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# I read somewhere that whenever you see the expression |x+y|=

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VP
Joined: 21 Jul 2006
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31 Jul 2008, 02:07
I read somewhere that whenever you see the expression |x+y|= |x| + |y|, then it means that both x and y have the same sign. So can someone then please explain what it means when you see the expression |x+y| |x| + |y|?

thanks

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Joined: 24 Jun 2008
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31 Jul 2008, 04:24
tarek99 wrote:
I read somewhere that whenever you see the expression |x+y|= |x| + |y|, then it means that both x and y have the same sign. So can someone then please explain what it means when you see the expression |x+y|< |x| + |y| and |x+y|> |x| + |y|?

thanks

If you see the expression |x+y| > |x| + |y|, that means you're in an alternate universe. This can never be true for real numbers x and y. If you see the expression |x+y| < |x| + |y|, then x and y have opposite signs. Lastly I'd note that if you see the expression |x+y| = |x| + |y|, that doesn't guarantee that x and y have the same sign- one of them could be equal to zero.
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VP
Joined: 21 Jul 2006
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Updated on: 31 Jul 2008, 06:30
IanStewart wrote:
tarek99 wrote:
I read somewhere that whenever you see the expression |x+y|= |x| + |y|, then it means that both x and y have the same sign. So can someone then please explain what it means when you see the expression |x+y|< |x| + |y| and |x+y|> |x| + |y|?

thanks

If you see the expression |x+y| > |x| + |y|, that means you're in an alternate universe. This can never be true for real numbers x and y. If you see the expression |x+y| < |x| + |y|, then x and y have opposite signs. Lastly I'd note that if you see the expression |x+y| = |x| + |y|, that doesn't guarantee that x and y have the same sign- one of them could be equal to zero.

That really cleared things up....thanks a lot! |x+y| = |x| + |y| would suggest that x and y have the same sign if only we were given an expression xy doesn't equal to zero.

however, if we were give |x+y| is less than or equal to |x| + |y|, then that would mean that either both have the same sign, or both have different signs, or one of them could be zero....correct?

Originally posted by tarek99 on 31 Jul 2008, 05:35.
Last edited by tarek99 on 31 Jul 2008, 06:30, edited 1 time in total.
VP
Joined: 21 Jul 2006
Posts: 1447

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31 Jul 2008, 05:41
Also, what does this expression mean when we have a minus sign?

|x-y| = |x| - |y| or

|x-y| = |y| - |x|

or, when we use the greater or less sign? for example:

|x-y| < |x| - |y|

|x-y| > |x| - |y|

I would really appreciate your input!
thanks
VP
Joined: 17 Jun 2008
Posts: 1279

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03 Aug 2008, 08:03
tarek99 wrote:
I read somewhere that whenever you see the expression |x+y|= |x| + |y|, then it means that both x and y have the same sign. So can someone then please explain what it means when you see the expression |x+y|< |x| + |y| and |x+y|> |x| + |y|?

thanks

the expr(1) : |x+y|= |x| + |y| means that mod of sum is equal to sum of mods.
When we say mod it means magnitude or abolute value irrespective of sign of a number
in expr (1) is true when both x and y are of same sign since value on LHS and RHS are equal => magnitude of -(x+y) = x+y sum of magnitudes : |-x| +|-y| = x+y again hence LHS RHS same

now say in expr (1) ,x and y have opp signs then in LHS : magnitude of x+y < x+y when both have same signs i.e < |x|+|y|

|x+y|< |x| + |y| => this implies x,y have opp signs

|x+y|>|x| + |y| => there is no condition called this one since sum of the magnitudes can never be greater than RHS of this expr,Value of RHS is independent of sign.

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VP
Joined: 17 Jun 2008
Posts: 1279

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03 Aug 2008, 08:15
tarek99 wrote:
Also, what does this expression mean when we have a minus sign?

|x-y| = |x| - |y| or

|x-y| = |y| - |x|

or, when we use the greater or less sign? for example:

|x-y| < |x| - |y|

|x-y| > |x| - |y|

I would really appreciate your input!
thanks

consider :
|x-y| = |x| - |y| or |x-y| = |y| - |x|
we are just taking difference in magnitudes or simple value (irrespective of sign )
if both x and y have same signs then above equation is true ,since the operation subtraction is subtraction if both variabl;es are of same sign.
|x-y| = |x| - |y| => this implies x,y same sign and x>y since LHS is +ve and hence RHS is +ve
similarly :
|x-y| = |y| - |x|=> this implies x,y same sign and y>x since LHS is +ve and hence RHS is +ve

consider eqn:
|x-y| > |x| - |y|
this says mod of difference > diff of modes
now diff of mods is least always since max of x and max of y are subtracted.
if in case x and y have opp signs then subtraction operation becomes addition hence mod of diff of x,y with opp signs is > than actual diff of mods.

hence |x-y| > |x| - |y| => this implies x,y opp sign and x>y since LHS is +ve and hence RHS is +ve

similarly

|x-y| > |y| - |x| => this implies x,y opp sign and y>x since LHS is +ve and hence RHS is +ve

|x-y| < |x| - |y| => this conmdition is not feasible since minimum diff between two numbers is the diff(in magnitude) between their mods.INVALID CONDN.
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cheers
Its Now Or Never

VP
Joined: 21 Jul 2006
Posts: 1447

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10 Aug 2008, 09:59
spriya wrote:
tarek99 wrote:
Also, what does this expression mean when we have a minus sign?

|x-y| = |x| - |y| or

|x-y| = |y| - |x|

or, when we use the greater or less sign? for example:

|x-y| < |x| - |y|

|x-y| > |x| - |y|

I would really appreciate your input!
thanks

consider :
|x-y| = |x| - |y| or |x-y| = |y| - |x|
we are just taking difference in magnitudes or simple value (irrespective of sign )
if both x and y have same signs then above equation is true ,since the operation subtraction is subtraction if both variabl;es are of same sign.
|x-y| = |x| - |y| => this implies x,y same sign and x>y since LHS is +ve and hence RHS is +ve
similarly :
|x-y| = |y| - |x|=> this implies x,y same sign and y>x since LHS is +ve and hence RHS is +ve

consider eqn:
|x-y| > |x| - |y|
this says mod of difference > diff of modes
now diff of mods is least always since max of x and max of y are subtracted.
if in case x and y have opp signs then subtraction operation becomes addition hence mod of diff of x,y with opp signs is > than actual diff of mods.

hence |x-y| > |x| - |y| => this implies x,y opp sign and x>y since LHS is +ve and hence RHS is +ve

similarly

|x-y| > |y| - |x| => this implies x,y opp sign and y>x since LHS is +ve and hence RHS is +ve

|x-y| < |x| - |y| => this conmdition is not feasible since minimum diff between two numbers is the diff(in magnitude) between their mods.INVALID CONDN.

great explanation. Just what do you mean by RHS and LHS?
thanks
Intern
Joined: 21 Sep 2006
Posts: 9

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31 Aug 2008, 22:22
RHS = Right Hand Side
LHS = Left Hand Side

When you simplify for an unknown value, LHS side of the equation should match the RHS, in general.

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Re: Math question &nbs [#permalink] 31 Aug 2008, 22:22
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# I read somewhere that whenever you see the expression |x+y|=

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