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Is my definition of division by zero problematic? [closed]
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I am a complete amateur, but I've attempted to define division by zero. This it's usually classified as undefined. After all, if 1/0 = b, where b is some non-zero number, then b times 0 would have to equal 1, and there exists no real number that, when multiplied by zero, equals 1.
Therefore, it was evident that my definition would necessitate the creation of a new kind of number. This number, which I call Z, is defined as Z=1/0. Z is thus the reciprocal of 0.
Playing with this new number for awhile less me to discover that 2/0=2Z, and so on where n/0=nZ. This leads to a whole set of numbers, Z numbers, similar to imaginary numbers. After all, imaginary numbers were created to define something that was formerly undefined, namely, a number whose square was -1. This leads to the whole set of imaginary / complex numbers.
Already I've discovered interesting facts about this number Z, such as the square of Z is equal to Z, since 1/0 times 1/0 equals 1/0.
My question is, is there some way in which my new set of numbers to define division by zero falls apart under closer scrutiny? I have yet to discover one, but I'm very new to mathematical exploration.
I apologize for not being familiar with the Math font used to write equations.
elementary-number-theory definition
asked Jul 30 at 4:07
Kevin H
1987
closed as unclear what you're asking by Peter, Mostafa Ayaz, Isaac Browne, José Carlos Santos, amWhy Jul 31 at 0:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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1
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I am a complete amateur, but I've attempted to define division by zero. This it's usually classified as undefined. After all, if 1/0 = b, where b is some non-zero number, then b times 0 would have to equal 1, and there exists no real number that, when multiplied by zero, equals 1.
Therefore, it was evident that my definition would necessitate the creation of a new kind of number. This number, which I call Z, is defined as Z=1/0. Z is thus the reciprocal of 0.
Playing with this new number for awhile less me to discover that 2/0=2Z, and so on where n/0=nZ. This leads to a whole set of numbers, Z numbers, similar to imaginary numbers. After all, imaginary numbers were created to define something that was formerly undefined, namely, a number whose square was -1. This leads to the whole set of imaginary / complex numbers.
Already I've discovered interesting facts about this number Z, such as the square of Z is equal to Z, since 1/0 times 1/0 equals 1/0.
My question is, is there some way in which my new set of numbers to define division by zero falls apart under closer scrutiny? I have yet to discover one, but I'm very new to mathematical exploration.
I apologize for not being familiar with the Math font used to write equations.
elementary-number-theory definition
asked Jul 30 at 4:07
Kevin H
1987
closed as unclear what you're asking by Peter, Mostafa Ayaz, Isaac Browne, José Carlos Santos, amWhy Jul 31 at 0:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 times 0^2 = Z times 0^2 = (Z times 0) times 0 = 1 times 0 = 0$, but on the other hand $Z^2 times 0^2 = (Z times 0)^2 = 1^2 = 1$.
– Gregory J. Puleo
Jul 30 at 4:14
3
This doesn't seem any different from the symbol $infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$
– saulspatz
Jul 30 at 4:15
2
@GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $,(-1)/0=1/(-0)=1/0=Z,$.
– dxiv
Jul 30 at 4:25
3
You would lose associativity since $2 = 0times(2times Z) ne (0times 2)times Z = 1$.
– Rahul
Jul 30 at 4:30
2
Why are you trying to divide by zero ?
– Rene Schipperus
Jul 30 at 4:49
 |Â
show 15 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am a complete amateur, but I've attempted to define division by zero. This it's usually classified as undefined. After all, if 1/0 = b, where b is some non-zero number, then b times 0 would have to equal 1, and there exists no real number that, when multiplied by zero, equals 1.
Therefore, it was evident that my definition would necessitate the creation of a new kind of number. This number, which I call Z, is defined as Z=1/0. Z is thus the reciprocal of 0.
Playing with this new number for awhile less me to discover that 2/0=2Z, and so on where n/0=nZ. This leads to a whole set of numbers, Z numbers, similar to imaginary numbers. After all, imaginary numbers were created to define something that was formerly undefined, namely, a number whose square was -1. This leads to the whole set of imaginary / complex numbers.
Already I've discovered interesting facts about this number Z, such as the square of Z is equal to Z, since 1/0 times 1/0 equals 1/0.
My question is, is there some way in which my new set of numbers to define division by zero falls apart under closer scrutiny? I have yet to discover one, but I'm very new to mathematical exploration.
I apologize for not being familiar with the Math font used to write equations.
elementary-number-theory definition
asked Jul 30 at 4:07
Kevin H
1987
I am a complete amateur, but I've attempted to define division by zero. This it's usually classified as undefined. After all, if 1/0 = b, where b is some non-zero number, then b times 0 would have to equal 1, and there exists no real number that, when multiplied by zero, equals 1.
Therefore, it was evident that my definition would necessitate the creation of a new kind of number. This number, which I call Z, is defined as Z=1/0. Z is thus the reciprocal of 0.
Playing with this new number for awhile less me to discover that 2/0=2Z, and so on where n/0=nZ. This leads to a whole set of numbers, Z numbers, similar to imaginary numbers. After all, imaginary numbers were created to define something that was formerly undefined, namely, a number whose square was -1. This leads to the whole set of imaginary / complex numbers.
Already I've discovered interesting facts about this number Z, such as the square of Z is equal to Z, since 1/0 times 1/0 equals 1/0.
My question is, is there some way in which my new set of numbers to define division by zero falls apart under closer scrutiny? I have yet to discover one, but I'm very new to mathematical exploration.
I apologize for not being familiar with the Math font used to write equations.
elementary-number-theory definition
asked Jul 30 at 4:07
Kevin H
1987
asked Jul 30 at 4:07
Kevin H
1987
asked Jul 30 at 4:07
Kevin H
1987
asked Jul 30 at 4:07
Kevin H
1987
1987
closed as unclear what you're asking by Peter, Mostafa Ayaz, Isaac Browne, José Carlos Santos, amWhy Jul 31 at 0:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Peter, Mostafa Ayaz, Isaac Browne, José Carlos Santos, amWhy Jul 31 at 0:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 times 0^2 = Z times 0^2 = (Z times 0) times 0 = 1 times 0 = 0$, but on the other hand $Z^2 times 0^2 = (Z times 0)^2 = 1^2 = 1$.
– Gregory J. Puleo
Jul 30 at 4:14
3
This doesn't seem any different from the symbol $infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$
– saulspatz
Jul 30 at 4:15
2
@GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $,(-1)/0=1/(-0)=1/0=Z,$.
– dxiv
Jul 30 at 4:25
3
You would lose associativity since $2 = 0times(2times Z) ne (0times 2)times Z = 1$.
– Rahul
Jul 30 at 4:30
2
Why are you trying to divide by zero ?
– Rene Schipperus
Jul 30 at 4:49
 |Â
show 15 more comments
4
I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 times 0^2 = Z times 0^2 = (Z times 0) times 0 = 1 times 0 = 0$, but on the other hand $Z^2 times 0^2 = (Z times 0)^2 = 1^2 = 1$.
– Gregory J. Puleo
Jul 30 at 4:14
3
This doesn't seem any different from the symbol $infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$
– saulspatz
Jul 30 at 4:15
2
@GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $,(-1)/0=1/(-0)=1/0=Z,$.
– dxiv
Jul 30 at 4:25
3
You would lose associativity since $2 = 0times(2times Z) ne (0times 2)times Z = 1$.
– Rahul
Jul 30 at 4:30
2
Why are you trying to divide by zero ?
– Rene Schipperus
Jul 30 at 4:49
4
4
I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 times 0^2 = Z times 0^2 = (Z times 0) times 0 = 1 times 0 = 0$, but on the other hand $Z^2 times 0^2 = (Z times 0)^2 = 1^2 = 1$.
– Gregory J. Puleo
Jul 30 at 4:14
I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 times 0^2 = Z times 0^2 = (Z times 0) times 0 = 1 times 0 = 0$, but on the other hand $Z^2 times 0^2 = (Z times 0)^2 = 1^2 = 1$.
– Gregory J. Puleo
Jul 30 at 4:14
3
3
This doesn't seem any different from the symbol $infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$
– saulspatz
Jul 30 at 4:15
This doesn't seem any different from the symbol $infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$
– saulspatz
Jul 30 at 4:15
2
2
@GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $,(-1)/0=1/(-0)=1/0=Z,$.
– dxiv
Jul 30 at 4:25
@GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $,(-1)/0=1/(-0)=1/0=Z,$.
– dxiv
Jul 30 at 4:25
3
3
You would lose associativity since $2 = 0times(2times Z) ne (0times 2)times Z = 1$.
– Rahul
Jul 30 at 4:30
You would lose associativity since $2 = 0times(2times Z) ne (0times 2)times Z = 1$.
– Rahul
Jul 30 at 4:30
2
2
Why are you trying to divide by zero ?
– Rene Schipperus
Jul 30 at 4:49
Why are you trying to divide by zero ?
– Rene Schipperus
Jul 30 at 4:49
 |Â
show 15 more comments
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With your definition of Z, we have $1/Z= 1/(1/0) =0$
Now we get $$2/Z = 2(0)=0 =1/Z$$ Multiply by $Z$ and we get $1=2$
Similarly you can prove $m=n$ for any two integers which we do not approve of.
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
With your definition of Z, we have $1/Z= 1/(1/0) =0$
Now we get $$2/Z = 2(0)=0 =1/Z$$ Multiply by $Z$ and we get $1=2$
Similarly you can prove $m=n$ for any two integers which we do not approve of.
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
4
down vote
accepted
With your definition of Z, we have $1/Z= 1/(1/0) =0$
Now we get $$2/Z = 2(0)=0 =1/Z$$ Multiply by $Z$ and we get $1=2$
Similarly you can prove $m=n$ for any two integers which we do not approve of.
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
With your definition of Z, we have $1/Z= 1/(1/0) =0$
Now we get $$2/Z = 2(0)=0 =1/Z$$ Multiply by $Z$ and we get $1=2$
Similarly you can prove $m=n$ for any two integers which we do not approve of.
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
With your definition of Z, we have $1/Z= 1/(1/0) =0$
Now we get $$2/Z = 2(0)=0 =1/Z$$ Multiply by $Z$ and we get $1=2$
Similarly you can prove $m=n$ for any two integers which we do not approve of.
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
edited Jul 30 at 4:49
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
answered Jul 30 at 4:28
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
4
I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 times 0^2 = Z times 0^2 = (Z times 0) times 0 = 1 times 0 = 0$, but on the other hand $Z^2 times 0^2 = (Z times 0)^2 = 1^2 = 1$.
– Gregory J. Puleo
Jul 30 at 4:14
3
This doesn't seem any different from the symbol $infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$
– saulspatz
Jul 30 at 4:15
2
@GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $,(-1)/0=1/(-0)=1/0=Z,$.
– dxiv
Jul 30 at 4:25
3
You would lose associativity since $2 = 0times(2times Z) ne (0times 2)times Z = 1$.
– Rahul
Jul 30 at 4:30
2
Why are you trying to divide by zero ?
– Rene Schipperus
Jul 30 at 4:49