Question regarding an inequality in Spivak's Calculus
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On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.
|x-3| < 1, or 2 < x < 4
I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!
inequality
asked Jan 15 at 17:25
Julian25Julian25
1104
$endgroup$
add a comment |
$begingroup$
On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.
|x-3| < 1, or 2 < x < 4
I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!
inequality
asked Jan 15 at 17:25
Julian25Julian25
1104
$endgroup$
$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27
add a comment |
$begingroup$
On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.
|x-3| < 1, or 2 < x < 4
I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!
inequality
asked Jan 15 at 17:25
Julian25Julian25
1104
$endgroup$
On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.
|x-3| < 1, or 2 < x < 4
I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!
inequality
inequality
asked Jan 15 at 17:25
Julian25Julian25
1104
asked Jan 15 at 17:25
Julian25Julian25
1104
asked Jan 15 at 17:25
Julian25Julian25
1104
asked Jan 15 at 17:25
Julian25Julian25
1104
asked Jan 15 at 17:25
Julian25Julian25
1104
1104
$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27
add a comment |
$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27
$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27
$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
add a comment |
$begingroup$
By definition,
$$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
Now add $3$ throughout.
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
add a comment |
$begingroup$
$$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
add a comment |
$begingroup$
$$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
$endgroup$
$$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
answered Jan 15 at 17:26
angryavianangryavian
42.5k23481
42.5k23481
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
add a comment |
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
$begingroup$
Ah, that way of reading it is really helpful. Thanks!
$endgroup$
– Julian25
Jan 15 at 17:28
add a comment |
$begingroup$
By definition,
$$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
Now add $3$ throughout.
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
$endgroup$
add a comment |
$begingroup$
By definition,
$$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
Now add $3$ throughout.
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
$endgroup$
add a comment |
$begingroup$
By definition,
$$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
Now add $3$ throughout.
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
$endgroup$
By definition,
$$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
Now add $3$ throughout.
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
answered Jan 15 at 17:26
MisterRiemannMisterRiemann
5,9151625
5,9151625
add a comment |
add a comment |
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$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27