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How do we know our definitions don't lead to contradictions
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How do we assure ourselves when defining an operation that it does not lead to contradictions? For example 0! := 1. I understand the practicality of why it is defined this way, but I am wary of what could happen if another such operation is created and defined in a way that leads to future contradictions that were not considered. This often happens when people come up with definitions for division by zero. How do we know it doesn’t happen with other definitions?
Note that I'm not asking about the consistency of our axioms for N or anything like that, but about definitions built upon those axioms.
logic foundations
asked 1 hour ago
Afa Juste
413
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up vote
4
down vote
favorite
How do we assure ourselves when defining an operation that it does not lead to contradictions? For example 0! := 1. I understand the practicality of why it is defined this way, but I am wary of what could happen if another such operation is created and defined in a way that leads to future contradictions that were not considered. This often happens when people come up with definitions for division by zero. How do we know it doesn’t happen with other definitions?
Note that I'm not asking about the consistency of our axioms for N or anything like that, but about definitions built upon those axioms.
logic foundations
asked 1 hour ago
Afa Juste
413
1
Formally speaking, a definition cannot be inconsistent. What it might do is make certain universal theorems we have proved about the operation before we extended it no longer true.
– spaceisdarkgreen
1 hour ago
$0 ne 1$ is not a definition; it is a theorem.
– Mauro ALLEGRANZA
1 hour ago
@Mauro that’s “zero factorial equals oneâ€Â
– spaceisdarkgreen
20 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
How do we assure ourselves when defining an operation that it does not lead to contradictions? For example 0! := 1. I understand the practicality of why it is defined this way, but I am wary of what could happen if another such operation is created and defined in a way that leads to future contradictions that were not considered. This often happens when people come up with definitions for division by zero. How do we know it doesn’t happen with other definitions?
Note that I'm not asking about the consistency of our axioms for N or anything like that, but about definitions built upon those axioms.
logic foundations
asked 1 hour ago
Afa Juste
413
How do we assure ourselves when defining an operation that it does not lead to contradictions? For example 0! := 1. I understand the practicality of why it is defined this way, but I am wary of what could happen if another such operation is created and defined in a way that leads to future contradictions that were not considered. This often happens when people come up with definitions for division by zero. How do we know it doesn’t happen with other definitions?
Note that I'm not asking about the consistency of our axioms for N or anything like that, but about definitions built upon those axioms.
logic foundations
logic foundations
asked 1 hour ago
Afa Juste
413
asked 1 hour ago
Afa Juste
413
asked 1 hour ago
Afa Juste
413
asked 1 hour ago
Afa Juste
413
asked 1 hour ago
Afa Juste
413
413
1
Formally speaking, a definition cannot be inconsistent. What it might do is make certain universal theorems we have proved about the operation before we extended it no longer true.
– spaceisdarkgreen
1 hour ago
$0 ne 1$ is not a definition; it is a theorem.
– Mauro ALLEGRANZA
1 hour ago
@Mauro that’s “zero factorial equals oneâ€Â
– spaceisdarkgreen
20 mins ago
add a comment |Â
1
Formally speaking, a definition cannot be inconsistent. What it might do is make certain universal theorems we have proved about the operation before we extended it no longer true.
– spaceisdarkgreen
1 hour ago
$0 ne 1$ is not a definition; it is a theorem.
– Mauro ALLEGRANZA
1 hour ago
@Mauro that’s “zero factorial equals oneâ€Â
– spaceisdarkgreen
20 mins ago
1
1
Formally speaking, a definition cannot be inconsistent. What it might do is make certain universal theorems we have proved about the operation before we extended it no longer true.
– spaceisdarkgreen
1 hour ago
Formally speaking, a definition cannot be inconsistent. What it might do is make certain universal theorems we have proved about the operation before we extended it no longer true.
– spaceisdarkgreen
1 hour ago
$0 ne 1$ is not a definition; it is a theorem.
– Mauro ALLEGRANZA
1 hour ago
$0 ne 1$ is not a definition; it is a theorem.
– Mauro ALLEGRANZA
1 hour ago
@Mauro that’s “zero factorial equals oneâ€Â
– spaceisdarkgreen
20 mins ago
@Mauro that’s “zero factorial equals oneâ€Â
– spaceisdarkgreen
20 mins ago
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
1
down vote
Certain types of definitions can lead to contradictions. It is up to the person proposing the definition to show that this can't happen. Look up well-defined.
answered 1 hour ago
Robert Israel
310k23202447
add a comment |Â
up vote
1
down vote
It depends what you mean by a definition. A definition such as
$$
P mathrel:= i in BbbN mid mbox$i$ is prime
$$
where the defining property of $P$ simply states that it is equal to something that already exists is completely unproblematic. But in addition to explicit definitions like this, we often use various kinds of implicit definition. A recursive definition such as:
$$
beginalign*
0! &= 1 \
(n+1)! & = n! times (n + 1)
endalign*
$$
is a fairly explicit form of implicit definition that can be shown to be consistent using the recursion theorem as discussed Marcel's answer.
A recursive "definition" (of a function $f : BbbN to BbbN$) such as
$$
beginalign*
f(0) &= 1 \
f(n+1) & = f(n+2)
endalign*
$$
would be considered undefined in programming languages, but from a mathematical perspective it is just a loose definition (i.e., an implicit definition that does not uniquely determine the defined object): it says that $f(0) = 1$ and $f(i) = f(j)$ for $i, j > 0$ and this system of equations has many solutions. If we had instead:
$$
beginalign*
f(0) &= 1 \
f(n+1) & = 2(f(n+2)+1)
endalign*
$$
then this "recursive" definition would be inconsistent, i.e., there are no solutions and asserting the existence of such an $f$ would lead to a contradiction (do you see why?).
Another form of implicit definition would be to define the function $exp : BbbR to BbbR$ as the solution of the following system of equations:
$$
beginalign*
exp(0) &= 1 \
exp'(x) &= exp(x)
endalign*
$$
Here an explicit construction of a witness (say using a power series) shows that the definition is consistent (and a simple argument using properties of the derivative shows that the definition characterises $exp$, i.e., the equations have only one solution).
In general, there are many forms of implicit definition. To show that it is consistent to take a property $phi(X)$ to be the defining property of a new object $X$, you have to prove $exists xphi(x)$. You will probably also want to prove that the existence is unique, i.e., $forall xforall y(phi(x) land phi(y) Rightarrow x = y)$, unless you really did intend your definition to be loose. If you are lucky, then something like the recursion theorem or the theory of differential equations will help with these proofs.
answered 28 mins ago
Rob Arthan
28k42865
add a comment |Â
up vote
0
down vote
A recursively defined function like the factorial function is well-defined by the recursion theorem, which can be proved by induction:
https://math.stackexchange.com/a/46760/29892
Henkin has a great article on recursion available here:
https://www.jstor.org/stable/2308975
answered 1 hour ago
Marcel Besixdouze
1,112717
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
add a comment |Â
up vote
0
down vote
A definition like that can't lead to a contradiction since it is just a definition: the meaning you've assigned to the symbols "$0!$".
A definition might lead to a problem with some rules of arithmetic you like. For example, if you chose to define $x^0 = 0$ instead of $1$ then the identlty
$$
x^a+b= x^ax^b
$$
would fail sometimes. So it would be a bad definition, but not a contradiction.
All the usual suggestions for how to define division by $0$ break some identity in ordinary arithmetic, which is why we don't use any of them.
Defining $0! = 1$ preserves the identity
$$
n! = n(n-1)!
$$
No other definition would do that.
answered 1 hour ago
Ethan Bolker
37.8k543100
add a comment |Â
up vote
0
down vote
About $0^0$,
A polynomial, in general is written as
$$a_0+a_1x+...a_nx^n=sum_i=0^na_ix^i$$
To be consistant, one will put $0^0=1$, to get the constant term $a_0$.
but
If you compute $$lim_xto+infty(e^-x)^frac-1ln(x)$$
it gives $0^0=+infty$.
answered 1 hour ago
hamam_Abdallah
35.2k21532
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Certain types of definitions can lead to contradictions. It is up to the person proposing the definition to show that this can't happen. Look up well-defined.
answered 1 hour ago
Robert Israel
310k23202447
add a comment |Â
up vote
1
down vote
Certain types of definitions can lead to contradictions. It is up to the person proposing the definition to show that this can't happen. Look up well-defined.
answered 1 hour ago
Robert Israel
310k23202447
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Certain types of definitions can lead to contradictions. It is up to the person proposing the definition to show that this can't happen. Look up well-defined.
answered 1 hour ago
Robert Israel
310k23202447
Certain types of definitions can lead to contradictions. It is up to the person proposing the definition to show that this can't happen. Look up well-defined.
answered 1 hour ago
Robert Israel
310k23202447
answered 1 hour ago
Robert Israel
310k23202447
answered 1 hour ago
Robert Israel
310k23202447
answered 1 hour ago
Robert Israel
310k23202447
310k23202447
add a comment |Â
add a comment |Â
up vote
1
down vote
It depends what you mean by a definition. A definition such as
$$
P mathrel:= i in BbbN mid mbox$i$ is prime
$$
where the defining property of $P$ simply states that it is equal to something that already exists is completely unproblematic. But in addition to explicit definitions like this, we often use various kinds of implicit definition. A recursive definition such as:
$$
beginalign*
0! &= 1 \
(n+1)! & = n! times (n + 1)
endalign*
$$
is a fairly explicit form of implicit definition that can be shown to be consistent using the recursion theorem as discussed Marcel's answer.
A recursive "definition" (of a function $f : BbbN to BbbN$) such as
$$
beginalign*
f(0) &= 1 \
f(n+1) & = f(n+2)
endalign*
$$
would be considered undefined in programming languages, but from a mathematical perspective it is just a loose definition (i.e., an implicit definition that does not uniquely determine the defined object): it says that $f(0) = 1$ and $f(i) = f(j)$ for $i, j > 0$ and this system of equations has many solutions. If we had instead:
$$
beginalign*
f(0) &= 1 \
f(n+1) & = 2(f(n+2)+1)
endalign*
$$
then this "recursive" definition would be inconsistent, i.e., there are no solutions and asserting the existence of such an $f$ would lead to a contradiction (do you see why?).
Another form of implicit definition would be to define the function $exp : BbbR to BbbR$ as the solution of the following system of equations:
$$
beginalign*
exp(0) &= 1 \
exp'(x) &= exp(x)
endalign*
$$
Here an explicit construction of a witness (say using a power series) shows that the definition is consistent (and a simple argument using properties of the derivative shows that the definition characterises $exp$, i.e., the equations have only one solution).
In general, there are many forms of implicit definition. To show that it is consistent to take a property $phi(X)$ to be the defining property of a new object $X$, you have to prove $exists xphi(x)$. You will probably also want to prove that the existence is unique, i.e., $forall xforall y(phi(x) land phi(y) Rightarrow x = y)$, unless you really did intend your definition to be loose. If you are lucky, then something like the recursion theorem or the theory of differential equations will help with these proofs.
answered 28 mins ago
Rob Arthan
28k42865
add a comment |Â
up vote
1
down vote
It depends what you mean by a definition. A definition such as
$$
P mathrel:= i in BbbN mid mbox$i$ is prime
$$
where the defining property of $P$ simply states that it is equal to something that already exists is completely unproblematic. But in addition to explicit definitions like this, we often use various kinds of implicit definition. A recursive definition such as:
$$
beginalign*
0! &= 1 \
(n+1)! & = n! times (n + 1)
endalign*
$$
is a fairly explicit form of implicit definition that can be shown to be consistent using the recursion theorem as discussed Marcel's answer.
A recursive "definition" (of a function $f : BbbN to BbbN$) such as
$$
beginalign*
f(0) &= 1 \
f(n+1) & = f(n+2)
endalign*
$$
would be considered undefined in programming languages, but from a mathematical perspective it is just a loose definition (i.e., an implicit definition that does not uniquely determine the defined object): it says that $f(0) = 1$ and $f(i) = f(j)$ for $i, j > 0$ and this system of equations has many solutions. If we had instead:
$$
beginalign*
f(0) &= 1 \
f(n+1) & = 2(f(n+2)+1)
endalign*
$$
then this "recursive" definition would be inconsistent, i.e., there are no solutions and asserting the existence of such an $f$ would lead to a contradiction (do you see why?).
Another form of implicit definition would be to define the function $exp : BbbR to BbbR$ as the solution of the following system of equations:
$$
beginalign*
exp(0) &= 1 \
exp'(x) &= exp(x)
endalign*
$$
Here an explicit construction of a witness (say using a power series) shows that the definition is consistent (and a simple argument using properties of the derivative shows that the definition characterises $exp$, i.e., the equations have only one solution).
In general, there are many forms of implicit definition. To show that it is consistent to take a property $phi(X)$ to be the defining property of a new object $X$, you have to prove $exists xphi(x)$. You will probably also want to prove that the existence is unique, i.e., $forall xforall y(phi(x) land phi(y) Rightarrow x = y)$, unless you really did intend your definition to be loose. If you are lucky, then something like the recursion theorem or the theory of differential equations will help with these proofs.
answered 28 mins ago
Rob Arthan
28k42865
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It depends what you mean by a definition. A definition such as
$$
P mathrel:= i in BbbN mid mbox$i$ is prime
$$
where the defining property of $P$ simply states that it is equal to something that already exists is completely unproblematic. But in addition to explicit definitions like this, we often use various kinds of implicit definition. A recursive definition such as:
$$
beginalign*
0! &= 1 \
(n+1)! & = n! times (n + 1)
endalign*
$$
is a fairly explicit form of implicit definition that can be shown to be consistent using the recursion theorem as discussed Marcel's answer.
A recursive "definition" (of a function $f : BbbN to BbbN$) such as
$$
beginalign*
f(0) &= 1 \
f(n+1) & = f(n+2)
endalign*
$$
would be considered undefined in programming languages, but from a mathematical perspective it is just a loose definition (i.e., an implicit definition that does not uniquely determine the defined object): it says that $f(0) = 1$ and $f(i) = f(j)$ for $i, j > 0$ and this system of equations has many solutions. If we had instead:
$$
beginalign*
f(0) &= 1 \
f(n+1) & = 2(f(n+2)+1)
endalign*
$$
then this "recursive" definition would be inconsistent, i.e., there are no solutions and asserting the existence of such an $f$ would lead to a contradiction (do you see why?).
Another form of implicit definition would be to define the function $exp : BbbR to BbbR$ as the solution of the following system of equations:
$$
beginalign*
exp(0) &= 1 \
exp'(x) &= exp(x)
endalign*
$$
Here an explicit construction of a witness (say using a power series) shows that the definition is consistent (and a simple argument using properties of the derivative shows that the definition characterises $exp$, i.e., the equations have only one solution).
In general, there are many forms of implicit definition. To show that it is consistent to take a property $phi(X)$ to be the defining property of a new object $X$, you have to prove $exists xphi(x)$. You will probably also want to prove that the existence is unique, i.e., $forall xforall y(phi(x) land phi(y) Rightarrow x = y)$, unless you really did intend your definition to be loose. If you are lucky, then something like the recursion theorem or the theory of differential equations will help with these proofs.
answered 28 mins ago
Rob Arthan
28k42865
It depends what you mean by a definition. A definition such as
$$
P mathrel:= i in BbbN mid mbox$i$ is prime
$$
where the defining property of $P$ simply states that it is equal to something that already exists is completely unproblematic. But in addition to explicit definitions like this, we often use various kinds of implicit definition. A recursive definition such as:
$$
beginalign*
0! &= 1 \
(n+1)! & = n! times (n + 1)
endalign*
$$
is a fairly explicit form of implicit definition that can be shown to be consistent using the recursion theorem as discussed Marcel's answer.
A recursive "definition" (of a function $f : BbbN to BbbN$) such as
$$
beginalign*
f(0) &= 1 \
f(n+1) & = f(n+2)
endalign*
$$
would be considered undefined in programming languages, but from a mathematical perspective it is just a loose definition (i.e., an implicit definition that does not uniquely determine the defined object): it says that $f(0) = 1$ and $f(i) = f(j)$ for $i, j > 0$ and this system of equations has many solutions. If we had instead:
$$
beginalign*
f(0) &= 1 \
f(n+1) & = 2(f(n+2)+1)
endalign*
$$
then this "recursive" definition would be inconsistent, i.e., there are no solutions and asserting the existence of such an $f$ would lead to a contradiction (do you see why?).
Another form of implicit definition would be to define the function $exp : BbbR to BbbR$ as the solution of the following system of equations:
$$
beginalign*
exp(0) &= 1 \
exp'(x) &= exp(x)
endalign*
$$
Here an explicit construction of a witness (say using a power series) shows that the definition is consistent (and a simple argument using properties of the derivative shows that the definition characterises $exp$, i.e., the equations have only one solution).
In general, there are many forms of implicit definition. To show that it is consistent to take a property $phi(X)$ to be the defining property of a new object $X$, you have to prove $exists xphi(x)$. You will probably also want to prove that the existence is unique, i.e., $forall xforall y(phi(x) land phi(y) Rightarrow x = y)$, unless you really did intend your definition to be loose. If you are lucky, then something like the recursion theorem or the theory of differential equations will help with these proofs.
answered 28 mins ago
Rob Arthan
28k42865
answered 28 mins ago
Rob Arthan
28k42865
answered 28 mins ago
Rob Arthan
28k42865
answered 28 mins ago
Rob Arthan
28k42865
28k42865
add a comment |Â
add a comment |Â
up vote
0
down vote
A recursively defined function like the factorial function is well-defined by the recursion theorem, which can be proved by induction:
https://math.stackexchange.com/a/46760/29892
Henkin has a great article on recursion available here:
https://www.jstor.org/stable/2308975
answered 1 hour ago
Marcel Besixdouze
1,112717
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
add a comment |Â
up vote
0
down vote
A recursively defined function like the factorial function is well-defined by the recursion theorem, which can be proved by induction:
https://math.stackexchange.com/a/46760/29892
Henkin has a great article on recursion available here:
https://www.jstor.org/stable/2308975
answered 1 hour ago
Marcel Besixdouze
1,112717
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A recursively defined function like the factorial function is well-defined by the recursion theorem, which can be proved by induction:
https://math.stackexchange.com/a/46760/29892
Henkin has a great article on recursion available here:
https://www.jstor.org/stable/2308975
answered 1 hour ago
Marcel Besixdouze
1,112717
A recursively defined function like the factorial function is well-defined by the recursion theorem, which can be proved by induction:
https://math.stackexchange.com/a/46760/29892
Henkin has a great article on recursion available here:
https://www.jstor.org/stable/2308975
answered 1 hour ago
Marcel Besixdouze
1,112717
answered 1 hour ago
Marcel Besixdouze
1,112717
answered 1 hour ago
Marcel Besixdouze
1,112717
answered 1 hour ago
Marcel Besixdouze
1,112717
1,112717
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
add a comment |Â
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
Here's an excellent video from Computerphile for a different perspective: youtube.com/watch?v=Mv9NEXX1VHc
– Marcel Besixdouze
1 hour ago
add a comment |Â
up vote
0
down vote
A definition like that can't lead to a contradiction since it is just a definition: the meaning you've assigned to the symbols "$0!$".
A definition might lead to a problem with some rules of arithmetic you like. For example, if you chose to define $x^0 = 0$ instead of $1$ then the identlty
$$
x^a+b= x^ax^b
$$
would fail sometimes. So it would be a bad definition, but not a contradiction.
All the usual suggestions for how to define division by $0$ break some identity in ordinary arithmetic, which is why we don't use any of them.
Defining $0! = 1$ preserves the identity
$$
n! = n(n-1)!
$$
No other definition would do that.
answered 1 hour ago
Ethan Bolker
37.8k543100
add a comment |Â
up vote
0
down vote
A definition like that can't lead to a contradiction since it is just a definition: the meaning you've assigned to the symbols "$0!$".
A definition might lead to a problem with some rules of arithmetic you like. For example, if you chose to define $x^0 = 0$ instead of $1$ then the identlty
$$
x^a+b= x^ax^b
$$
would fail sometimes. So it would be a bad definition, but not a contradiction.
All the usual suggestions for how to define division by $0$ break some identity in ordinary arithmetic, which is why we don't use any of them.
Defining $0! = 1$ preserves the identity
$$
n! = n(n-1)!
$$
No other definition would do that.
answered 1 hour ago
Ethan Bolker
37.8k543100
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A definition like that can't lead to a contradiction since it is just a definition: the meaning you've assigned to the symbols "$0!$".
A definition might lead to a problem with some rules of arithmetic you like. For example, if you chose to define $x^0 = 0$ instead of $1$ then the identlty
$$
x^a+b= x^ax^b
$$
would fail sometimes. So it would be a bad definition, but not a contradiction.
All the usual suggestions for how to define division by $0$ break some identity in ordinary arithmetic, which is why we don't use any of them.
Defining $0! = 1$ preserves the identity
$$
n! = n(n-1)!
$$
No other definition would do that.
answered 1 hour ago
Ethan Bolker
37.8k543100
A definition like that can't lead to a contradiction since it is just a definition: the meaning you've assigned to the symbols "$0!$".
A definition might lead to a problem with some rules of arithmetic you like. For example, if you chose to define $x^0 = 0$ instead of $1$ then the identlty
$$
x^a+b= x^ax^b
$$
would fail sometimes. So it would be a bad definition, but not a contradiction.
All the usual suggestions for how to define division by $0$ break some identity in ordinary arithmetic, which is why we don't use any of them.
Defining $0! = 1$ preserves the identity
$$
n! = n(n-1)!
$$
No other definition would do that.
answered 1 hour ago
Ethan Bolker
37.8k543100
answered 1 hour ago
Ethan Bolker
37.8k543100
answered 1 hour ago
Ethan Bolker
37.8k543100
answered 1 hour ago
Ethan Bolker
37.8k543100
37.8k543100
add a comment |Â
add a comment |Â
up vote
0
down vote
About $0^0$,
A polynomial, in general is written as
$$a_0+a_1x+...a_nx^n=sum_i=0^na_ix^i$$
To be consistant, one will put $0^0=1$, to get the constant term $a_0$.
but
If you compute $$lim_xto+infty(e^-x)^frac-1ln(x)$$
it gives $0^0=+infty$.
answered 1 hour ago
hamam_Abdallah
35.2k21532
add a comment |Â
up vote
0
down vote
About $0^0$,
A polynomial, in general is written as
$$a_0+a_1x+...a_nx^n=sum_i=0^na_ix^i$$
To be consistant, one will put $0^0=1$, to get the constant term $a_0$.
but
If you compute $$lim_xto+infty(e^-x)^frac-1ln(x)$$
it gives $0^0=+infty$.
answered 1 hour ago
hamam_Abdallah
35.2k21532
add a comment |Â
up vote
0
down vote
up vote
0
down vote
About $0^0$,
A polynomial, in general is written as
$$a_0+a_1x+...a_nx^n=sum_i=0^na_ix^i$$
To be consistant, one will put $0^0=1$, to get the constant term $a_0$.
but
If you compute $$lim_xto+infty(e^-x)^frac-1ln(x)$$
it gives $0^0=+infty$.
answered 1 hour ago
hamam_Abdallah
35.2k21532
About $0^0$,
A polynomial, in general is written as
$$a_0+a_1x+...a_nx^n=sum_i=0^na_ix^i$$
To be consistant, one will put $0^0=1$, to get the constant term $a_0$.
but
If you compute $$lim_xto+infty(e^-x)^frac-1ln(x)$$
it gives $0^0=+infty$.
answered 1 hour ago
hamam_Abdallah
35.2k21532
answered 1 hour ago
hamam_Abdallah
35.2k21532
answered 1 hour ago
hamam_Abdallah
35.2k21532
answered 1 hour ago
hamam_Abdallah
35.2k21532
35.2k21532
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1
Formally speaking, a definition cannot be inconsistent. What it might do is make certain universal theorems we have proved about the operation before we extended it no longer true.
– spaceisdarkgreen
1 hour ago
$0 ne 1$ is not a definition; it is a theorem.
– Mauro ALLEGRANZA
1 hour ago
@Mauro that’s “zero factorial equals oneâ€Â
– spaceisdarkgreen
20 mins ago