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Verify the rule that for two real numbers X and Y then

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  • #1
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Homework Statement



Verify the rule that for two real numbers X and Y then |X+Y|≤|X|+|Y|



Homework Equations


The Attempt at a Solution



1. When all the variables are positive:

|X+Y|=(X+Y) because (X+Y)>0

|X|=X because X>0

|Y|=Y because Y>0

So we got

(X+Y)=X+Y

2. When all the variables are negative:

|-X+(-Y)|= -(-[X+Y])= (X+Y) because -(X+Y)<0

|-X|= -(-X)=X because -X<0

|-Y|= -(-Y)=Y because -Y<0

Now we got

(X+Y)=X+Y

3. When the two variables have opposite signs:

|X+(-Y)|= (X-Y) because (X-Y)>0

|X|=X because X>0

|-Y|= -(-Y)=Y because Y<0

Finally we got:

(X-Y)< X+Y

==================================================

So that solution was shot down as a bit shaky. How about the one below then?

|X+Y|≤|X|+|Y|

|X+Y|-|X|-|Y|≤0

If we remove the variables from the modulus we get

X+Y-X-Y≤0

Thanks.
 

Answers and Replies

  • #2
8
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Not sure if this works

|X+Y|≤|X|+|Y|

|X+Y|-|X|-|Y|≤0

0≤ X+Y-X-Y≤ 0

0≤ 0≤ 0
 
  • #3
33,270
4,975

Homework Statement



Verify the rule that for two real numbers X and Y then |X+Y|≤|X|+|Y|



Homework Equations


The Attempt at a Solution



1. When all the variables are positive:

|X+Y|=(X+Y) because (X+Y)>0

|X|=X because X>0

|Y|=Y because Y>0

So we got

(X+Y)=X+Y
Here's a better way to say this.
If x >= 0, and y >= 0,

|x + y| = x + y = |x| + |y|
2. When all the variables are negative:

|-X+(-Y)|= -(-[X+Y])= (X+Y) because -(X+Y)<0

|-X|= -(-X)=X because -X<0

|-Y|= -(-Y)=Y because -Y<0

Now we got

(X+Y)=X+Y
You need to show, when x < 0 and y < 0, that |x + y| <= |x| + |y|. It looks like you showed (proved) that (x + y) = x + y. This is true, but not what you're supposed to show.

Start with |x + y| and show that this equals |x| + |y|.

3. When the two variables have opposite signs:
Similar comments as above in your case 2. It is in these two cases (3 and 4) that you'll end up with actual inequalities.
|X+(-Y)|= (X-Y) because (X-Y)>0
The thing is, you're not being asked about |x + (-y)| - you need to show that |x + y| < |x| + |y|.
|X|=X because X>0

|-Y|= -(-Y)=Y because Y<0

Finally we got:

(X-Y)< X+Y

==================================================

So that solution was shot down as a bit shaky. How about the one below then?

|X+Y|≤|X|+|Y|

|X+Y|-|X|-|Y|≤0

If we remove the variables from the modulus we get

X+Y-X-Y≤0
 

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