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 $|a| geq 2|b| implies |a+b| geq |b|$
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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
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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 |
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$begingroup$
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 |
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LfQ,kbOAItv EUrhojjhifOZ
$begingroup$
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