Another proof that dividing by 0 does not exist — is it right? The Next CEO of Stack OverflowHow do you explain to a 5th grader why division by zero is meaningless?Proof that $Bbb Z$ has no other subring than itselfProof that odd perfect numbers cannot consist of single unique factors?showing that no repunit is a square - proof verificationUsing induction, prove that $(3^2^n -1)$ is divisible by $2^n+2$ but not by $2^n+3$.Proof that $sqrt2$ is irrationalProve that between two unequal rational numbers there is another rationalMultiplicative inverse questionsIs my proof of $sqrt2 + sqrt3 + sqrt5$ is an irrational number valid?Proof of even numbersLeft and Right inverses - Proof by contradiction
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Another proof that dividing by 0 does not exist — is it right?
The Next CEO of Stack OverflowHow do you explain to a 5th grader why division by zero is meaningless?Proof that $Bbb Z$ has no other subring than itselfProof that odd perfect numbers cannot consist of single unique factors?showing that no repunit is a square - proof verificationUsing induction, prove that $(3^2^n -1)$ is divisible by $2^n+2$ but not by $2^n+3$.Proof that $sqrt2$ is irrationalProve that between two unequal rational numbers there is another rationalMultiplicative inverse questionsIs my proof of $sqrt2 + sqrt3 + sqrt5$ is an irrational number valid?Proof of even numbersLeft and Right inverses - Proof by contradiction
$begingroup$
Ok I am in grade 9 and I am maybe too young for this.
But I thought about this, why dividing by 0 is impossible.
Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.
So if we multiply a number by 0 then by 1/0 we get the same number.
But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible
Is this right?
proof-verification
New contributor
$endgroup$
add a comment |
$begingroup$
Ok I am in grade 9 and I am maybe too young for this.
But I thought about this, why dividing by 0 is impossible.
Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.
So if we multiply a number by 0 then by 1/0 we get the same number.
But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible
Is this right?
proof-verification
New contributor
$endgroup$
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago
11
$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago
add a comment |
$begingroup$
Ok I am in grade 9 and I am maybe too young for this.
But I thought about this, why dividing by 0 is impossible.
Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.
So if we multiply a number by 0 then by 1/0 we get the same number.
But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible
Is this right?
proof-verification
New contributor
$endgroup$
Ok I am in grade 9 and I am maybe too young for this.
But I thought about this, why dividing by 0 is impossible.
Dividing by 0 is possible would mean 1/0 is possible, which would mean 0 has a multiplicative inverse.
So if we multiply a number by 0 then by 1/0 we get the same number.
But thats impossible because all numbers multiplied by 0 gives 0 therefore we can’t have an inverse for 0, as that gives us the initial number and thus division by 0 is impossible
Is this right?
proof-verification
proof-verification
New contributor
New contributor
edited 2 hours ago
H Huang
401111
401111
New contributor
asked 8 hours ago
Selim Jean ElliehSelim Jean Ellieh
965
965
New contributor
New contributor
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago
11
$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago
add a comment |
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago
11
$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago
11
11
$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago
$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):
- What $1$ means ($1cdot a = a$ for any $a$)
- What $0$ means ($0 cdot a = 0$ for any $a$)
- What division means ($frac ab = c$ means $a = ccdot b$)
$endgroup$
add a comment |
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
$endgroup$
10
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
5
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
|
show 3 more comments
$begingroup$
That is quite right. However, I would like you to have a higher point of view.
Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.
A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(atimes b)times c=atimes (btimes c)$,
- $e_timestimes a=a$,
- there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.
Now verify that the rationals and the reals are fields.
Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.
$endgroup$
add a comment |
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3 Answers
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3 Answers
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active
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votes
$begingroup$
That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):
- What $1$ means ($1cdot a = a$ for any $a$)
- What $0$ means ($0 cdot a = 0$ for any $a$)
- What division means ($frac ab = c$ means $a = ccdot b$)
$endgroup$
add a comment |
$begingroup$
That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):
- What $1$ means ($1cdot a = a$ for any $a$)
- What $0$ means ($0 cdot a = 0$ for any $a$)
- What division means ($frac ab = c$ means $a = ccdot b$)
$endgroup$
add a comment |
$begingroup$
That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):
- What $1$ means ($1cdot a = a$ for any $a$)
- What $0$ means ($0 cdot a = 0$ for any $a$)
- What division means ($frac ab = c$ means $a = ccdot b$)
$endgroup$
That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):
- What $1$ means ($1cdot a = a$ for any $a$)
- What $0$ means ($0 cdot a = 0$ for any $a$)
- What division means ($frac ab = c$ means $a = ccdot b$)
answered 8 hours ago
ArthurArthur
121k7121208
121k7121208
add a comment |
add a comment |
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
$endgroup$
10
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
5
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
|
show 3 more comments
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
$endgroup$
10
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
5
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
|
show 3 more comments
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
$endgroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
edited 8 hours ago
answered 8 hours ago
ShaunShaun
9,933113684
9,933113684
10
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
5
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
|
show 3 more comments
10
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
1
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
1
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
5
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
10
10
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
$begingroup$
You think this is very relevant for a ninth grader? I mean, it might be cool to know it's out there, but does this really answer the question that is asked?
$endgroup$
– Arthur
8 hours ago
1
1
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
8 hours ago
1
1
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
8 hours ago
1
1
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
8 hours ago
5
5
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
$begingroup$
@Arthur I think this is extremely relevant. It shows that division by zero isn't some sort of sacred thing that we must not touch, it's just contradictory to the three Very Nice things in your post, and there are systems of "multiplication" and "division" out there where we are allowed to divide by zero. +1 for this answer.
$endgroup$
– YiFan
5 hours ago
|
show 3 more comments
$begingroup$
That is quite right. However, I would like you to have a higher point of view.
Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.
A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(atimes b)times c=atimes (btimes c)$,
- $e_timestimes a=a$,
- there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.
Now verify that the rationals and the reals are fields.
Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.
$endgroup$
add a comment |
$begingroup$
That is quite right. However, I would like you to have a higher point of view.
Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.
A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(atimes b)times c=atimes (btimes c)$,
- $e_timestimes a=a$,
- there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.
Now verify that the rationals and the reals are fields.
Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.
$endgroup$
add a comment |
$begingroup$
That is quite right. However, I would like you to have a higher point of view.
Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.
A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(atimes b)times c=atimes (btimes c)$,
- $e_timestimes a=a$,
- there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.
Now verify that the rationals and the reals are fields.
Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.
$endgroup$
That is quite right. However, I would like you to have a higher point of view.
Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.
A field is a set $F$ equipped with two binary operations $+,times$, such that there exists $e_+, e_times$, such that for all $a,b,cin F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(atimes b)times c=atimes (btimes c)$,
- $e_timestimes a=a$,
- there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.
Now verify that the rationals and the reals are fields.
Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.
answered 56 mins ago
TreborTrebor
95815
95815
add a comment |
add a comment |
Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.
Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.
Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.
Selim Jean Ellieh is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
8 hours ago
11
$begingroup$
Your answer is 100% correct and you should probably become a mathematician. These kinds of answers (mathematicians also call them proofs) are what mathematicians do all day long.
$endgroup$
– ErotemeObelus
4 hours ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 hours ago