assumption we have $ab=0$, hence
$(b^-1a^-1)(ab)= (b^-1a^-1)0=0.$
Hence $1=0$, a contradiction.
answered Jul 23 at 7:46
Peter Szilas
7,9252617
Correct me if wrong.
Proof by contradiction
Assume:
$ab= 0$ implies $a not =0$ and $b not = 0$.
If $a not =0$, and $b not =0$, $a, b$ have inverses $a^-1,b^-1$.
$(b^-1a^-1)(ab)=b^-1(a^-1a)b=$
$ b^-1(1b)= b^-1b=1;$
By assumption we have $ab=0$, hence
$(b^-1a^-1)(ab)= (b^-1a^-1)0=0.$
Hence $1=0$, a contradiction.
answered Jul 23 at 7:46
Peter Szilas
7,9252617
answered Jul 23 at 7:46
Peter Szilas
7,9252617
answered Jul 23 at 7:46
Peter Szilas
7,9252617
answered Jul 23 at 7:46
Peter Szilas
7,9252617
7,9252617
add a comment |Â
add a comment |Â
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No, this isn't a proof of $ab=0$, remotely.
– Kenny Lau
Jul 23 at 1:59