Mixing
How to solve the equality $|x−y|=|x|−|y|$ given that both $x$ and $y$ are of same sign and $|x|>|y|$
2
I have tried using the method of
$|x| = |x-y+y| < |x-y| + |y|$
so $|x| - |y| < |x - y|$
But the above statement not right coz the greater than sign comes in because of Triangle inequality
inequality
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
asked Nov 20 '18 at 19:43
RiRi
143
add a comment |
2
I have tried using the method of
$|x| = |x-y+y| < |x-y| + |y|$
so $|x| - |y| < |x - y|$
But the above statement not right coz the greater than sign comes in because of Triangle inequality
inequality
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
asked Nov 20 '18 at 19:43
RiRi
143
The title presents an equality.
– gimusi
Nov 20 '18 at 19:46
thank you sir. I fixed it
– RiRi
Nov 20 '18 at 19:50
add a comment |
2
2
2
I have tried using the method of
$|x| = |x-y+y| < |x-y| + |y|$
so $|x| - |y| < |x - y|$
But the above statement not right coz the greater than sign comes in because of Triangle inequality
inequality
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
asked Nov 20 '18 at 19:43
RiRi
143
I have tried using the method of
$|x| = |x-y+y| < |x-y| + |y|$
so $|x| - |y| < |x - y|$
But the above statement not right coz the greater than sign comes in because of Triangle inequality
inequality
inequality
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
asked Nov 20 '18 at 19:43
RiRi
143
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
asked Nov 20 '18 at 19:43
RiRi
143
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
edited Nov 20 '18 at 19:52
Key Flex
7,53441232
7,53441232
asked Nov 20 '18 at 19:43
RiRi
143
asked Nov 20 '18 at 19:43
RiRi
143
asked Nov 20 '18 at 19:43
RiRi
143
143
The title presents an equality.
– gimusi
Nov 20 '18 at 19:46
thank you sir. I fixed it
– RiRi
Nov 20 '18 at 19:50
add a comment |
The title presents an equality.
– gimusi
Nov 20 '18 at 19:46
thank you sir. I fixed it
– RiRi
Nov 20 '18 at 19:50
The title presents an equality.
– gimusi
Nov 20 '18 at 19:46
The title presents an equality.
– gimusi
Nov 20 '18 at 19:46
thank you sir. I fixed it
– RiRi
Nov 20 '18 at 19:50
thank you sir. I fixed it
– RiRi
Nov 20 '18 at 19:50
add a comment |
2 Answers
2
active
oldest
votes
1
We can assume that $xge 0$ and $yge 0$.
then
$$|x|>|y|implies x>y$$
$$ implies |x-y|=x-y=|x|-|y|$$
it is always satisfied.
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
add a comment |
1
We have two cases
$x,y>0 quad |x|>|y| implies x>y$
$$|x−y|=x-y=|x|-|y|$$
$x,y<0 quad |x|>|y| implies x<y$
$$|x−y|=y-x=|x|-|y|$$
answered Nov 20 '18 at 19:51
gimusi
1
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
|
show 2 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
We can assume that $xge 0$ and $yge 0$.
then
$$|x|>|y|implies x>y$$
$$ implies |x-y|=x-y=|x|-|y|$$
it is always satisfied.
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
add a comment |
1
We can assume that $xge 0$ and $yge 0$.
then
$$|x|>|y|implies x>y$$
$$ implies |x-y|=x-y=|x|-|y|$$
it is always satisfied.
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
add a comment |
1
1
1
We can assume that $xge 0$ and $yge 0$.
then
$$|x|>|y|implies x>y$$
$$ implies |x-y|=x-y=|x|-|y|$$
it is always satisfied.
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
We can assume that $xge 0$ and $yge 0$.
then
$$|x|>|y|implies x>y$$
$$ implies |x-y|=x-y=|x|-|y|$$
it is always satisfied.
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 19:48
hamam_Abdallah
37.9k21634
37.9k21634
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
add a comment |
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
Thank you very much!
– RiRi
Nov 20 '18 at 20:08
add a comment |
1
We have two cases
$x,y>0 quad |x|>|y| implies x>y$
$$|x−y|=x-y=|x|-|y|$$
$x,y<0 quad |x|>|y| implies x<y$
$$|x−y|=y-x=|x|-|y|$$
answered Nov 20 '18 at 19:51
gimusi
1
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
|
show 2 more comments
1
We have two cases
$x,y>0 quad |x|>|y| implies x>y$
$$|x−y|=x-y=|x|-|y|$$
$x,y<0 quad |x|>|y| implies x<y$
$$|x−y|=y-x=|x|-|y|$$
answered Nov 20 '18 at 19:51
gimusi
1
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
|
show 2 more comments
1
1
1
We have two cases
$x,y>0 quad |x|>|y| implies x>y$
$$|x−y|=x-y=|x|-|y|$$
$x,y<0 quad |x|>|y| implies x<y$
$$|x−y|=y-x=|x|-|y|$$
answered Nov 20 '18 at 19:51
gimusi
1
We have two cases
$x,y>0 quad |x|>|y| implies x>y$
$$|x−y|=x-y=|x|-|y|$$
$x,y<0 quad |x|>|y| implies x<y$
$$|x−y|=y-x=|x|-|y|$$
answered Nov 20 '18 at 19:51
gimusi
1
edited Nov 20 '18 at 20:04
answered Nov 20 '18 at 19:51
gimusi
1
answered Nov 20 '18 at 19:51
gimusi
1
answered Nov 20 '18 at 19:51
gimusi
1
1
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
|
show 2 more comments
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
Hi sir thank you for answering. I just wanted to know will it not give a different outcome since |x| - |y| is not equal to |y| - |x| right?
– RiRi
Nov 20 '18 at 20:00
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi Ops sorry there was a typo for the second one!
– gimusi
Nov 20 '18 at 20:04
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
@RiRi The fact is that $|x-y|=|y-x|>0$ and for $x$ and $y$ with the same sign assuming $|x|>|y|$ we have $|x-y|=|x|-|y|$.
– gimusi
Nov 20 '18 at 20:07
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
Aaah i see now, thank you very much sir
– RiRi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
@RiRi To convince yourself let try with $x=4$ and $y=3$ for the first case and $x=-4$ and $y=-3$ for the second one.
– gimusi
Nov 20 '18 at 20:08
|
show 2 more comments
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The title presents an equality.
– gimusi
Nov 20 '18 at 19:46
thank you sir. I fixed it
– RiRi
Nov 20 '18 at 19:50