Question on exclusive or vs inclusive or
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1
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Here's what my textbook asks me to prove:
From algebra, recall that, if a prime $p$ divides a product $ab$, then
$p$ must divide either $a$ or $b$. Use this to prove that $sqrt{p}$
is irrational for any prime $p$.
I know the proof follows exactly like the proof of irrationality of $sqrt{2}$. But my question is, what if $b=a$? Then the statement would follow like "if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" which is false.
I wonder if I am doing some mistake here because the following video https://youtu.be/uQ6KSt94jVY seems to prove the same statement. I think it should be an "inclusive or" there.
logic
asked Nov 22 at 10:26
Ashish K
775513
add a comment |
up vote
1
down vote
favorite
Here's what my textbook asks me to prove:
From algebra, recall that, if a prime $p$ divides a product $ab$, then
$p$ must divide either $a$ or $b$. Use this to prove that $sqrt{p}$
is irrational for any prime $p$.
I know the proof follows exactly like the proof of irrationality of $sqrt{2}$. But my question is, what if $b=a$? Then the statement would follow like "if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" which is false.
I wonder if I am doing some mistake here because the following video https://youtu.be/uQ6KSt94jVY seems to prove the same statement. I think it should be an "inclusive or" there.
logic
asked Nov 22 at 10:26
Ashish K
775513
How is the statement false? $ p | a lor p | a equiv p |a $. And $p | a^2 equiv p |a$ (for p prime, of course)
– F.Carette
Nov 22 at 10:31
@F.Carette it's an exclusive or.. "exclusive or" is true if exactly one of the statements are true. In this particular case, both would be true.
– Ashish K
Nov 22 at 10:34
I see no reason to think that it's an exclusive or.
– Andreas Blass
Nov 22 at 17:37
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's what my textbook asks me to prove:
From algebra, recall that, if a prime $p$ divides a product $ab$, then
$p$ must divide either $a$ or $b$. Use this to prove that $sqrt{p}$
is irrational for any prime $p$.
I know the proof follows exactly like the proof of irrationality of $sqrt{2}$. But my question is, what if $b=a$? Then the statement would follow like "if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" which is false.
I wonder if I am doing some mistake here because the following video https://youtu.be/uQ6KSt94jVY seems to prove the same statement. I think it should be an "inclusive or" there.
logic
asked Nov 22 at 10:26
Ashish K
775513
Here's what my textbook asks me to prove:
From algebra, recall that, if a prime $p$ divides a product $ab$, then
$p$ must divide either $a$ or $b$. Use this to prove that $sqrt{p}$
is irrational for any prime $p$.
I know the proof follows exactly like the proof of irrationality of $sqrt{2}$. But my question is, what if $b=a$? Then the statement would follow like "if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" which is false.
I wonder if I am doing some mistake here because the following video https://youtu.be/uQ6KSt94jVY seems to prove the same statement. I think it should be an "inclusive or" there.
logic
logic
asked Nov 22 at 10:26
Ashish K
775513
asked Nov 22 at 10:26
Ashish K
775513
asked Nov 22 at 10:26
Ashish K
775513
asked Nov 22 at 10:26
Ashish K
775513
asked Nov 22 at 10:26
Ashish K
775513
775513
How is the statement false? $ p | a lor p | a equiv p |a $. And $p | a^2 equiv p |a$ (for p prime, of course)
– F.Carette
Nov 22 at 10:31
@F.Carette it's an exclusive or.. "exclusive or" is true if exactly one of the statements are true. In this particular case, both would be true.
– Ashish K
Nov 22 at 10:34
I see no reason to think that it's an exclusive or.
– Andreas Blass
Nov 22 at 17:37
add a comment |
How is the statement false? $ p | a lor p | a equiv p |a $. And $p | a^2 equiv p |a$ (for p prime, of course)
– F.Carette
Nov 22 at 10:31
@F.Carette it's an exclusive or.. "exclusive or" is true if exactly one of the statements are true. In this particular case, both would be true.
– Ashish K
Nov 22 at 10:34
I see no reason to think that it's an exclusive or.
– Andreas Blass
Nov 22 at 17:37
How is the statement false? $ p | a lor p | a equiv p |a $. And $p | a^2 equiv p |a$ (for p prime, of course)
– F.Carette
Nov 22 at 10:31
How is the statement false? $ p | a lor p | a equiv p |a $. And $p | a^2 equiv p |a$ (for p prime, of course)
– F.Carette
Nov 22 at 10:31
@F.Carette it's an exclusive or.. "exclusive or" is true if exactly one of the statements are true. In this particular case, both would be true.
– Ashish K
Nov 22 at 10:34
@F.Carette it's an exclusive or.. "exclusive or" is true if exactly one of the statements are true. In this particular case, both would be true.
– Ashish K
Nov 22 at 10:34
I see no reason to think that it's an exclusive or.
– Andreas Blass
Nov 22 at 17:37
I see no reason to think that it's an exclusive or.
– Andreas Blass
Nov 22 at 17:37
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Usually, in mathematical and logical contexts, "or" has to be interpreted as an inclusive disjunction, unless it is indicated otherwise.
Note that $3$ (a prime number) divides not only $27 cdot 9$, but also $27$ and $9$. Hence, it is clear that the correct interpretation of the "or" in the statement "if a prime $p$ divides a product $ab$, then $p$ must divide either $a$ or $b$" is the "inclusive or", even when you consider the case where $a neq b$.
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
add a comment |
up vote
0
down vote
"if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" is correct if and only if the or is an inclusive "or". Hence you are correct.
answered Nov 22 at 10:32
Vee Hua Zhi
776124
add a comment |
up vote
0
down vote
The statement " if $p$ divides $a^2$ then $p$ divides $a$ or $p$ divides $a$ "
is not problematic.
for example $7$ divides $35^2$ so $7$ divides $35$ or $7$ divides $35$ is a true statement. This OR is inclusive and as you know $Plor P equiv P$.
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Usually, in mathematical and logical contexts, "or" has to be interpreted as an inclusive disjunction, unless it is indicated otherwise.
Note that $3$ (a prime number) divides not only $27 cdot 9$, but also $27$ and $9$. Hence, it is clear that the correct interpretation of the "or" in the statement "if a prime $p$ divides a product $ab$, then $p$ must divide either $a$ or $b$" is the "inclusive or", even when you consider the case where $a neq b$.
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
add a comment |
up vote
0
down vote
accepted
Usually, in mathematical and logical contexts, "or" has to be interpreted as an inclusive disjunction, unless it is indicated otherwise.
Note that $3$ (a prime number) divides not only $27 cdot 9$, but also $27$ and $9$. Hence, it is clear that the correct interpretation of the "or" in the statement "if a prime $p$ divides a product $ab$, then $p$ must divide either $a$ or $b$" is the "inclusive or", even when you consider the case where $a neq b$.
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Usually, in mathematical and logical contexts, "or" has to be interpreted as an inclusive disjunction, unless it is indicated otherwise.
Note that $3$ (a prime number) divides not only $27 cdot 9$, but also $27$ and $9$. Hence, it is clear that the correct interpretation of the "or" in the statement "if a prime $p$ divides a product $ab$, then $p$ must divide either $a$ or $b$" is the "inclusive or", even when you consider the case where $a neq b$.
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
Usually, in mathematical and logical contexts, "or" has to be interpreted as an inclusive disjunction, unless it is indicated otherwise.
Note that $3$ (a prime number) divides not only $27 cdot 9$, but also $27$ and $9$. Hence, it is clear that the correct interpretation of the "or" in the statement "if a prime $p$ divides a product $ab$, then $p$ must divide either $a$ or $b$" is the "inclusive or", even when you consider the case where $a neq b$.
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
answered Nov 22 at 10:38
Taroccoesbrocco
4,92761838
4,92761838
add a comment |
add a comment |
up vote
0
down vote
"if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" is correct if and only if the or is an inclusive "or". Hence you are correct.
answered Nov 22 at 10:32
Vee Hua Zhi
776124
add a comment |
up vote
0
down vote
"if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" is correct if and only if the or is an inclusive "or". Hence you are correct.
answered Nov 22 at 10:32
Vee Hua Zhi
776124
add a comment |
up vote
0
down vote
up vote
0
down vote
"if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" is correct if and only if the or is an inclusive "or". Hence you are correct.
answered Nov 22 at 10:32
Vee Hua Zhi
776124
"if $p$ divides $a^2$ then either $p$ divides $a$ or $p$ divides $a$" is correct if and only if the or is an inclusive "or". Hence you are correct.
answered Nov 22 at 10:32
Vee Hua Zhi
776124
answered Nov 22 at 10:32
Vee Hua Zhi
776124
answered Nov 22 at 10:32
Vee Hua Zhi
776124
answered Nov 22 at 10:32
Vee Hua Zhi
776124
776124
add a comment |
add a comment |
up vote
0
down vote
The statement " if $p$ divides $a^2$ then $p$ divides $a$ or $p$ divides $a$ "
is not problematic.
for example $7$ divides $35^2$ so $7$ divides $35$ or $7$ divides $35$ is a true statement. This OR is inclusive and as you know $Plor P equiv P$.
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
add a comment |
up vote
0
down vote
The statement " if $p$ divides $a^2$ then $p$ divides $a$ or $p$ divides $a$ "
is not problematic.
for example $7$ divides $35^2$ so $7$ divides $35$ or $7$ divides $35$ is a true statement. This OR is inclusive and as you know $Plor P equiv P$.
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
add a comment |
up vote
0
down vote
up vote
0
down vote
The statement " if $p$ divides $a^2$ then $p$ divides $a$ or $p$ divides $a$ "
is not problematic.
for example $7$ divides $35^2$ so $7$ divides $35$ or $7$ divides $35$ is a true statement. This OR is inclusive and as you know $Plor P equiv P$.
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
The statement " if $p$ divides $a^2$ then $p$ divides $a$ or $p$ divides $a$ "
is not problematic.
for example $7$ divides $35^2$ so $7$ divides $35$ or $7$ divides $35$ is a true statement. This OR is inclusive and as you know $Plor P equiv P$.
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
answered Nov 22 at 10:38
Mohammad Riazi-Kermani
40.4k41958
40.4k41958
add a comment |
add a comment |
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How is the statement false? $ p | a lor p | a equiv p |a $. And $p | a^2 equiv p |a$ (for p prime, of course)
– F.Carette
Nov 22 at 10:31
@F.Carette it's an exclusive or.. "exclusive or" is true if exactly one of the statements are true. In this particular case, both would be true.
– Ashish K
Nov 22 at 10:34
I see no reason to think that it's an exclusive or.
– Andreas Blass
Nov 22 at 17:37