Prove that, for every natural number $n$, $10^n > n^2$Proof writing: how to write a clear induction proof?Equality such that induction step is valid but basis is not?Prove inequality by inductionProve that $2n+1 leq 2^n$ for $n geq 3$ using mathematical induction.Induction Proof Check: For a binary tree T, Prove that the number of full nodes in T is always one less than the number of leaves in T.Prove that $n!>n^2$ for all integers $n geq 4$.When I have 2 variables and 2 base cases, can I use mathematical induction multiple times to form a proof?Induction Proof Verification via Binomial TheoremProving $1^3 + 2^3 + 3^3 + … + n^3 = (1/4)n^4 + (1/2)n^3 + (1/4)n^2$ by induction.Prove: $2^n+2ge n^3$ induction proofDiscrete Mathematics induction proof: $1! + 2! + 3! + ldots + n! < (n + 1)!$
Reversed Sudoku
In the late 1940’s to early 1950’s what technology was available that could melt a LOT of ice?
Can one live in the U.S. and not use a credit card?
Signed and unsigned numbers
'The literal of type int is out of range' con número enteros pequeños (2 dígitos)
What are the threaded holes in Manfrotto camera brackets?
Vocabulary for giving just numbers, not a full answer
Does a warlock using the Darkness/Devil's Sight combo still have advantage on ranged attacks against a target outside the Darkness?
Should I take out a loan for a friend to invest on my behalf?
What problems would a superhuman have whose skin is constantly hot?
Are all players supposed to be able to see each others' character sheets?
An alternative proof of an application of Hahn-Banach
How do I express some one as a black person?
Conservation of Mass and Energy
Makefile strange variable substitution
Should I tell my boss the work he did was worthless
Recommendation letter by significant other if you worked with them professionally?
They call me Inspector Morse
How can The Temple of Elementary Evil reliably protect itself against kinetic bombardment?
Accountant/ lawyer will not return my call
UART pins to unpowered MCU?
Examples of a statistic that is not independent of sample's distribution?
Virginia employer terminated employee and wants signing bonus returned
Does this video of collapsing warehouse shelves show a real incident?
Prove that, for every natural number $n$, $10^n > n^2$
Proof writing: how to write a clear induction proof?Equality such that induction step is valid but basis is not?Prove inequality by inductionProve that $2n+1 leq 2^n$ for $n geq 3$ using mathematical induction.Induction Proof Check: For a binary tree T, Prove that the number of full nodes in T is always one less than the number of leaves in T.Prove that $n!>n^2$ for all integers $n geq 4$.When I have 2 variables and 2 base cases, can I use mathematical induction multiple times to form a proof?Induction Proof Verification via Binomial TheoremProving $1^3 + 2^3 + 3^3 + … + n^3 = (1/4)n^4 + (1/2)n^3 + (1/4)n^2$ by induction.Prove: $2^n+2ge n^3$ induction proofDiscrete Mathematics induction proof: $1! + 2! + 3! + ldots + n! < (n + 1)!$
$begingroup$
I attempted to solve the proof but looking at another proof but I got stuck. Here's what I have so far.
Let $n = 1$, $10 > 1$.
Assume valid for $k$ greater than or equal to $1$.
Then $n = k + 1$.
$10^k > k^2$ .... Case(1) = $100 > 10$
And then what is after? Am I even doing the proof properly? I feel that I did a step incorrectly which led me to a dead-end.
EDIT: There was a hint in the back of my book, "$10n^2 = n^2 + 2n^2 + 7n^2$, now prove that $2n^2 geq 2n$ and $7n^2 geq 1$.
induction
edited 2 days ago
Alan Muniz
2,5961830
asked 2 days ago
Mark ParkMark Park
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I attempted to solve the proof but looking at another proof but I got stuck. Here's what I have so far.
Let $n = 1$, $10 > 1$.
Assume valid for $k$ greater than or equal to $1$.
Then $n = k + 1$.
$10^k > k^2$ .... Case(1) = $100 > 10$
And then what is after? Am I even doing the proof properly? I feel that I did a step incorrectly which led me to a dead-end.
EDIT: There was a hint in the back of my book, "$10n^2 = n^2 + 2n^2 + 7n^2$, now prove that $2n^2 geq 2n$ and $7n^2 geq 1$.
induction
edited 2 days ago
Alan Muniz
2,5961830
asked 2 days ago
Mark ParkMark Park
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Suggested reading: How to write a clear induction proof
$endgroup$
– JMoravitz
2 days ago
add a comment |
$begingroup$
I attempted to solve the proof but looking at another proof but I got stuck. Here's what I have so far.
Let $n = 1$, $10 > 1$.
Assume valid for $k$ greater than or equal to $1$.
Then $n = k + 1$.
$10^k > k^2$ .... Case(1) = $100 > 10$
And then what is after? Am I even doing the proof properly? I feel that I did a step incorrectly which led me to a dead-end.
EDIT: There was a hint in the back of my book, "$10n^2 = n^2 + 2n^2 + 7n^2$, now prove that $2n^2 geq 2n$ and $7n^2 geq 1$.
induction
edited 2 days ago
Alan Muniz
2,5961830
asked 2 days ago
Mark ParkMark Park
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I attempted to solve the proof but looking at another proof but I got stuck. Here's what I have so far.
Let $n = 1$, $10 > 1$.
Assume valid for $k$ greater than or equal to $1$.
Then $n = k + 1$.
$10^k > k^2$ .... Case(1) = $100 > 10$
And then what is after? Am I even doing the proof properly? I feel that I did a step incorrectly which led me to a dead-end.
EDIT: There was a hint in the back of my book, "$10n^2 = n^2 + 2n^2 + 7n^2$, now prove that $2n^2 geq 2n$ and $7n^2 geq 1$.
induction
induction
edited 2 days ago
Alan Muniz
2,5961830
asked 2 days ago
Mark ParkMark Park
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Alan Muniz
2,5961830
asked 2 days ago
Mark ParkMark Park
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Alan Muniz
2,5961830
edited 2 days ago
Alan Muniz
2,5961830
edited 2 days ago
Alan Muniz
2,5961830
2,5961830
asked 2 days ago
Mark ParkMark Park
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Mark ParkMark Park
1
asked 2 days ago
Mark ParkMark Park
1
1
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mark Park is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Suggested reading: How to write a clear induction proof
$endgroup$
– JMoravitz
2 days ago
add a comment |
$begingroup$
Suggested reading: How to write a clear induction proof
$endgroup$
– JMoravitz
2 days ago
$begingroup$
Suggested reading: How to write a clear induction proof
$endgroup$
– JMoravitz
2 days ago
$begingroup$
Suggested reading: How to write a clear induction proof
$endgroup$
– JMoravitz
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have the right general concept, but you should be much clearer about your strategy. You want to show that if the statement holds for some $k geq 1$, then it also holds for $k+1$. So assume that $10^k > k^2$ for some $k geq 1$. Then, $10^k+1 = 10^k cdot 10 > 10k^2$. All you have left to do is check that $10k^2 > (k+1)^2$ for $k geq 1$, and I'll leave this part to you.
answered 2 days ago
JavaManJavaMan
11.1k12755
$endgroup$
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
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
);
);
Mark Park is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3141696%2fprove-that-for-every-natural-number-n-10n-n2%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
$begingroup$
You have the right general concept, but you should be much clearer about your strategy. You want to show that if the statement holds for some $k geq 1$, then it also holds for $k+1$. So assume that $10^k > k^2$ for some $k geq 1$. Then, $10^k+1 = 10^k cdot 10 > 10k^2$. All you have left to do is check that $10k^2 > (k+1)^2$ for $k geq 1$, and I'll leave this part to you.
answered 2 days ago
JavaManJavaMan
11.1k12755
$endgroup$
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
add a comment |
$begingroup$
You have the right general concept, but you should be much clearer about your strategy. You want to show that if the statement holds for some $k geq 1$, then it also holds for $k+1$. So assume that $10^k > k^2$ for some $k geq 1$. Then, $10^k+1 = 10^k cdot 10 > 10k^2$. All you have left to do is check that $10k^2 > (k+1)^2$ for $k geq 1$, and I'll leave this part to you.
answered 2 days ago
JavaManJavaMan
11.1k12755
$endgroup$
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
add a comment |
$begingroup$
You have the right general concept, but you should be much clearer about your strategy. You want to show that if the statement holds for some $k geq 1$, then it also holds for $k+1$. So assume that $10^k > k^2$ for some $k geq 1$. Then, $10^k+1 = 10^k cdot 10 > 10k^2$. All you have left to do is check that $10k^2 > (k+1)^2$ for $k geq 1$, and I'll leave this part to you.
answered 2 days ago
JavaManJavaMan
11.1k12755
$endgroup$
You have the right general concept, but you should be much clearer about your strategy. You want to show that if the statement holds for some $k geq 1$, then it also holds for $k+1$. So assume that $10^k > k^2$ for some $k geq 1$. Then, $10^k+1 = 10^k cdot 10 > 10k^2$. All you have left to do is check that $10k^2 > (k+1)^2$ for $k geq 1$, and I'll leave this part to you.
answered 2 days ago
JavaManJavaMan
11.1k12755
answered 2 days ago
JavaManJavaMan
11.1k12755
answered 2 days ago
JavaManJavaMan
11.1k12755
answered 2 days ago
JavaManJavaMan
11.1k12755
11.1k12755
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
add a comment |
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
$begingroup$
I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan.
$endgroup$
– Mark Park
2 days ago
add a comment |
Mark Park is a new contributor. Be nice, and check out our Code of Conduct.
Mark Park is a new contributor. Be nice, and check out our Code of Conduct.
Mark Park is a new contributor. Be nice, and check out our Code of Conduct.
Mark Park is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3141696%2fprove-that-for-every-natural-number-n-10n-n2%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
$begingroup$
Suggested reading: How to write a clear induction proof
$endgroup$
– JMoravitz
2 days ago