Proving by Contrapositive (specific question within)
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I'm having issues coming up with a contrapositive proof for the following question.
As far as I know, a proof by contraposition is based on the following :
$overline Q to overline P equiv P to Q$
or that's where I'm mistaken?
The question is:
X,Y,Z are natural numbers.
if $X^3+Y^3=Z^3$ , then at least one of them is divisible by 3.
Provide a proof by contrapositive.
I've tried to substitude X,Y,Z with (3A-1),(3B-1),(3C-1) , essentially trying to get to a point where:
$overline Q$ : none is divisble by 3 (3N-1).
$overline P$ : $X^3+Y^3neq Z^3$
$overline Q$ -> $overline P$ $equiv$ P -> Q
I tried to simplify the equation but it doesn't look like I got anywhere, where am I being wrong?
Thanks for helping!
discrete-mathematics logic proof-explanation
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
asked Nov 15 at 14:00
Idan Botbol
6
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Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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show 9 more comments
up vote
0
down vote
favorite
I'm having issues coming up with a contrapositive proof for the following question.
As far as I know, a proof by contraposition is based on the following :
$overline Q to overline P equiv P to Q$
or that's where I'm mistaken?
The question is:
X,Y,Z are natural numbers.
if $X^3+Y^3=Z^3$ , then at least one of them is divisible by 3.
Provide a proof by contrapositive.
I've tried to substitude X,Y,Z with (3A-1),(3B-1),(3C-1) , essentially trying to get to a point where:
$overline Q$ : none is divisble by 3 (3N-1).
$overline P$ : $X^3+Y^3neq Z^3$
$overline Q$ -> $overline P$ $equiv$ P -> Q
I tried to simplify the equation but it doesn't look like I got anywhere, where am I being wrong?
Thanks for helping!
discrete-mathematics logic proof-explanation
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
asked Nov 15 at 14:00
Idan Botbol
6
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
You assumed that a number not divisible by $3$ must be of the form $3N-1$. What about $4$ ?
– Mauro ALLEGRANZA
Nov 15 at 14:07
@MauroALLEGRANZA, I should have written it specifically, my professor gave me that idea, what other route can I go to solve this? Thanks for the quick reply.
– Idan Botbol
Nov 15 at 14:10
I think if you consider x,y,z individually in form that is not divisible by 3 i.e., 3n-1 and 3n-2, you would have total 27 permutations of these numbers and you would have 27 cases to prove. So I think you should think of a different approach.
– Jimmy
Nov 15 at 14:16
Have you learned about the modulo notation in your course?
– F.Carette
Nov 15 at 14:18
1
Maybe this will help. You can write $X^3=Xmod(3)$, whenever $X$ is not divisible by 3.
– Jimmy
Nov 15 at 15:13
|
show 9 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having issues coming up with a contrapositive proof for the following question.
As far as I know, a proof by contraposition is based on the following :
$overline Q to overline P equiv P to Q$
or that's where I'm mistaken?
The question is:
X,Y,Z are natural numbers.
if $X^3+Y^3=Z^3$ , then at least one of them is divisible by 3.
Provide a proof by contrapositive.
I've tried to substitude X,Y,Z with (3A-1),(3B-1),(3C-1) , essentially trying to get to a point where:
$overline Q$ : none is divisble by 3 (3N-1).
$overline P$ : $X^3+Y^3neq Z^3$
$overline Q$ -> $overline P$ $equiv$ P -> Q
I tried to simplify the equation but it doesn't look like I got anywhere, where am I being wrong?
Thanks for helping!
discrete-mathematics logic proof-explanation
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
asked Nov 15 at 14:00
Idan Botbol
6
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm having issues coming up with a contrapositive proof for the following question.
As far as I know, a proof by contraposition is based on the following :
$overline Q to overline P equiv P to Q$
or that's where I'm mistaken?
The question is:
X,Y,Z are natural numbers.
if $X^3+Y^3=Z^3$ , then at least one of them is divisible by 3.
Provide a proof by contrapositive.
I've tried to substitude X,Y,Z with (3A-1),(3B-1),(3C-1) , essentially trying to get to a point where:
$overline Q$ : none is divisble by 3 (3N-1).
$overline P$ : $X^3+Y^3neq Z^3$
$overline Q$ -> $overline P$ $equiv$ P -> Q
I tried to simplify the equation but it doesn't look like I got anywhere, where am I being wrong?
Thanks for helping!
discrete-mathematics logic proof-explanation
discrete-mathematics logic proof-explanation
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
asked Nov 15 at 14:00
Idan Botbol
6
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
asked Nov 15 at 14:00
Idan Botbol
6
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.4k448110
63.4k448110
asked Nov 15 at 14:00
Idan Botbol
6
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 14:00
Idan Botbol
6
asked Nov 15 at 14:00
Idan Botbol
6
6
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Idan Botbol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
You assumed that a number not divisible by $3$ must be of the form $3N-1$. What about $4$ ?
– Mauro ALLEGRANZA
Nov 15 at 14:07
@MauroALLEGRANZA, I should have written it specifically, my professor gave me that idea, what other route can I go to solve this? Thanks for the quick reply.
– Idan Botbol
Nov 15 at 14:10
I think if you consider x,y,z individually in form that is not divisible by 3 i.e., 3n-1 and 3n-2, you would have total 27 permutations of these numbers and you would have 27 cases to prove. So I think you should think of a different approach.
– Jimmy
Nov 15 at 14:16
Have you learned about the modulo notation in your course?
– F.Carette
Nov 15 at 14:18
1
Maybe this will help. You can write $X^3=Xmod(3)$, whenever $X$ is not divisible by 3.
– Jimmy
Nov 15 at 15:13
|
show 9 more comments
1
You assumed that a number not divisible by $3$ must be of the form $3N-1$. What about $4$ ?
– Mauro ALLEGRANZA
Nov 15 at 14:07
@MauroALLEGRANZA, I should have written it specifically, my professor gave me that idea, what other route can I go to solve this? Thanks for the quick reply.
– Idan Botbol
Nov 15 at 14:10
I think if you consider x,y,z individually in form that is not divisible by 3 i.e., 3n-1 and 3n-2, you would have total 27 permutations of these numbers and you would have 27 cases to prove. So I think you should think of a different approach.
– Jimmy
Nov 15 at 14:16
Have you learned about the modulo notation in your course?
– F.Carette
Nov 15 at 14:18
1
Maybe this will help. You can write $X^3=Xmod(3)$, whenever $X$ is not divisible by 3.
– Jimmy
Nov 15 at 15:13
1
1
You assumed that a number not divisible by $3$ must be of the form $3N-1$. What about $4$ ?
– Mauro ALLEGRANZA
Nov 15 at 14:07
You assumed that a number not divisible by $3$ must be of the form $3N-1$. What about $4$ ?
– Mauro ALLEGRANZA
Nov 15 at 14:07
@MauroALLEGRANZA, I should have written it specifically, my professor gave me that idea, what other route can I go to solve this? Thanks for the quick reply.
– Idan Botbol
Nov 15 at 14:10
@MauroALLEGRANZA, I should have written it specifically, my professor gave me that idea, what other route can I go to solve this? Thanks for the quick reply.
– Idan Botbol
Nov 15 at 14:10
I think if you consider x,y,z individually in form that is not divisible by 3 i.e., 3n-1 and 3n-2, you would have total 27 permutations of these numbers and you would have 27 cases to prove. So I think you should think of a different approach.
– Jimmy
Nov 15 at 14:16
I think if you consider x,y,z individually in form that is not divisible by 3 i.e., 3n-1 and 3n-2, you would have total 27 permutations of these numbers and you would have 27 cases to prove. So I think you should think of a different approach.
– Jimmy
Nov 15 at 14:16
Have you learned about the modulo notation in your course?
– F.Carette
Nov 15 at 14:18
Have you learned about the modulo notation in your course?
– F.Carette
Nov 15 at 14:18
1
1
Maybe this will help. You can write $X^3=Xmod(3)$, whenever $X$ is not divisible by 3.
– Jimmy
Nov 15 at 15:13
Maybe this will help. You can write $X^3=Xmod(3)$, whenever $X$ is not divisible by 3.
– Jimmy
Nov 15 at 15:13
|
show 9 more comments
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Idan Botbol is a new contributor. Be nice, and check out our Code of Conduct.
Idan Botbol is a new contributor. Be nice, and check out our Code of Conduct.
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1
You assumed that a number not divisible by $3$ must be of the form $3N-1$. What about $4$ ?
– Mauro ALLEGRANZA
Nov 15 at 14:07
@MauroALLEGRANZA, I should have written it specifically, my professor gave me that idea, what other route can I go to solve this? Thanks for the quick reply.
– Idan Botbol
Nov 15 at 14:10
I think if you consider x,y,z individually in form that is not divisible by 3 i.e., 3n-1 and 3n-2, you would have total 27 permutations of these numbers and you would have 27 cases to prove. So I think you should think of a different approach.
– Jimmy
Nov 15 at 14:16
Have you learned about the modulo notation in your course?
– F.Carette
Nov 15 at 14:18
1
Maybe this will help. You can write $X^3=Xmod(3)$, whenever $X$ is not divisible by 3.
– Jimmy
Nov 15 at 15:13