Proof by contradiction - Getting my head around it
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Hey there Math community!
I have a general question on contradiction and it's getting difficult to get my head around it.
Notes:
I have some background in math and I have read several proofs by contradiction already
For the sake of the argument, let us assume the fundamental theorem of arithmetic
As per the general strategy for the proof, we assume the opposite of something that we wish to prove to begin with.
i.e A number that does NOT have a unique decomposition of primes.
We then proceed by a logical sequence of steps to show that this leads to a contradiction.
**Thus our original assumption was untenable and hence we have proved that all numbers have a unique decomposition of primes.
I have a problem understanding the star marked step.
It's like saying, if we want to prove the man is happy, let us assume the man is unhappy.
A logical sequence of steps leads to a contradiction.
Hence, our initial assumption is flawed, so the man is 'happy'?!?
What ensures that 'NOT unhappy' means 'happy' in the realm of math?
Thank you for your time :)
formal-proofs
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
asked Feb 24 at 6:32
hargun3045hargun3045
7918
$endgroup$
add a comment |
$begingroup$
Hey there Math community!
I have a general question on contradiction and it's getting difficult to get my head around it.
Notes:
I have some background in math and I have read several proofs by contradiction already
For the sake of the argument, let us assume the fundamental theorem of arithmetic
As per the general strategy for the proof, we assume the opposite of something that we wish to prove to begin with.
i.e A number that does NOT have a unique decomposition of primes.
We then proceed by a logical sequence of steps to show that this leads to a contradiction.
**Thus our original assumption was untenable and hence we have proved that all numbers have a unique decomposition of primes.
I have a problem understanding the star marked step.
It's like saying, if we want to prove the man is happy, let us assume the man is unhappy.
A logical sequence of steps leads to a contradiction.
Hence, our initial assumption is flawed, so the man is 'happy'?!?
What ensures that 'NOT unhappy' means 'happy' in the realm of math?
Thank you for your time :)
formal-proofs
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
asked Feb 24 at 6:32
hargun3045hargun3045
7918
$endgroup$
add a comment |
$begingroup$
Hey there Math community!
I have a general question on contradiction and it's getting difficult to get my head around it.
Notes:
I have some background in math and I have read several proofs by contradiction already
For the sake of the argument, let us assume the fundamental theorem of arithmetic
As per the general strategy for the proof, we assume the opposite of something that we wish to prove to begin with.
i.e A number that does NOT have a unique decomposition of primes.
We then proceed by a logical sequence of steps to show that this leads to a contradiction.
**Thus our original assumption was untenable and hence we have proved that all numbers have a unique decomposition of primes.
I have a problem understanding the star marked step.
It's like saying, if we want to prove the man is happy, let us assume the man is unhappy.
A logical sequence of steps leads to a contradiction.
Hence, our initial assumption is flawed, so the man is 'happy'?!?
What ensures that 'NOT unhappy' means 'happy' in the realm of math?
Thank you for your time :)
formal-proofs
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
asked Feb 24 at 6:32
hargun3045hargun3045
7918
$endgroup$
Hey there Math community!
I have a general question on contradiction and it's getting difficult to get my head around it.
Notes:
I have some background in math and I have read several proofs by contradiction already
For the sake of the argument, let us assume the fundamental theorem of arithmetic
As per the general strategy for the proof, we assume the opposite of something that we wish to prove to begin with.
i.e A number that does NOT have a unique decomposition of primes.
We then proceed by a logical sequence of steps to show that this leads to a contradiction.
**Thus our original assumption was untenable and hence we have proved that all numbers have a unique decomposition of primes.
I have a problem understanding the star marked step.
It's like saying, if we want to prove the man is happy, let us assume the man is unhappy.
A logical sequence of steps leads to a contradiction.
Hence, our initial assumption is flawed, so the man is 'happy'?!?
What ensures that 'NOT unhappy' means 'happy' in the realm of math?
Thank you for your time :)
formal-proofs
formal-proofs
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
asked Feb 24 at 6:32
hargun3045hargun3045
7918
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
asked Feb 24 at 6:32
hargun3045hargun3045
7918
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
edited Feb 24 at 7:21
J. W. Tanner
3,5831320
3,5831320
asked Feb 24 at 6:32
hargun3045hargun3045
7918
asked Feb 24 at 6:32
hargun3045hargun3045
7918
asked Feb 24 at 6:32
hargun3045hargun3045
7918
7918
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is called the law of excluded middle, and it is a kind of "meta-axiom" that most mathematicians accept.
answered Feb 24 at 6:37
AGFAGF
15516
$endgroup$
add a comment |
$begingroup$
The basic principle of proof by contradiction is that any proposition must be either true or false. If you assume a proposition is true and by logical steps arrive at the conclusion that the proposition is false, then you know there is an inconsistency in your argument. If you have not made any mistakes in the reasoning following your original assumption, this can only mean that the original assumption rendered your argument inconsistent.
In your example, if you assume that a man is unhappy and by logical steps arrive at the conclusion that the man is happy, then we know there is an inconsistency in the argument somewhere, since it is impossible to be both happy and unhappy at the same time. Since the only thing you have assumed is that he is unhappy, this is what created the inconsistency. It follows that he must be happy. (Note that the example can be confusing since "being happy" is somewhat subjective and could be argued not a a valid proposition in the first place)
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
$endgroup$
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',
autoActivateHeartbeat: false,
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%2f3124522%2fproof-by-contradiction-getting-my-head-around-it%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is called the law of excluded middle, and it is a kind of "meta-axiom" that most mathematicians accept.
answered Feb 24 at 6:37
AGFAGF
15516
$endgroup$
add a comment |
$begingroup$
This is called the law of excluded middle, and it is a kind of "meta-axiom" that most mathematicians accept.
answered Feb 24 at 6:37
AGFAGF
15516
$endgroup$
add a comment |
$begingroup$
This is called the law of excluded middle, and it is a kind of "meta-axiom" that most mathematicians accept.
answered Feb 24 at 6:37
AGFAGF
15516
$endgroup$
This is called the law of excluded middle, and it is a kind of "meta-axiom" that most mathematicians accept.
answered Feb 24 at 6:37
AGFAGF
15516
answered Feb 24 at 6:37
AGFAGF
15516
answered Feb 24 at 6:37
AGFAGF
15516
answered Feb 24 at 6:37
AGFAGF
15516
15516
add a comment |
add a comment |
$begingroup$
The basic principle of proof by contradiction is that any proposition must be either true or false. If you assume a proposition is true and by logical steps arrive at the conclusion that the proposition is false, then you know there is an inconsistency in your argument. If you have not made any mistakes in the reasoning following your original assumption, this can only mean that the original assumption rendered your argument inconsistent.
In your example, if you assume that a man is unhappy and by logical steps arrive at the conclusion that the man is happy, then we know there is an inconsistency in the argument somewhere, since it is impossible to be both happy and unhappy at the same time. Since the only thing you have assumed is that he is unhappy, this is what created the inconsistency. It follows that he must be happy. (Note that the example can be confusing since "being happy" is somewhat subjective and could be argued not a a valid proposition in the first place)
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
$endgroup$
add a comment |
$begingroup$
The basic principle of proof by contradiction is that any proposition must be either true or false. If you assume a proposition is true and by logical steps arrive at the conclusion that the proposition is false, then you know there is an inconsistency in your argument. If you have not made any mistakes in the reasoning following your original assumption, this can only mean that the original assumption rendered your argument inconsistent.
In your example, if you assume that a man is unhappy and by logical steps arrive at the conclusion that the man is happy, then we know there is an inconsistency in the argument somewhere, since it is impossible to be both happy and unhappy at the same time. Since the only thing you have assumed is that he is unhappy, this is what created the inconsistency. It follows that he must be happy. (Note that the example can be confusing since "being happy" is somewhat subjective and could be argued not a a valid proposition in the first place)
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
$endgroup$
add a comment |
$begingroup$
The basic principle of proof by contradiction is that any proposition must be either true or false. If you assume a proposition is true and by logical steps arrive at the conclusion that the proposition is false, then you know there is an inconsistency in your argument. If you have not made any mistakes in the reasoning following your original assumption, this can only mean that the original assumption rendered your argument inconsistent.
In your example, if you assume that a man is unhappy and by logical steps arrive at the conclusion that the man is happy, then we know there is an inconsistency in the argument somewhere, since it is impossible to be both happy and unhappy at the same time. Since the only thing you have assumed is that he is unhappy, this is what created the inconsistency. It follows that he must be happy. (Note that the example can be confusing since "being happy" is somewhat subjective and could be argued not a a valid proposition in the first place)
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
$endgroup$
The basic principle of proof by contradiction is that any proposition must be either true or false. If you assume a proposition is true and by logical steps arrive at the conclusion that the proposition is false, then you know there is an inconsistency in your argument. If you have not made any mistakes in the reasoning following your original assumption, this can only mean that the original assumption rendered your argument inconsistent.
In your example, if you assume that a man is unhappy and by logical steps arrive at the conclusion that the man is happy, then we know there is an inconsistency in the argument somewhere, since it is impossible to be both happy and unhappy at the same time. Since the only thing you have assumed is that he is unhappy, this is what created the inconsistency. It follows that he must be happy. (Note that the example can be confusing since "being happy" is somewhat subjective and could be argued not a a valid proposition in the first place)
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
answered Feb 24 at 10:41
Thomas FjærvikThomas Fjærvik
2859
2859
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.
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%2f3124522%2fproof-by-contradiction-getting-my-head-around-it%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