Absolute value less than some value [closed]
$begingroup$
This is a noob question. If,
$$biggm| frac{1}{2} - e biggm | le n$$
Then how do I get the following?
$$e le frac{1}{2} + n$$
inequality fractions absolute-value
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
$endgroup$
closed as off-topic by Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol. Dec 13 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol.
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
This is a noob question. If,
$$biggm| frac{1}{2} - e biggm | le n$$
Then how do I get the following?
$$e le frac{1}{2} + n$$
inequality fractions absolute-value
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
$endgroup$
closed as off-topic by Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol. Dec 13 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol.
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
This is a noob question. If,
$$biggm| frac{1}{2} - e biggm | le n$$
Then how do I get the following?
$$e le frac{1}{2} + n$$
inequality fractions absolute-value
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
$endgroup$
This is a noob question. If,
$$biggm| frac{1}{2} - e biggm | le n$$
Then how do I get the following?
$$e le frac{1}{2} + n$$
inequality fractions absolute-value
inequality fractions absolute-value
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
asked Dec 12 '18 at 23:21
Hasan IqbalHasan Iqbal
1267
1267
closed as off-topic by Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol. Dec 13 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol.
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol. Dec 13 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Chris Custer, KReiser, Leucippus, Shailesh, Cyclohexanol.
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Having $lvert a - b rvert leq c$ is basically the same as $lvert b - a rvert leq c$.
You can also think of distances. The distance between $a$ and $b$ is less than $c$. If you imagine dots on the number line and keep the distance thingy in mind it should be very easy to see.
answered Dec 12 '18 at 23:29
drixdrix
364
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add a comment |
$begingroup$
Observe that if $dfrac{1}{2} - n leq e$, the first equation holds.
Plot the graph of the absolut value function. All values of $x$ such that $|x|leq n$ lie in the interval $[-n,n]$, i.e.
$$xin[-n,n] iff |x|leq n.$$
Substitute the right hand side of the equation where $x$ is to get
$$| e + dfrac{1}{2}| leq n iff e + dfrac{1}{2} in [-n,n] iff -nleq e + dfrac{1}{2} leq n.$$ For solving it, you need to solve both inequalities:
$$-nleq e + dfrac{1}{2}$$
$$e + dfrac{1}{2} leq n.$$
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
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add a comment |
$begingroup$
You have to know that
$$[a-b|le ciff b-cle ale b+c,$$
and remener tha $|a-b|=|b-a|.
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Having $lvert a - b rvert leq c$ is basically the same as $lvert b - a rvert leq c$.
You can also think of distances. The distance between $a$ and $b$ is less than $c$. If you imagine dots on the number line and keep the distance thingy in mind it should be very easy to see.
answered Dec 12 '18 at 23:29
drixdrix
364
$endgroup$
add a comment |
$begingroup$
Having $lvert a - b rvert leq c$ is basically the same as $lvert b - a rvert leq c$.
You can also think of distances. The distance between $a$ and $b$ is less than $c$. If you imagine dots on the number line and keep the distance thingy in mind it should be very easy to see.
answered Dec 12 '18 at 23:29
drixdrix
364
$endgroup$
add a comment |
$begingroup$
Having $lvert a - b rvert leq c$ is basically the same as $lvert b - a rvert leq c$.
You can also think of distances. The distance between $a$ and $b$ is less than $c$. If you imagine dots on the number line and keep the distance thingy in mind it should be very easy to see.
answered Dec 12 '18 at 23:29
drixdrix
364
$endgroup$
Having $lvert a - b rvert leq c$ is basically the same as $lvert b - a rvert leq c$.
You can also think of distances. The distance between $a$ and $b$ is less than $c$. If you imagine dots on the number line and keep the distance thingy in mind it should be very easy to see.
answered Dec 12 '18 at 23:29
drixdrix
364
answered Dec 12 '18 at 23:29
drixdrix
364
answered Dec 12 '18 at 23:29
drixdrix
364
answered Dec 12 '18 at 23:29
drixdrix
364
364
add a comment |
add a comment |
$begingroup$
Observe that if $dfrac{1}{2} - n leq e$, the first equation holds.
Plot the graph of the absolut value function. All values of $x$ such that $|x|leq n$ lie in the interval $[-n,n]$, i.e.
$$xin[-n,n] iff |x|leq n.$$
Substitute the right hand side of the equation where $x$ is to get
$$| e + dfrac{1}{2}| leq n iff e + dfrac{1}{2} in [-n,n] iff -nleq e + dfrac{1}{2} leq n.$$ For solving it, you need to solve both inequalities:
$$-nleq e + dfrac{1}{2}$$
$$e + dfrac{1}{2} leq n.$$
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
$endgroup$
add a comment |
$begingroup$
Observe that if $dfrac{1}{2} - n leq e$, the first equation holds.
Plot the graph of the absolut value function. All values of $x$ such that $|x|leq n$ lie in the interval $[-n,n]$, i.e.
$$xin[-n,n] iff |x|leq n.$$
Substitute the right hand side of the equation where $x$ is to get
$$| e + dfrac{1}{2}| leq n iff e + dfrac{1}{2} in [-n,n] iff -nleq e + dfrac{1}{2} leq n.$$ For solving it, you need to solve both inequalities:
$$-nleq e + dfrac{1}{2}$$
$$e + dfrac{1}{2} leq n.$$
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
$endgroup$
add a comment |
$begingroup$
Observe that if $dfrac{1}{2} - n leq e$, the first equation holds.
Plot the graph of the absolut value function. All values of $x$ such that $|x|leq n$ lie in the interval $[-n,n]$, i.e.
$$xin[-n,n] iff |x|leq n.$$
Substitute the right hand side of the equation where $x$ is to get
$$| e + dfrac{1}{2}| leq n iff e + dfrac{1}{2} in [-n,n] iff -nleq e + dfrac{1}{2} leq n.$$ For solving it, you need to solve both inequalities:
$$-nleq e + dfrac{1}{2}$$
$$e + dfrac{1}{2} leq n.$$
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
$endgroup$
Observe that if $dfrac{1}{2} - n leq e$, the first equation holds.
Plot the graph of the absolut value function. All values of $x$ such that $|x|leq n$ lie in the interval $[-n,n]$, i.e.
$$xin[-n,n] iff |x|leq n.$$
Substitute the right hand side of the equation where $x$ is to get
$$| e + dfrac{1}{2}| leq n iff e + dfrac{1}{2} in [-n,n] iff -nleq e + dfrac{1}{2} leq n.$$ For solving it, you need to solve both inequalities:
$$-nleq e + dfrac{1}{2}$$
$$e + dfrac{1}{2} leq n.$$
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
answered Dec 12 '18 at 23:34
Dr PotatoDr Potato
375
375
add a comment |
add a comment |
$begingroup$
You have to know that
$$[a-b|le ciff b-cle ale b+c,$$
and remener tha $|a-b|=|b-a|.
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
You have to know that
$$[a-b|le ciff b-cle ale b+c,$$
and remener tha $|a-b|=|b-a|.
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
You have to know that
$$[a-b|le ciff b-cle ale b+c,$$
and remener tha $|a-b|=|b-a|.
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
$endgroup$
You have to know that
$$[a-b|le ciff b-cle ale b+c,$$
and remener tha $|a-b|=|b-a|.
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
answered Dec 12 '18 at 23:42
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
