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Another proof that dividing by $0$ does not exist — is it right?
How do you explain to a 5th grader why division by zero is meaningless?Please help I'm in grade 9 and excellent at maths but I keep asking why does it workDoes a negative number really exist?Resource for low level maths explained in high level perspectivesHow to define the operation of division apart from the inverse of multiplication?Non-trivial “I know what number you're thinking of”showing that no repunit is a square - proof verificationProve that between two unequal rational numbers there is another rationalIntuitively, if addition can be interpreted as combining sets, then what can multiplication and division be interpreted as?Multiplicative inverse questionsIs my proof of $sqrt2 + sqrt3 + sqrt5$ is an irrational number valid?Are positive numbers somehow more “fundamental” than negative numbers?
$begingroup$
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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question
edited yesterday
Jack
27.7k1782204
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
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Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question
edited yesterday
Jack
27.7k1782204
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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
Apr 1 at 19:57
20
$begingroup$
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
Apr 1 at 23:44
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
Apr 2 at 1:28
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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question
edited yesterday
Jack
27.7k1782204
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 that's impossible because all numbers multiplied by $0$ give $0$ 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 soft-question
proof-verification soft-question
edited yesterday
Jack
27.7k1782204
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jack
27.7k1782204
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jack
27.7k1782204
edited yesterday
Jack
27.7k1782204
edited yesterday
Jack
27.7k1782204
27.7k1782204
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
asked Apr 1 at 19:54
Selim Jean ElliehSelim Jean Ellieh
19618
19618
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Selim Jean Ellieh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
Apr 1 at 19:57
20
$begingroup$
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
Apr 1 at 23:44
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
Apr 2 at 1:28
add a comment |
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
Apr 1 at 19:57
20
$begingroup$
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
Apr 1 at 23:44
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
Apr 2 at 1:28
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
Apr 1 at 19:57
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
Apr 1 at 19:57
20
20
$begingroup$
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
Apr 1 at 23:44
$begingroup$
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
Apr 1 at 23:44
1
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
Apr 2 at 1:28
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
Apr 2 at 1:28
add a comment |
4 Answers
4
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 $a = ccdot b$)
answered Apr 1 at 20:00
ArthurArthur
122k7122210
$endgroup$
8
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
add a comment |
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac00=:bot$.
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
$endgroup$
23
$begingroup$
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
Apr 1 at 20:02
3
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
10
$begingroup$
@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
Apr 1 at 22:34
|
show 6 more comments
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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 Apr 2 at 3:12
TreborTrebor
99815
$endgroup$
8
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
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– Chef Cyanide
Apr 2 at 4:49
5
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
2
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
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– Henning Makholm
2 days ago
1
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@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
1
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days ago
|
show 3 more comments
$begingroup$
You are quite right.
There is a simpler way, though (which spares the concept of multiplicative inverse):
By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:
$$0cdot q=d.$$
But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
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add a comment |
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4 Answers
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4 Answers
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$begingroup$
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 $a = ccdot b$)
answered Apr 1 at 20:00
ArthurArthur
122k7122210
$endgroup$
8
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
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 $a = ccdot b$)
answered Apr 1 at 20:00
ArthurArthur
122k7122210
$endgroup$
8
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
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 $a = ccdot b$)
answered Apr 1 at 20:00
ArthurArthur
122k7122210
$endgroup$
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 Apr 1 at 20:00
ArthurArthur
122k7122210
answered Apr 1 at 20:00
ArthurArthur
122k7122210
answered Apr 1 at 20:00
ArthurArthur
122k7122210
answered Apr 1 at 20:00
ArthurArthur
122k7122210
122k7122210
8
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
add a comment |
8
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
8
8
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
2 days ago
add a comment |
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac00=:bot$.
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
$endgroup$
23
$begingroup$
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
Apr 1 at 20:02
3
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
10
$begingroup$
@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
Apr 1 at 22:34
|
show 6 more comments
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac00=:bot$.
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
$endgroup$
23
$begingroup$
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
Apr 1 at 20:02
3
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
10
$begingroup$
@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
Apr 1 at 22:34
|
show 6 more comments
$begingroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac00=:bot$.
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
$endgroup$
Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_bot^infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $frac00=:bot$.
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
edited 2 days ago
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
answered Apr 1 at 19:59
ShaunShaun
10.3k113686
10.3k113686
23
$begingroup$
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
Apr 1 at 20:02
3
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
10
$begingroup$
@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
Apr 1 at 22:34
|
show 6 more comments
23
$begingroup$
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
Apr 1 at 20:02
3
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
10
$begingroup$
@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
Apr 1 at 22:34
23
23
$begingroup$
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
Apr 1 at 20:02
$begingroup$
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
Apr 1 at 20:02
3
3
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
Apr 1 at 20:03
2
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
Apr 1 at 20:04
2
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
Apr 1 at 20:06
10
10
$begingroup$
@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
Apr 1 at 22:34
$begingroup$
@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
Apr 1 at 22:34
|
show 6 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.
answered Apr 2 at 3:12
TreborTrebor
99815
$endgroup$
8
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
$endgroup$
– Chef Cyanide
Apr 2 at 4:49
5
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
2
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
1
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days 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.
answered Apr 2 at 3:12
TreborTrebor
99815
$endgroup$
8
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
$endgroup$
– Chef Cyanide
Apr 2 at 4:49
5
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
2
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
1
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days 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.
answered Apr 2 at 3:12
TreborTrebor
99815
$endgroup$
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 Apr 2 at 3:12
TreborTrebor
99815
answered Apr 2 at 3:12
TreborTrebor
99815
answered Apr 2 at 3:12
TreborTrebor
99815
answered Apr 2 at 3:12
TreborTrebor
99815
99815
8
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
$endgroup$
– Chef Cyanide
Apr 2 at 4:49
5
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
2
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
1
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days ago
|
show 3 more comments
8
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
$endgroup$
– Chef Cyanide
Apr 2 at 4:49
5
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
2
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
1
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days ago
8
8
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
$endgroup$
– Chef Cyanide
Apr 2 at 4:49
$begingroup$
While I think the mathematical theory behind fields is definitely a good point to bring up, I'd like to suggest that this answer be simplified considerably (after all, OP is in 9th grade, and this is generally considered a good ways above the level of mathematics taught in most High Schools).
$endgroup$
– Chef Cyanide
Apr 2 at 4:49
5
5
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
$begingroup$
In particular, jargon such as "such that", "there exists", and "for all" are probably unfamiliar to a 9th grader. Additionally, all non arithmetic symbols (like "∈") are probably off the table. Finally, 9th graders probably won't be familiar with conventions like "$e_*$" meaning the identity element with respect to $*$.
$endgroup$
– Vaelus
Apr 2 at 5:40
2
2
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
$endgroup$
– Henning Makholm
2 days ago
$begingroup$
@Vaelus: That would be a relevant consideration if we were talking to a randomly picked 9th-grader. However, here we're dealing with a 9th-grader who is inquisitive and mathematically minded enough to come up with their own proofs just out of curiosity. I don't think a bit of mathematics jargon will scare them away; they'll learn it sooner or later anyway. At university, 13th-graders are expected to absorb the lingo mostly by imitation and examples; for this OP getting a four-year head start on that will not harm.
$endgroup$
– Henning Makholm
2 days ago
1
1
$begingroup$
@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
$begingroup$
@HenningMakholm I don’t think that just throwing around jargon will help, even for an inquisitive 9th grader. At the very least, such jargon should be carefully introduced and motivated. I too was once an inquisitive 9th grader, but even then I might have been intimidated by an excessive amount of unfamiliar jargon which was not defined and simply assumed as known.
$endgroup$
– Sir Jective
2 days ago
1
1
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days ago
$begingroup$
@HenningMakholm In particular, one of the most frustrating experiences for me in college was being expected to “absorb the lingo mostly by imitation and examples”; it felt like it sapped away any of the intuition or motivation that should have been there. Why were we building up things this way, and not some other? It’s like this sort of inquisitiveness was being actively stifled.
$endgroup$
– Sir Jective
2 days ago
|
show 3 more comments
$begingroup$
You are quite right.
There is a simpler way, though (which spares the concept of multiplicative inverse):
By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:
$$0cdot q=d.$$
But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
You are quite right.
There is a simpler way, though (which spares the concept of multiplicative inverse):
By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:
$$0cdot q=d.$$
But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
You are quite right.
There is a simpler way, though (which spares the concept of multiplicative inverse):
By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:
$$0cdot q=d.$$
But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
$endgroup$
You are quite right.
There is a simpler way, though (which spares the concept of multiplicative inverse):
By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:
$$0cdot q=d.$$
But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
answered 10 hours ago
Yves DaoustYves Daoust
132k676230
132k676230
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
Apr 1 at 19:57
20
$begingroup$
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
Apr 1 at 23:44
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
Apr 2 at 1:28