Is this a typo in Section 1.8.1 Mathematics for Computer Science?How can I learn about proofs for computer...
"My boss was furious with me and I have been fired" vs. "My boss was furious with me and I was fired"
What is /etc/mtab in Linux?
An array in a equation with curly braces in both sides
How to not starve gigantic beasts
Can I criticise the more senior developers around me for not writing clean code?
Nails holding drywall
Zonal Statistics is returning null values in ArcGIS
How bug prioritization works in agile projects vs non agile
My bank got bought out, am I now going to have to start filing tax returns in a different state?
Can a stored procedure reference the database in which it is stored?
A faster way to compute the largest prime factor
What was Apollo 13's "Little Jolt" after MECO?
How do I reattach a shelf to the wall when it ripped out of the wall?
How can I get rid of an unhelpful parallel branch when unpivoting a single row?
What is the best way to deal with NPC-NPC combat?
Philosophical question on logistic regression: why isn't the optimal threshold value trained?
How exactly does Hawking radiation decrease the mass of black holes?
Can someone publish a story that happened to you?
Mistake in years of experience in resume?
Find a stone which is not the lightest one
Retract an already submitted recommendation letter (written for an undergrad student)
Contradiction proof for inequality of P and NP?
What is the term for a person whose job is to place products on shelves in stores?
Where was the County of Thurn und Taxis located?
Is this a typo in Section 1.8.1 Mathematics for Computer Science?
How can I learn about proofs for computer science?How can I start to learn proof theory?Proving the existence of a proof without actually giving a proofProof by well ordering: Every positive integer greater than one can be factored as a product of primes.Gentzen and computer scienceElegant demonstration with minimum category theory knowledgeDifferent definitions of Natural Numbers and Proof by ContradictionProve $3^{2n}-5$ is a multiple of $4$Defining new symbols in a proof, when is this justified?Prove convergence rate to zero increases as n increases
$begingroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
asked 5 hours ago
doctopusdoctopus
1413
$endgroup$
add a comment |
$begingroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
asked 5 hours ago
doctopusdoctopus
1413
$endgroup$
$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago
add a comment |
$begingroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
asked 5 hours ago
doctopusdoctopus
1413
$endgroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
proof-explanation proof-theory
asked 5 hours ago
doctopusdoctopus
1413
asked 5 hours ago
doctopusdoctopus
1413
asked 5 hours ago
doctopusdoctopus
1413
asked 5 hours ago
doctopusdoctopus
1413
asked 5 hours ago
doctopusdoctopus
1413
1413
$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago
add a comment |
$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago
$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago
$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered 4 hours ago
RandallRandall
10.9k11431
$endgroup$
add a comment |
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered 4 hours ago
John DoeJohn Doe
12.2k11341
$endgroup$
add a comment |
Your Answer
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%2f3202666%2fis-this-a-typo-in-section-1-8-1-mathematics-for-computer-science%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
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered 4 hours ago
RandallRandall
10.9k11431
$endgroup$
add a comment |
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered 4 hours ago
RandallRandall
10.9k11431
$endgroup$
add a comment |
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered 4 hours ago
RandallRandall
10.9k11431
$endgroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered 4 hours ago
RandallRandall
10.9k11431
answered 4 hours ago
RandallRandall
10.9k11431
answered 4 hours ago
RandallRandall
10.9k11431
answered 4 hours ago
RandallRandall
10.9k11431
10.9k11431
add a comment |
add a comment |
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered 4 hours ago
John DoeJohn Doe
12.2k11341
$endgroup$
add a comment |
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered 4 hours ago
John DoeJohn Doe
12.2k11341
$endgroup$
add a comment |
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered 4 hours ago
John DoeJohn Doe
12.2k11341
$endgroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered 4 hours ago
John DoeJohn Doe
12.2k11341
answered 4 hours ago
John DoeJohn Doe
12.2k11341
answered 4 hours ago
John DoeJohn Doe
12.2k11341
answered 4 hours ago
John DoeJohn Doe
12.2k11341
12.2k11341
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%2f3202666%2fis-this-a-typo-in-section-1-8-1-mathematics-for-computer-science%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
$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago