What is $frac{1}{|{x}|}-frac{x^2}{|x|^3}$?
up vote
8
down vote
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What's the result of:
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}$$
Is it
$$frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|x|}=0$$
or
$$frac{1}{|{x}|}-frac{x^2}{|x|^2x}=frac{1}{|{x}|}-frac{1}{x}frac{x^2}{|x|^2}=frac{1}{|{x}|}-frac{1}{x}=left{begin{matrix}frac{2}{x},x<0\ 0,x>0 end{matrix}right.$$
algebra-precalculus absolute-value
edited Nov 24 at 13:36
user21820
38.6k542152
asked Nov 23 at 21:07
BinaryBurst
359110
add a comment |
up vote
8
down vote
favorite
What's the result of:
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}$$
Is it
$$frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|x|}=0$$
or
$$frac{1}{|{x}|}-frac{x^2}{|x|^2x}=frac{1}{|{x}|}-frac{1}{x}frac{x^2}{|x|^2}=frac{1}{|{x}|}-frac{1}{x}=left{begin{matrix}frac{2}{x},x<0\ 0,x>0 end{matrix}right.$$
algebra-precalculus absolute-value
edited Nov 24 at 13:36
user21820
38.6k542152
asked Nov 23 at 21:07
BinaryBurst
359110
7
First one is correct $|x|^2=x^2$
– Yadati Kiran
Nov 23 at 21:08
4
Just insert $-1$. Which of the expressions are equal? (Note that this by itself is not a proof, as you have to rule out that both expressions are false, but it lets you quickly see that one of them simply cannot be right).
– celtschk
Nov 23 at 22:17
1
Please also note that none of this would work with complex numbers ($xinmathbb C$).
– Martin Rosenau
Nov 24 at 13:38
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
What's the result of:
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}$$
Is it
$$frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|x|}=0$$
or
$$frac{1}{|{x}|}-frac{x^2}{|x|^2x}=frac{1}{|{x}|}-frac{1}{x}frac{x^2}{|x|^2}=frac{1}{|{x}|}-frac{1}{x}=left{begin{matrix}frac{2}{x},x<0\ 0,x>0 end{matrix}right.$$
algebra-precalculus absolute-value
edited Nov 24 at 13:36
user21820
38.6k542152
asked Nov 23 at 21:07
BinaryBurst
359110
What's the result of:
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}$$
Is it
$$frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|x|}=0$$
or
$$frac{1}{|{x}|}-frac{x^2}{|x|^2x}=frac{1}{|{x}|}-frac{1}{x}frac{x^2}{|x|^2}=frac{1}{|{x}|}-frac{1}{x}=left{begin{matrix}frac{2}{x},x<0\ 0,x>0 end{matrix}right.$$
algebra-precalculus absolute-value
algebra-precalculus absolute-value
edited Nov 24 at 13:36
user21820
38.6k542152
asked Nov 23 at 21:07
BinaryBurst
359110
edited Nov 24 at 13:36
user21820
38.6k542152
asked Nov 23 at 21:07
BinaryBurst
359110
edited Nov 24 at 13:36
user21820
38.6k542152
edited Nov 24 at 13:36
user21820
38.6k542152
edited Nov 24 at 13:36
user21820
38.6k542152
38.6k542152
asked Nov 23 at 21:07
BinaryBurst
359110
asked Nov 23 at 21:07
BinaryBurst
359110
asked Nov 23 at 21:07
BinaryBurst
359110
359110
7
First one is correct $|x|^2=x^2$
– Yadati Kiran
Nov 23 at 21:08
4
Just insert $-1$. Which of the expressions are equal? (Note that this by itself is not a proof, as you have to rule out that both expressions are false, but it lets you quickly see that one of them simply cannot be right).
– celtschk
Nov 23 at 22:17
1
Please also note that none of this would work with complex numbers ($xinmathbb C$).
– Martin Rosenau
Nov 24 at 13:38
add a comment |
7
First one is correct $|x|^2=x^2$
– Yadati Kiran
Nov 23 at 21:08
4
Just insert $-1$. Which of the expressions are equal? (Note that this by itself is not a proof, as you have to rule out that both expressions are false, but it lets you quickly see that one of them simply cannot be right).
– celtschk
Nov 23 at 22:17
1
Please also note that none of this would work with complex numbers ($xinmathbb C$).
– Martin Rosenau
Nov 24 at 13:38
7
7
First one is correct $|x|^2=x^2$
– Yadati Kiran
Nov 23 at 21:08
First one is correct $|x|^2=x^2$
– Yadati Kiran
Nov 23 at 21:08
4
4
Just insert $-1$. Which of the expressions are equal? (Note that this by itself is not a proof, as you have to rule out that both expressions are false, but it lets you quickly see that one of them simply cannot be right).
– celtschk
Nov 23 at 22:17
Just insert $-1$. Which of the expressions are equal? (Note that this by itself is not a proof, as you have to rule out that both expressions are false, but it lets you quickly see that one of them simply cannot be right).
– celtschk
Nov 23 at 22:17
1
1
Please also note that none of this would work with complex numbers ($xinmathbb C$).
– Martin Rosenau
Nov 24 at 13:38
Please also note that none of this would work with complex numbers ($xinmathbb C$).
– Martin Rosenau
Nov 24 at 13:38
add a comment |
4 Answers
4
active
oldest
votes
up vote
21
down vote
accepted
As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.
The error in the second proof is in the very beginning, namely $frac{1}{|{x}|}-frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.
answered Nov 23 at 21:14
Scientifica
6,34141333
add a comment |
up vote
5
down vote
Your first conclusion is right since $$|x|^3=|x|^2cdot |x|=x^2cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3ne x^3= |x|^2cdot x$$
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
add a comment |
up vote
2
down vote
Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}=frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
answered Nov 23 at 21:12
gimusi
1
add a comment |
up vote
1
down vote
As the other answers already told you, the first one is true.
But maybe this way of thinking gives you an intuition: Both terms $frac{1}{|x|}$ and $frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.
answered Nov 24 at 10:18
Fritz Hefter
113
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
21
down vote
accepted
As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.
The error in the second proof is in the very beginning, namely $frac{1}{|{x}|}-frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.
answered Nov 23 at 21:14
Scientifica
6,34141333
add a comment |
up vote
21
down vote
accepted
As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.
The error in the second proof is in the very beginning, namely $frac{1}{|{x}|}-frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.
answered Nov 23 at 21:14
Scientifica
6,34141333
add a comment |
up vote
21
down vote
accepted
up vote
21
down vote
accepted
As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.
The error in the second proof is in the very beginning, namely $frac{1}{|{x}|}-frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.
answered Nov 23 at 21:14
Scientifica
6,34141333
As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.
The error in the second proof is in the very beginning, namely $frac{1}{|{x}|}-frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.
answered Nov 23 at 21:14
Scientifica
6,34141333
answered Nov 23 at 21:14
Scientifica
6,34141333
answered Nov 23 at 21:14
Scientifica
6,34141333
answered Nov 23 at 21:14
Scientifica
6,34141333
6,34141333
add a comment |
add a comment |
up vote
5
down vote
Your first conclusion is right since $$|x|^3=|x|^2cdot |x|=x^2cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3ne x^3= |x|^2cdot x$$
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
add a comment |
up vote
5
down vote
Your first conclusion is right since $$|x|^3=|x|^2cdot |x|=x^2cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3ne x^3= |x|^2cdot x$$
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
add a comment |
up vote
5
down vote
up vote
5
down vote
Your first conclusion is right since $$|x|^3=|x|^2cdot |x|=x^2cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3ne x^3= |x|^2cdot x$$
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
Your first conclusion is right since $$|x|^3=|x|^2cdot |x|=x^2cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3ne x^3= |x|^2cdot x$$
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
answered Nov 23 at 21:13
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
add a comment |
up vote
2
down vote
Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}=frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
answered Nov 23 at 21:12
gimusi
1
add a comment |
up vote
2
down vote
Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}=frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
answered Nov 23 at 21:12
gimusi
1
add a comment |
up vote
2
down vote
up vote
2
down vote
Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}=frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
answered Nov 23 at 21:12
gimusi
1
Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}=frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
answered Nov 23 at 21:12
gimusi
1
answered Nov 23 at 21:12
gimusi
1
answered Nov 23 at 21:12
gimusi
1
answered Nov 23 at 21:12
gimusi
1
1
add a comment |
add a comment |
up vote
1
down vote
As the other answers already told you, the first one is true.
But maybe this way of thinking gives you an intuition: Both terms $frac{1}{|x|}$ and $frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.
answered Nov 24 at 10:18
Fritz Hefter
113
add a comment |
up vote
1
down vote
As the other answers already told you, the first one is true.
But maybe this way of thinking gives you an intuition: Both terms $frac{1}{|x|}$ and $frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.
answered Nov 24 at 10:18
Fritz Hefter
113
add a comment |
up vote
1
down vote
up vote
1
down vote
As the other answers already told you, the first one is true.
But maybe this way of thinking gives you an intuition: Both terms $frac{1}{|x|}$ and $frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.
answered Nov 24 at 10:18
Fritz Hefter
113
As the other answers already told you, the first one is true.
But maybe this way of thinking gives you an intuition: Both terms $frac{1}{|x|}$ and $frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.
answered Nov 24 at 10:18
Fritz Hefter
113
answered Nov 24 at 10:18
Fritz Hefter
113
answered Nov 24 at 10:18
Fritz Hefter
113
answered Nov 24 at 10:18
Fritz Hefter
113
113
add a comment |
add a comment |
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7
First one is correct $|x|^2=x^2$
– Yadati Kiran
Nov 23 at 21:08
4
Just insert $-1$. Which of the expressions are equal? (Note that this by itself is not a proof, as you have to rule out that both expressions are false, but it lets you quickly see that one of them simply cannot be right).
– celtschk
Nov 23 at 22:17
1
Please also note that none of this would work with complex numbers ($xinmathbb C$).
– Martin Rosenau
Nov 24 at 13:38