Mixing
Is x*(1/x)=1 propositional logic?
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My book says variables cannot be used in propositional logic. Is this an exception? It can't be false so I know it is a tautology, but my book doesn't categorize tautology into a type of logic.
discrete-mathematics
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
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add a comment |
$begingroup$
My book says variables cannot be used in propositional logic. Is this an exception? It can't be false so I know it is a tautology, but my book doesn't categorize tautology into a type of logic.
discrete-mathematics
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
$endgroup$
$begingroup$
It is not propositional logic and it is not a tautology. It is a formula of arithmetic or algebra expressible in the language of predicate logic : $forall x (x * frac 1 x = 1)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 14 at 8:48
add a comment |
$begingroup$
My book says variables cannot be used in propositional logic. Is this an exception? It can't be false so I know it is a tautology, but my book doesn't categorize tautology into a type of logic.
discrete-mathematics
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
$endgroup$
My book says variables cannot be used in propositional logic. Is this an exception? It can't be false so I know it is a tautology, but my book doesn't categorize tautology into a type of logic.
discrete-mathematics
discrete-mathematics
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
asked Jan 13 at 21:58
newtodiscretenewtodiscrete
31
31
$begingroup$
It is not propositional logic and it is not a tautology. It is a formula of arithmetic or algebra expressible in the language of predicate logic : $forall x (x * frac 1 x = 1)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 14 at 8:48
add a comment |
$begingroup$
It is not propositional logic and it is not a tautology. It is a formula of arithmetic or algebra expressible in the language of predicate logic : $forall x (x * frac 1 x = 1)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 14 at 8:48
$begingroup$
It is not propositional logic and it is not a tautology. It is a formula of arithmetic or algebra expressible in the language of predicate logic : $forall x (x * frac 1 x = 1)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 14 at 8:48
$begingroup$
It is not propositional logic and it is not a tautology. It is a formula of arithmetic or algebra expressible in the language of predicate logic : $forall x (x * frac 1 x = 1)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 14 at 8:48
add a comment |
3 Answers
3
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There are multiple things going on here, so I'm going to unpack this.
$x cdot (frac{1}{x}) = 1$ is not necessarily true. If $x = 0$, then the left-hand side is nonsense.- When developing basic propositional logic, numbers don't yet exist. The numbers are (traditionally) built out of sets or directly out of axioms, but both of these approaches are dependent on logic. We have to define logic first and then define the numbers on top of it.
Of course, once the numbers are defined, you can "go back" and mix basic logic with the "new" machinery of numbers, so that it is perfectly reasonable to write $x < 3 vee x > 7$, for example. But then you're not working in "pure" propositional logic any more, because you've added axioms describing the numbers.
answered Jan 13 at 22:11
KevinKevin
1,650722
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add a comment |
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No, this is not a formula of propositional logic.
In propositional logic, the formulas are always either propositional letters or built from smaller formulas using logical connectives. That doesn't match here because your formula is something different.
(Note that in the jargon of mathematical logic, something like $xcdot(1/x)$ is not a "formula" but a "term". Formulas are just things that can be true or false).
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
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add a comment |
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This is not a propositional logic. It would be a propositional logic if it was:
$$xneq0 Rightarrow xfrac{1}{x} = 1$$
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
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add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are multiple things going on here, so I'm going to unpack this.
$x cdot (frac{1}{x}) = 1$ is not necessarily true. If $x = 0$, then the left-hand side is nonsense.- When developing basic propositional logic, numbers don't yet exist. The numbers are (traditionally) built out of sets or directly out of axioms, but both of these approaches are dependent on logic. We have to define logic first and then define the numbers on top of it.
Of course, once the numbers are defined, you can "go back" and mix basic logic with the "new" machinery of numbers, so that it is perfectly reasonable to write $x < 3 vee x > 7$, for example. But then you're not working in "pure" propositional logic any more, because you've added axioms describing the numbers.
answered Jan 13 at 22:11
KevinKevin
1,650722
$endgroup$
add a comment |
$begingroup$
There are multiple things going on here, so I'm going to unpack this.
$x cdot (frac{1}{x}) = 1$ is not necessarily true. If $x = 0$, then the left-hand side is nonsense.- When developing basic propositional logic, numbers don't yet exist. The numbers are (traditionally) built out of sets or directly out of axioms, but both of these approaches are dependent on logic. We have to define logic first and then define the numbers on top of it.
Of course, once the numbers are defined, you can "go back" and mix basic logic with the "new" machinery of numbers, so that it is perfectly reasonable to write $x < 3 vee x > 7$, for example. But then you're not working in "pure" propositional logic any more, because you've added axioms describing the numbers.
answered Jan 13 at 22:11
KevinKevin
1,650722
$endgroup$
add a comment |
$begingroup$
There are multiple things going on here, so I'm going to unpack this.
$x cdot (frac{1}{x}) = 1$ is not necessarily true. If $x = 0$, then the left-hand side is nonsense.- When developing basic propositional logic, numbers don't yet exist. The numbers are (traditionally) built out of sets or directly out of axioms, but both of these approaches are dependent on logic. We have to define logic first and then define the numbers on top of it.
Of course, once the numbers are defined, you can "go back" and mix basic logic with the "new" machinery of numbers, so that it is perfectly reasonable to write $x < 3 vee x > 7$, for example. But then you're not working in "pure" propositional logic any more, because you've added axioms describing the numbers.
answered Jan 13 at 22:11
KevinKevin
1,650722
$endgroup$
There are multiple things going on here, so I'm going to unpack this.
$x cdot (frac{1}{x}) = 1$ is not necessarily true. If $x = 0$, then the left-hand side is nonsense.- When developing basic propositional logic, numbers don't yet exist. The numbers are (traditionally) built out of sets or directly out of axioms, but both of these approaches are dependent on logic. We have to define logic first and then define the numbers on top of it.
Of course, once the numbers are defined, you can "go back" and mix basic logic with the "new" machinery of numbers, so that it is perfectly reasonable to write $x < 3 vee x > 7$, for example. But then you're not working in "pure" propositional logic any more, because you've added axioms describing the numbers.
answered Jan 13 at 22:11
KevinKevin
1,650722
answered Jan 13 at 22:11
KevinKevin
1,650722
answered Jan 13 at 22:11
KevinKevin
1,650722
answered Jan 13 at 22:11
KevinKevin
1,650722
1,650722
add a comment |
add a comment |
$begingroup$
No, this is not a formula of propositional logic.
In propositional logic, the formulas are always either propositional letters or built from smaller formulas using logical connectives. That doesn't match here because your formula is something different.
(Note that in the jargon of mathematical logic, something like $xcdot(1/x)$ is not a "formula" but a "term". Formulas are just things that can be true or false).
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
$endgroup$
add a comment |
$begingroup$
No, this is not a formula of propositional logic.
In propositional logic, the formulas are always either propositional letters or built from smaller formulas using logical connectives. That doesn't match here because your formula is something different.
(Note that in the jargon of mathematical logic, something like $xcdot(1/x)$ is not a "formula" but a "term". Formulas are just things that can be true or false).
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
$endgroup$
add a comment |
$begingroup$
No, this is not a formula of propositional logic.
In propositional logic, the formulas are always either propositional letters or built from smaller formulas using logical connectives. That doesn't match here because your formula is something different.
(Note that in the jargon of mathematical logic, something like $xcdot(1/x)$ is not a "formula" but a "term". Formulas are just things that can be true or false).
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
$endgroup$
No, this is not a formula of propositional logic.
In propositional logic, the formulas are always either propositional letters or built from smaller formulas using logical connectives. That doesn't match here because your formula is something different.
(Note that in the jargon of mathematical logic, something like $xcdot(1/x)$ is not a "formula" but a "term". Formulas are just things that can be true or false).
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
answered Jan 13 at 22:04
Henning MakholmHenning Makholm
240k17306543
240k17306543
add a comment |
add a comment |
$begingroup$
This is not a propositional logic. It would be a propositional logic if it was:
$$xneq0 Rightarrow xfrac{1}{x} = 1$$
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
$endgroup$
add a comment |
$begingroup$
This is not a propositional logic. It would be a propositional logic if it was:
$$xneq0 Rightarrow xfrac{1}{x} = 1$$
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
$endgroup$
add a comment |
$begingroup$
This is not a propositional logic. It would be a propositional logic if it was:
$$xneq0 Rightarrow xfrac{1}{x} = 1$$
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
$endgroup$
This is not a propositional logic. It would be a propositional logic if it was:
$$xneq0 Rightarrow xfrac{1}{x} = 1$$
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
answered Jan 13 at 22:18
Liuc ScaiLiuc Scai
11
11
add a comment |
add a comment |
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$begingroup$
It is not propositional logic and it is not a tautology. It is a formula of arithmetic or algebra expressible in the language of predicate logic : $forall x (x * frac 1 x = 1)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 14 at 8:48