If $x+h0$, why does $|x+h|-|x|=-h$?

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0
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If $x+h<0$ and $h>0$

Why does

$|x+h|-|x|=-h$

I think I'm having difficulties with the absolute operator. How do I get rid of it to move on? And how does knowing that $x+h<0$ change how I can solve this?

edited Jul 18 at 0:43

asked Jul 18 at 0:36

Cedric Martens

286211

3

It doesn't, necessarily. For example, $|1 - 2| - |1| neq -(-2)$.
â€“Â Theo Bendit
Jul 18 at 0:39

1

Hi Theo, I did a mistake that was pointed out by Arnaud. I edited my question. Thank you
â€“Â Cedric Martens
Jul 18 at 0:44

add a commentÂ |Â

up vote
0
down vote

favorite

If $x+h<0$ and $h>0$

Why does

$|x+h|-|x|=-h$

I think I'm having difficulties with the absolute operator. How do I get rid of it to move on? And how does knowing that $x+h<0$ change how I can solve this?

edited Jul 18 at 0:43

asked Jul 18 at 0:36

Cedric Martens

286211

3

It doesn't, necessarily. For example, $|1 - 2| - |1| neq -(-2)$.
â€“Â Theo Bendit
Jul 18 at 0:39

1

Hi Theo, I did a mistake that was pointed out by Arnaud. I edited my question. Thank you
â€“Â Cedric Martens
Jul 18 at 0:44

add a commentÂ |Â

up vote
0
down vote

favorite

If $x+h<0$ and $h>0$

Why does

$|x+h|-|x|=-h$

I think I'm having difficulties with the absolute operator. How do I get rid of it to move on? And how does knowing that $x+h<0$ change how I can solve this?

edited Jul 18 at 0:43

asked Jul 18 at 0:36

Cedric Martens

286211

If $x+h<0$ and $h>0$

Why does

$|x+h|-|x|=-h$

I think I'm having difficulties with the absolute operator. How do I get rid of it to move on? And how does knowing that $x+h<0$ change how I can solve this?

edited Jul 18 at 0:43

asked Jul 18 at 0:36

Cedric Martens

286211

edited Jul 18 at 0:43

asked Jul 18 at 0:36

Cedric Martens

286211

asked Jul 18 at 0:36

Cedric Martens

286211

asked Jul 18 at 0:36

Cedric Martens

286211

3

It doesn't, necessarily. For example, $|1 - 2| - |1| neq -(-2)$.
â€“Â Theo Bendit
Jul 18 at 0:39

1

Hi Theo, I did a mistake that was pointed out by Arnaud. I edited my question. Thank you
â€“Â Cedric Martens
Jul 18 at 0:44

add a commentÂ |Â

3

It doesn't, necessarily. For example, $|1 - 2| - |1| neq -(-2)$.
â€“Â Theo Bendit
Jul 18 at 0:39

1

Hi Theo, I did a mistake that was pointed out by Arnaud. I edited my question. Thank you
â€“Â Cedric Martens
Jul 18 at 0:44

It doesn't, necessarily. For example, $|1 - 2| - |1| neq -(-2)$.
â€“Â Theo Bendit
Jul 18 at 0:39

Hi Theo, I did a mistake that was pointed out by Arnaud. I edited my question. Thank you
â€“Â Cedric Martens
Jul 18 at 0:44

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
3
down vote

accepted

The absolute operator is defined as follows:

$$left|xright| = begincases
x & textif x geq 0 \
-x & textif x lt 0
endcases$$

In other words, if $x$ is positive (or zero), it stays the same. If $x$ is negative, then multiply it by $-1$ to make it positive.

When $h > 0$ and $x + h < 0$ (which happens when $|x| > h$, i.e. $x$ is "more negative" than $h$ is positive), you have $|x + h| = -(x + h) = -x - h$, and so $|x + h| - |x| = (-x - h) - (-x) = -x - h + x = -h$).

answered Jul 18 at 0:45

ConMan

6,9451324

$$left|xright| = begincases x & textif x geq 0 \ -x & textif x lt 0 endcases$$ Although I knew what the absolute operator did, this really helped me understand why $|x+h| = -(x+h)$
â€“Â Cedric Martens
Jul 18 at 0:58

Glad to have helped!
â€“Â ConMan
Jul 18 at 3:20

add a commentÂ |Â

up vote
1
down vote

If $x+h<0$ then all you can say is that $|x+h|=-x-h$.

Therefore what you are saying is that if $x+h<0$ then $-x-|x|=0$, or in other words $|x|=-x$.

Equivalently, you are saying that

if $x+h<0$ then $x<0$.

Can you see now why it is wrong?

It becomes true, however, if you add the requirement that $h>0$.

edited Jul 18 at 0:43

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

Yes I did omit that requirement of $h>0$, I will edit my question
â€“Â Cedric Martens
Jul 18 at 0:43

@CedricMartens If $h>0$, can you prove that $x+h<0implies x<0$?
â€“Â Arnaud Mortier
Jul 18 at 0:45

add a commentÂ |Â

up vote
0
down vote

This is a follow-on previous question. Please read about absolute value. When $x+h lt 0, |x+h|=-(x+h)$ directly from the definition. With the condition $h gt 0$ the fact that $x+h lt 0$ guarantees that $x lt 0$ and $|x|gt h$. In that case $$|x+h|-|x|=-(x+h)-(-x)=-x-h+x=-h$$

answered Jul 18 at 0:47

Ross Millikan

276k21187352

add a commentÂ |Â

up vote
0
down vote

We can get rid of some absolute since we know if it's positive/negative.

We have $-x-h-|x|=-h$ is what you're trying to prove which means

$x=-|x|.$ This simply means $x$ is not positive.

We have $|x-h|-|x|=-hiffx=-.$

This can be proved. Since $x+h<0$, $h>0,$ we have to have a positive plus a negative is less than $0,$ since a positive plus a nonnegative is still positive.

Therefore, we have proven that $x$ is negative, so $x=-|x|$ which proves $|x-h|-|x|=-h$ $boxedtextQ.E.D.$

answered Jul 18 at 0:52

Jason Kim

789

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
3
down vote

accepted

The absolute operator is defined as follows:

$$left|xright| = begincases
x & textif x geq 0 \
-x & textif x lt 0
endcases$$

In other words, if $x$ is positive (or zero), it stays the same. If $x$ is negative, then multiply it by $-1$ to make it positive.

answered Jul 18 at 0:45

ConMan

6,9451324

$$left|xright| = begincases x & textif x geq 0 \ -x & textif x lt 0 endcases$$ Although I knew what the absolute operator did, this really helped me understand why $|x+h| = -(x+h)$
â€“Â Cedric Martens
Jul 18 at 0:58

Glad to have helped!
â€“Â ConMan
Jul 18 at 3:20

add a commentÂ |Â

up vote
3
down vote

accepted

The absolute operator is defined as follows:

$$left|xright| = begincases
x & textif x geq 0 \
-x & textif x lt 0
endcases$$

In other words, if $x$ is positive (or zero), it stays the same. If $x$ is negative, then multiply it by $-1$ to make it positive.

answered Jul 18 at 0:45

ConMan

6,9451324

$$left|xright| = begincases x & textif x geq 0 \ -x & textif x lt 0 endcases$$ Although I knew what the absolute operator did, this really helped me understand why $|x+h| = -(x+h)$
â€“Â Cedric Martens
Jul 18 at 0:58

Glad to have helped!
â€“Â ConMan
Jul 18 at 3:20

add a commentÂ |Â

up vote
3
down vote

accepted

The absolute operator is defined as follows:

$$left|xright| = begincases
x & textif x geq 0 \
-x & textif x lt 0
endcases$$

In other words, if $x$ is positive (or zero), it stays the same. If $x$ is negative, then multiply it by $-1$ to make it positive.

answered Jul 18 at 0:45

ConMan

6,9451324

The absolute operator is defined as follows:

$$left|xright| = begincases
x & textif x geq 0 \
-x & textif x lt 0
endcases$$

In other words, if $x$ is positive (or zero), it stays the same. If $x$ is negative, then multiply it by $-1$ to make it positive.

answered Jul 18 at 0:45

ConMan

6,9451324

answered Jul 18 at 0:45

ConMan

6,9451324

answered Jul 18 at 0:45

ConMan

6,9451324

answered Jul 18 at 0:45

ConMan

6,9451324

$$left|xright| = begincases x & textif x geq 0 \ -x & textif x lt 0 endcases$$ Although I knew what the absolute operator did, this really helped me understand why $|x+h| = -(x+h)$
â€“Â Cedric Martens
Jul 18 at 0:58

Glad to have helped!
â€“Â ConMan
Jul 18 at 3:20

add a commentÂ |Â

$$left|xright| = begincases x & textif x geq 0 \ -x & textif x lt 0 endcases$$ Although I knew what the absolute operator did, this really helped me understand why $|x+h| = -(x+h)$
â€“Â Cedric Martens
Jul 18 at 0:58

Glad to have helped!
â€“Â ConMan
Jul 18 at 3:20

$$left|xright| = begincases x & textif x geq 0 \ -x & textif x lt 0 endcases$$ Although I knew what the absolute operator did, this really helped me understand why $|x+h| = -(x+h)$
â€“Â Cedric Martens
Jul 18 at 0:58

Glad to have helped!
â€“Â ConMan
Jul 18 at 3:20

add a commentÂ |Â

up vote
1
down vote

If $x+h<0$ then all you can say is that $|x+h|=-x-h$.

Therefore what you are saying is that if $x+h<0$ then $-x-|x|=0$, or in other words $|x|=-x$.

Equivalently, you are saying that

if $x+h<0$ then $x<0$.

Can you see now why it is wrong?

It becomes true, however, if you add the requirement that $h>0$.

edited Jul 18 at 0:43

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

Yes I did omit that requirement of $h>0$, I will edit my question
â€“Â Cedric Martens
Jul 18 at 0:43

@CedricMartens If $h>0$, can you prove that $x+h<0implies x<0$?
â€“Â Arnaud Mortier
Jul 18 at 0:45

add a commentÂ |Â

up vote
1
down vote

If $x+h<0$ then all you can say is that $|x+h|=-x-h$.

Therefore what you are saying is that if $x+h<0$ then $-x-|x|=0$, or in other words $|x|=-x$.

Equivalently, you are saying that

if $x+h<0$ then $x<0$.

Can you see now why it is wrong?

It becomes true, however, if you add the requirement that $h>0$.

edited Jul 18 at 0:43

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

Yes I did omit that requirement of $h>0$, I will edit my question
â€“Â Cedric Martens
Jul 18 at 0:43

@CedricMartens If $h>0$, can you prove that $x+h<0implies x<0$?
â€“Â Arnaud Mortier
Jul 18 at 0:45

add a commentÂ |Â

up vote
1
down vote

If $x+h<0$ then all you can say is that $|x+h|=-x-h$.

Therefore what you are saying is that if $x+h<0$ then $-x-|x|=0$, or in other words $|x|=-x$.

Equivalently, you are saying that

if $x+h<0$ then $x<0$.

Can you see now why it is wrong?

It becomes true, however, if you add the requirement that $h>0$.

edited Jul 18 at 0:43

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

If $x+h<0$ then all you can say is that $|x+h|=-x-h$.

Therefore what you are saying is that if $x+h<0$ then $-x-|x|=0$, or in other words $|x|=-x$.

Equivalently, you are saying that

if $x+h<0$ then $x<0$.

Can you see now why it is wrong?

It becomes true, however, if you add the requirement that $h>0$.

edited Jul 18 at 0:43

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

edited Jul 18 at 0:43

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

answered Jul 18 at 0:40

Arnaud Mortier

19.1k22159

Yes I did omit that requirement of $h>0$, I will edit my question
â€“Â Cedric Martens
Jul 18 at 0:43

@CedricMartens If $h>0$, can you prove that $x+h<0implies x<0$?
â€“Â Arnaud Mortier
Jul 18 at 0:45

add a commentÂ |Â

Yes I did omit that requirement of $h>0$, I will edit my question
â€“Â Cedric Martens
Jul 18 at 0:43

@CedricMartens If $h>0$, can you prove that $x+h<0implies x<0$?
â€“Â Arnaud Mortier
Jul 18 at 0:45

Yes I did omit that requirement of $h>0$, I will edit my question
â€“Â Cedric Martens
Jul 18 at 0:43

@CedricMartens If $h>0$, can you prove that $x+h<0implies x<0$?
â€“Â Arnaud Mortier
Jul 18 at 0:45

add a commentÂ |Â

up vote
0
down vote

answered Jul 18 at 0:47

Ross Millikan

276k21187352

add a commentÂ |Â

up vote
0
down vote

answered Jul 18 at 0:47

Ross Millikan

276k21187352

add a commentÂ |Â

up vote
0
down vote

answered Jul 18 at 0:47

Ross Millikan

276k21187352

answered Jul 18 at 0:47

Ross Millikan

276k21187352

answered Jul 18 at 0:47

Ross Millikan

276k21187352

answered Jul 18 at 0:47

Ross Millikan

276k21187352

answered Jul 18 at 0:47

Ross Millikan

276k21187352

add a commentÂ |Â

up vote
0
down vote

We can get rid of some absolute since we know if it's positive/negative.

We have $-x-h-|x|=-h$ is what you're trying to prove which means

$x=-|x|.$ This simply means $x$ is not positive.

We have $|x-h|-|x|=-hiffx=-.$

This can be proved. Since $x+h<0$, $h>0,$ we have to have a positive plus a negative is less than $0,$ since a positive plus a nonnegative is still positive.

Therefore, we have proven that $x$ is negative, so $x=-|x|$ which proves $|x-h|-|x|=-h$ $boxedtextQ.E.D.$

answered Jul 18 at 0:52

Jason Kim

789

add a commentÂ |Â

up vote
0
down vote

We can get rid of some absolute since we know if it's positive/negative.

We have $-x-h-|x|=-h$ is what you're trying to prove which means

$x=-|x|.$ This simply means $x$ is not positive.

We have $|x-h|-|x|=-hiffx=-.$

This can be proved. Since $x+h<0$, $h>0,$ we have to have a positive plus a negative is less than $0,$ since a positive plus a nonnegative is still positive.

Therefore, we have proven that $x$ is negative, so $x=-|x|$ which proves $|x-h|-|x|=-h$ $boxedtextQ.E.D.$

answered Jul 18 at 0:52

Jason Kim

789

add a commentÂ |Â

up vote
0
down vote

We can get rid of some absolute since we know if it's positive/negative.

We have $-x-h-|x|=-h$ is what you're trying to prove which means

$x=-|x|.$ This simply means $x$ is not positive.

We have $|x-h|-|x|=-hiffx=-.$

This can be proved. Since $x+h<0$, $h>0,$ we have to have a positive plus a negative is less than $0,$ since a positive plus a nonnegative is still positive.

Therefore, we have proven that $x$ is negative, so $x=-|x|$ which proves $|x-h|-|x|=-h$ $boxedtextQ.E.D.$

answered Jul 18 at 0:52

Jason Kim

789

We can get rid of some absolute since we know if it's positive/negative.

We have $-x-h-|x|=-h$ is what you're trying to prove which means

$x=-|x|.$ This simply means $x$ is not positive.

We have $|x-h|-|x|=-hiffx=-.$

This can be proved. Since $x+h<0$, $h>0,$ we have to have a positive plus a negative is less than $0,$ since a positive plus a nonnegative is still positive.

Therefore, we have proven that $x$ is negative, so $x=-|x|$ which proves $|x-h|-|x|=-h$ $boxedtextQ.E.D.$

answered Jul 18 at 0:52

Jason Kim

789

answered Jul 18 at 0:52

Jason Kim

789

answered Jul 18 at 0:52

Jason Kim

789

answered Jul 18 at 0:52

Jason Kim

789

add a commentÂ |Â

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