Proving by Contrapositive (specific question within)
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
63.9k448110
asked Nov 15 at 14:00
Idan Botbol
6
|
show 10 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
63.9k448110
asked Nov 15 at 14:00
Idan Botbol
6
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 10 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
63.9k448110
asked Nov 15 at 14:00
Idan Botbol
6
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
63.9k448110
asked Nov 15 at 14:00
Idan Botbol
6
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.9k448110
asked Nov 15 at 14:00
Idan Botbol
6
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.9k448110
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.9k448110
edited Nov 15 at 14:04
Mauro ALLEGRANZA
63.9k448110
63.9k448110
asked Nov 15 at 14:00
Idan Botbol
6
asked Nov 15 at 14:00
Idan Botbol
6
asked Nov 15 at 14:00
Idan Botbol
6
6
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 10 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 10 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
Following the comments, if we have $X, Y$ and $Z$ none of which are divisible by 3 and such that $X^3 + Y^3 = Z^3$, we may assume that $X,Y equiv 1$ (mod 3) and $Z equiv 2$ (mod 3). So, we may assume that there are $a$ $b$ and $c$ such that $X = 3a+1$, $Y = 3b+1$ and $Z = 3c+2$. By applying the binomial expansion to $X^3$ and $Y^3$, it turns out that $X^3 equiv 1$ (mod 9) and $Y^3 equiv 1$ (mod 9), so that $X^3 + Y^3 equiv 2$ (mod 9). But if you apply the binomial expansion to $Z^3 = (3c+2)^3$, it turns out that $Z^3 equiv 8$ (mod 9). Thus, the equality $X^3 + Y^3 = Z^3$ cannot hold.
answered Nov 23 at 17:51
Tony Dean
512
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999739%2fproving-by-contrapositive-specific-question-within%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Following the comments, if we have $X, Y$ and $Z$ none of which are divisible by 3 and such that $X^3 + Y^3 = Z^3$, we may assume that $X,Y equiv 1$ (mod 3) and $Z equiv 2$ (mod 3). So, we may assume that there are $a$ $b$ and $c$ such that $X = 3a+1$, $Y = 3b+1$ and $Z = 3c+2$. By applying the binomial expansion to $X^3$ and $Y^3$, it turns out that $X^3 equiv 1$ (mod 9) and $Y^3 equiv 1$ (mod 9), so that $X^3 + Y^3 equiv 2$ (mod 9). But if you apply the binomial expansion to $Z^3 = (3c+2)^3$, it turns out that $Z^3 equiv 8$ (mod 9). Thus, the equality $X^3 + Y^3 = Z^3$ cannot hold.
answered Nov 23 at 17:51
Tony Dean
512
add a comment |
up vote
0
down vote
Following the comments, if we have $X, Y$ and $Z$ none of which are divisible by 3 and such that $X^3 + Y^3 = Z^3$, we may assume that $X,Y equiv 1$ (mod 3) and $Z equiv 2$ (mod 3). So, we may assume that there are $a$ $b$ and $c$ such that $X = 3a+1$, $Y = 3b+1$ and $Z = 3c+2$. By applying the binomial expansion to $X^3$ and $Y^3$, it turns out that $X^3 equiv 1$ (mod 9) and $Y^3 equiv 1$ (mod 9), so that $X^3 + Y^3 equiv 2$ (mod 9). But if you apply the binomial expansion to $Z^3 = (3c+2)^3$, it turns out that $Z^3 equiv 8$ (mod 9). Thus, the equality $X^3 + Y^3 = Z^3$ cannot hold.
answered Nov 23 at 17:51
Tony Dean
512
add a comment |
up vote
0
down vote
up vote
0
down vote
Following the comments, if we have $X, Y$ and $Z$ none of which are divisible by 3 and such that $X^3 + Y^3 = Z^3$, we may assume that $X,Y equiv 1$ (mod 3) and $Z equiv 2$ (mod 3). So, we may assume that there are $a$ $b$ and $c$ such that $X = 3a+1$, $Y = 3b+1$ and $Z = 3c+2$. By applying the binomial expansion to $X^3$ and $Y^3$, it turns out that $X^3 equiv 1$ (mod 9) and $Y^3 equiv 1$ (mod 9), so that $X^3 + Y^3 equiv 2$ (mod 9). But if you apply the binomial expansion to $Z^3 = (3c+2)^3$, it turns out that $Z^3 equiv 8$ (mod 9). Thus, the equality $X^3 + Y^3 = Z^3$ cannot hold.
answered Nov 23 at 17:51
Tony Dean
512
Following the comments, if we have $X, Y$ and $Z$ none of which are divisible by 3 and such that $X^3 + Y^3 = Z^3$, we may assume that $X,Y equiv 1$ (mod 3) and $Z equiv 2$ (mod 3). So, we may assume that there are $a$ $b$ and $c$ such that $X = 3a+1$, $Y = 3b+1$ and $Z = 3c+2$. By applying the binomial expansion to $X^3$ and $Y^3$, it turns out that $X^3 equiv 1$ (mod 9) and $Y^3 equiv 1$ (mod 9), so that $X^3 + Y^3 equiv 2$ (mod 9). But if you apply the binomial expansion to $Z^3 = (3c+2)^3$, it turns out that $Z^3 equiv 8$ (mod 9). Thus, the equality $X^3 + Y^3 = Z^3$ cannot hold.
answered Nov 23 at 17:51
Tony Dean
512
answered Nov 23 at 17:51
Tony Dean
512
answered Nov 23 at 17:51
Tony Dean
512
answered Nov 23 at 17:51
Tony Dean
512
512
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999739%2fproving-by-contrapositive-specific-question-within%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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