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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 16 hours ago
Jack
27.7k1782204
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
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.
$endgroup$
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 16 hours ago
Jack
27.7k1782204
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
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.
$endgroup$
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
2 days ago
19
$begingroup$
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
2 days ago
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 days 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 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 16 hours ago
Jack
27.7k1782204
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
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 16 hours ago
Jack
27.7k1782204
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
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 16 hours ago
Jack
27.7k1782204
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
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 16 hours ago
Jack
27.7k1782204
edited 16 hours ago
Jack
27.7k1782204
edited 16 hours ago
Jack
27.7k1782204
27.7k1782204
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
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 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
asked 2 days ago
Selim Jean ElliehSelim Jean Ellieh
18117
18117
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
2 days ago
19
$begingroup$
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
2 days ago
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 days ago
add a comment |
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
2 days ago
19
$begingroup$
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
2 days ago
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 days ago
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
2 days ago
$begingroup$
Sometimes division by zero is defined, such as in the extended complex plane.
$endgroup$
– Shaun
2 days ago
19
19
$begingroup$
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
2 days 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
2 days ago
1
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 days ago
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 days 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 $a = ccdot b$)
answered 2 days ago
ArthurArthur
121k7122209
$endgroup$
7
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
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 2 days ago
ShaunShaun
10.1k113685
$endgroup$
20
$begingroup$
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
2 days ago
2
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
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
2 days ago
|
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 2 days ago
TreborTrebor
1,00015
$endgroup$
6
$begingroup$
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
2 days ago
4
$begingroup$
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
2 days ago
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
yesterday
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
yesterday
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
yesterday
|
show 3 more comments
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3 Answers
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3 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 2 days ago
ArthurArthur
121k7122209
$endgroup$
7
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
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 2 days ago
ArthurArthur
121k7122209
$endgroup$
7
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
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 2 days ago
ArthurArthur
121k7122209
$endgroup$
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 2 days ago
ArthurArthur
121k7122209
answered 2 days ago
ArthurArthur
121k7122209
answered 2 days ago
ArthurArthur
121k7122209
answered 2 days ago
ArthurArthur
121k7122209
121k7122209
7
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
add a comment |
7
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
7
7
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
$begingroup$
+1, but usually "what $0$ means" is $a+0=a$, and one gets $0a=0$ via distributivity (which very desirable).
$endgroup$
– Martin Argerami
yesterday
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 2 days ago
ShaunShaun
10.1k113685
$endgroup$
20
$begingroup$
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
2 days ago
2
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
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
2 days ago
|
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 2 days ago
ShaunShaun
10.1k113685
$endgroup$
20
$begingroup$
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
2 days ago
2
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
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
2 days ago
|
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 2 days ago
ShaunShaun
10.1k113685
$endgroup$
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 2 days ago
ShaunShaun
10.1k113685
edited yesterday
answered 2 days ago
ShaunShaun
10.1k113685
answered 2 days ago
ShaunShaun
10.1k113685
answered 2 days ago
ShaunShaun
10.1k113685
10.1k113685
20
$begingroup$
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
2 days ago
2
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
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
2 days ago
|
show 6 more comments
20
$begingroup$
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
2 days ago
2
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
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
2 days ago
20
20
$begingroup$
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
2 days 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
2 days ago
2
2
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
$begingroup$
That's a fair comment, @Arthur. Thank you for the feedback.
$endgroup$
– Shaun
2 days ago
2
2
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
$begingroup$
What d'you think, @SelimJeanEllieh?
$endgroup$
– Shaun
2 days ago
2
2
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
$begingroup$
Oh: The OP has insufficient rep to comment. Nevermind.
$endgroup$
– Shaun
2 days ago
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
2 days 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
2 days ago
|
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 2 days ago
TreborTrebor
1,00015
$endgroup$
6
$begingroup$
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
2 days ago
4
$begingroup$
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
2 days ago
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
yesterday
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
yesterday
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
yesterday
|
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 2 days ago
TreborTrebor
1,00015
$endgroup$
6
$begingroup$
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
2 days ago
4
$begingroup$
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
2 days ago
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
yesterday
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
yesterday
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
yesterday
|
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 2 days ago
TreborTrebor
1,00015
$endgroup$
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 2 days ago
TreborTrebor
1,00015
answered 2 days ago
TreborTrebor
1,00015
answered 2 days ago
TreborTrebor
1,00015
answered 2 days ago
TreborTrebor
1,00015
1,00015
6
$begingroup$
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
2 days ago
4
$begingroup$
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
2 days ago
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
yesterday
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
yesterday
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
yesterday
|
show 3 more comments
6
$begingroup$
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
2 days ago
4
$begingroup$
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
2 days ago
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
yesterday
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
yesterday
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
yesterday
6
6
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
4
4
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
|
show 3 more comments
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
2 days ago
19
$begingroup$
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
2 days ago
1
$begingroup$
math.stackexchange.com/questions/2883450/…
$endgroup$
– Maria Mazur
2 days ago