If $1 > a-b > 0$ then $[a]geq[b]$If $a$ and $b$ are positive real numbers, then $a + b geq 2 sqrtab$.Proof if $n_k < n_k+1$ for all $k in mathbbN$, then $n_k geq k$ for all $k in mathbbN$.If $f(n) geq g(n)$ for all $n geq n_0$ then $f(n) geq cg(n)$ for all $n geq n_1$Inequality $2a^nb^nc^n+1geq a^2n+b^2n+c^2n$Suppose $0<a<b$. Prove for all $ngeq 2$, $0< sqrt[n]a< sqrt[n]b$.On some iterated inequalities and $x geq 5$Show that $2^n geq (n +2)^2$ for all $n geq 6$Induction for $3^ngeq n^2$What about $f$ if $fracdleft(xf(x)-xright)dx geq 0$Prove or disprove $|a| geq 2|b| implies |a+b| geq |b|$
Multi tool use
How difficult is it to simply disable/disengage the MCAS on Boeing 737 Max 8 & 9 Aircraft?
What options are left, if Britain cannot decide?
What is the significance behind "40 days" that often appears in the Bible?
Gantt Chart like rectangles with log scale
Why do Australian milk farmers need to protest supermarkets' milk price?
Why would a flight no longer considered airworthy be redirected like this?
Why one should not leave fingerprints on bulbs and plugs?
How can I track script which gives me "command not found" right after the login?
Is it possible to upcast ritual spells?
If curse and magic is two sides of the same coin, why the former is forbidden?
What exactly is this small puffer fish doing and how did it manage to accomplish such a feat?
Gravity magic - How does it work?
How to write cleanly even if my character uses expletive language?
What has been your most complicated TikZ drawing?
Define, (actually define) the "stability" and "energy" of a compound
Unexpected result from ArcLength
Is a party consisting of only a bard, a cleric, and a warlock functional long-term?
Bach's Toccata and Fugue in D minor breaks the "no parallel octaves" rule?
Error in Twin Prime Conjecture
Dice rolling probability game
Continuity of Linear Operator Between Hilbert Spaces
Provisioning profile doesn't include the application-identifier and keychain-access-groups entitlements
How do I hide Chekhov's Gun?
Credit cards used everywhere in Singapore or Malaysia?
If $1 > a-b > 0$ then $[a]geq[b]$
If $a$ and $b$ are positive real numbers, then $a + b geq 2 sqrtab$.Proof if $n_k < n_k+1$ for all $k in mathbbN$, then $n_k geq k$ for all $k in mathbbN$.If $f(n) geq g(n)$ for all $n geq n_0$ then $f(n) geq cg(n)$ for all $n geq n_1$Inequality $2a^nb^nc^n+1geq a^2n+b^2n+c^2n$Suppose $0<a<b$. Prove for all $ngeq 2$, $0< sqrt[n]a< sqrt[n]b$.On some iterated inequalities and $x geq 5$Show that $2^n geq (n +2)^2$ for all $n geq 6$Induction for $3^ngeq n^2$What about $f$ if $fracdleft(xf(x)-xright)dx geq 0$Prove or disprove $|a| geq 2|b| implies |a+b| geq |b|$
$begingroup$
what I can do for The case $[a]<[b]$.
I know that $[a]>[b]Rightarrow [a]geq [b]+1$
EDIT:
I have realized that the Initial assumption $[a]=[b]$ is false therefore I have changed it to $[a]geq [b]$
inequality
asked Mar 11 at 20:59
New2MathNew2Math
11812
$endgroup$
|
show 1 more comment
$begingroup$
what I can do for The case $[a]<[b]$.
I know that $[a]>[b]Rightarrow [a]geq [b]+1$
EDIT:
I have realized that the Initial assumption $[a]=[b]$ is false therefore I have changed it to $[a]geq [b]$
inequality
asked Mar 11 at 20:59
New2MathNew2Math
11812
$endgroup$
$begingroup$
For clarity, what does $[x]$ mean in this context? I've usually seen this to refer to the rounding function, i.e. rounding $x$ to the nearest integer, but a few counterexamples immediately come to mind in that instance
$endgroup$
– Eevee Trainer
Mar 11 at 21:02
$begingroup$
$[x]$ is the biggest integer for which we have $leq x$. Can you give me an counterexample?
$endgroup$
– New2Math
Mar 11 at 21:04
3
$begingroup$
What if $a=1.1$ and $b=0.9$?
$endgroup$
– clark
Mar 11 at 21:06
$begingroup$
Okay so you're referring to the floor function then @New2Math - so my counterexample would be irrelevant. (Were it rounding and not floor, I was thinking $a=0.6,b=0.4$.) clark's counterexample works for the floor function though.
$endgroup$
– Eevee Trainer
Mar 11 at 21:10
$begingroup$
@clark I see now that this assumption is false but the assumption $a-b>1Rightarrow [a]>[b]$ is true right, also If I would have changed the assumption in the question to $geq$ then it would be right, right?
$endgroup$
– New2Math
Mar 11 at 21:10
|
show 1 more comment
$begingroup$
what I can do for The case $[a]<[b]$.
I know that $[a]>[b]Rightarrow [a]geq [b]+1$
EDIT:
I have realized that the Initial assumption $[a]=[b]$ is false therefore I have changed it to $[a]geq [b]$
inequality
asked Mar 11 at 20:59
New2MathNew2Math
11812
$endgroup$
what I can do for The case $[a]<[b]$.
I know that $[a]>[b]Rightarrow [a]geq [b]+1$
EDIT:
I have realized that the Initial assumption $[a]=[b]$ is false therefore I have changed it to $[a]geq [b]$
inequality
inequality
asked Mar 11 at 20:59
New2MathNew2Math
11812
asked Mar 11 at 20:59
New2MathNew2Math
11812
edited Mar 11 at 21:16
New2Math
asked Mar 11 at 20:59
New2MathNew2Math
11812
asked Mar 11 at 20:59
New2MathNew2Math
11812
asked Mar 11 at 20:59
New2MathNew2Math
11812
11812
$begingroup$
For clarity, what does $[x]$ mean in this context? I've usually seen this to refer to the rounding function, i.e. rounding $x$ to the nearest integer, but a few counterexamples immediately come to mind in that instance
$endgroup$
– Eevee Trainer
Mar 11 at 21:02
$begingroup$
$[x]$ is the biggest integer for which we have $leq x$. Can you give me an counterexample?
$endgroup$
– New2Math
Mar 11 at 21:04
3
$begingroup$
What if $a=1.1$ and $b=0.9$?
$endgroup$
– clark
Mar 11 at 21:06
$begingroup$
Okay so you're referring to the floor function then @New2Math - so my counterexample would be irrelevant. (Were it rounding and not floor, I was thinking $a=0.6,b=0.4$.) clark's counterexample works for the floor function though.
$endgroup$
– Eevee Trainer
Mar 11 at 21:10
$begingroup$
@clark I see now that this assumption is false but the assumption $a-b>1Rightarrow [a]>[b]$ is true right, also If I would have changed the assumption in the question to $geq$ then it would be right, right?
$endgroup$
– New2Math
Mar 11 at 21:10
|
show 1 more comment
$begingroup$
For clarity, what does $[x]$ mean in this context? I've usually seen this to refer to the rounding function, i.e. rounding $x$ to the nearest integer, but a few counterexamples immediately come to mind in that instance
$endgroup$
– Eevee Trainer
Mar 11 at 21:02
$begingroup$
$[x]$ is the biggest integer for which we have $leq x$. Can you give me an counterexample?
$endgroup$
– New2Math
Mar 11 at 21:04
3
$begingroup$
What if $a=1.1$ and $b=0.9$?
$endgroup$
– clark
Mar 11 at 21:06
$begingroup$
Okay so you're referring to the floor function then @New2Math - so my counterexample would be irrelevant. (Were it rounding and not floor, I was thinking $a=0.6,b=0.4$.) clark's counterexample works for the floor function though.
$endgroup$
– Eevee Trainer
Mar 11 at 21:10
$begingroup$
@clark I see now that this assumption is false but the assumption $a-b>1Rightarrow [a]>[b]$ is true right, also If I would have changed the assumption in the question to $geq$ then it would be right, right?
$endgroup$
– New2Math
Mar 11 at 21:10
$begingroup$
For clarity, what does $[x]$ mean in this context? I've usually seen this to refer to the rounding function, i.e. rounding $x$ to the nearest integer, but a few counterexamples immediately come to mind in that instance
$endgroup$
– Eevee Trainer
Mar 11 at 21:02
$begingroup$
For clarity, what does $[x]$ mean in this context? I've usually seen this to refer to the rounding function, i.e. rounding $x$ to the nearest integer, but a few counterexamples immediately come to mind in that instance
$endgroup$
– Eevee Trainer
Mar 11 at 21:02
$begingroup$
$[x]$ is the biggest integer for which we have $leq x$. Can you give me an counterexample?
$endgroup$
– New2Math
Mar 11 at 21:04
$begingroup$
$[x]$ is the biggest integer for which we have $leq x$. Can you give me an counterexample?
$endgroup$
– New2Math
Mar 11 at 21:04
3
3
$begingroup$
What if $a=1.1$ and $b=0.9$?
$endgroup$
– clark
Mar 11 at 21:06
$begingroup$
What if $a=1.1$ and $b=0.9$?
$endgroup$
– clark
Mar 11 at 21:06
$begingroup$
Okay so you're referring to the floor function then @New2Math - so my counterexample would be irrelevant. (Were it rounding and not floor, I was thinking $a=0.6,b=0.4$.) clark's counterexample works for the floor function though.
$endgroup$
– Eevee Trainer
Mar 11 at 21:10
$begingroup$
Okay so you're referring to the floor function then @New2Math - so my counterexample would be irrelevant. (Were it rounding and not floor, I was thinking $a=0.6,b=0.4$.) clark's counterexample works for the floor function though.
$endgroup$
– Eevee Trainer
Mar 11 at 21:10
$begingroup$
@clark I see now that this assumption is false but the assumption $a-b>1Rightarrow [a]>[b]$ is true right, also If I would have changed the assumption in the question to $geq$ then it would be right, right?
$endgroup$
– New2Math
Mar 11 at 21:10
$begingroup$
@clark I see now that this assumption is false but the assumption $a-b>1Rightarrow [a]>[b]$ is true right, also If I would have changed the assumption in the question to $geq$ then it would be right, right?
$endgroup$
– New2Math
Mar 11 at 21:10
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If $a geq b$ then it immediately follows that $lfloor a rfloor geq lfloor brfloor$. So we don't even need the upper bound.
(All this says is that the floor function is monotone increasing.)
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
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%2f3144238%2fif-1-a-b-0-then-a-geqb%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
$begingroup$
If $a geq b$ then it immediately follows that $lfloor a rfloor geq lfloor brfloor$. So we don't even need the upper bound.
(All this says is that the floor function is monotone increasing.)
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
add a comment |
$begingroup$
If $a geq b$ then it immediately follows that $lfloor a rfloor geq lfloor brfloor$. So we don't even need the upper bound.
(All this says is that the floor function is monotone increasing.)
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
add a comment |
$begingroup$
If $a geq b$ then it immediately follows that $lfloor a rfloor geq lfloor brfloor$. So we don't even need the upper bound.
(All this says is that the floor function is monotone increasing.)
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
If $a geq b$ then it immediately follows that $lfloor a rfloor geq lfloor brfloor$. So we don't even need the upper bound.
(All this says is that the floor function is monotone increasing.)
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
answered Mar 11 at 21:27
Daniel McLauryDaniel McLaury
15.9k32981
15.9k32981
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%2f3144238%2fif-1-a-b-0-then-a-geqb%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
yTRz0Ajy1PrAeBAx9cgCZ,oTn1UMmvz64Dwp,Ku9WmOf6TPXQN6km e cNAs,a,kba,JSEqI30chhT4QF2D,Rq
$begingroup$
For clarity, what does $[x]$ mean in this context? I've usually seen this to refer to the rounding function, i.e. rounding $x$ to the nearest integer, but a few counterexamples immediately come to mind in that instance
$endgroup$
– Eevee Trainer
Mar 11 at 21:02
$begingroup$
$[x]$ is the biggest integer for which we have $leq x$. Can you give me an counterexample?
$endgroup$
– New2Math
Mar 11 at 21:04
3
$begingroup$
What if $a=1.1$ and $b=0.9$?
$endgroup$
– clark
Mar 11 at 21:06
$begingroup$
Okay so you're referring to the floor function then @New2Math - so my counterexample would be irrelevant. (Were it rounding and not floor, I was thinking $a=0.6,b=0.4$.) clark's counterexample works for the floor function though.
$endgroup$
– Eevee Trainer
Mar 11 at 21:10
$begingroup$
@clark I see now that this assumption is false but the assumption $a-b>1Rightarrow [a]>[b]$ is true right, also If I would have changed the assumption in the question to $geq$ then it would be right, right?
$endgroup$
– New2Math
Mar 11 at 21:10