For all real numbers x and y, |x+y| < |x|+|y|. This resultis called the triangle

Question

For all real numbers x and y, |x+y| < |x|+|y|. This resultis called the triangle inequality. -------------------------------------------------------------------------------------------------------------
This is what I got,
Case 1.
x <= |x| y <= |y|
so x+y<= |x|+|y| if x+y >= 0 then x+y= |x+y| and then |x+y| <= |x| + |y|proving this case.
I know there is 4 total cases but I got stuck. I need help.Please. -------------------------------------------------------------------------------------------------------------
This is what I got,
Case 1.
x <= |x| y <= |y|
so x+y<= |x|+|y| if x+y >= 0 then x+y= |x+y| and then |x+y| <= |x| + |y|proving this case.
I know there is 4 total cases but I got stuck. I need help.Please.

Explanation / Answer

Question Details: For all real numbers x andy, |x+y| < |x|+|y|. This result is called the triangleinequality. -------------------------------------------------------------------------------------------------------------
This is what I got, -------------------------------------------------------------------------------------------------------------
Case 1.
x <= |x| y <= |y|
so x+y<= |x|+|y| if x+y >= 0 then x+y= |x+y| and then |x+y| <= |x| + |y|proving this case.
I know there is 4 total cases but I got stuck. I need help.Please.
===============NO...TAKE THE CASES AS FOLLOWS ============
CASE 1
X>=0 AND Y>=0...THAT IS |X|=X AND |Y|=Y
CASE 2
X<0 AND Y<0...THAT IS |X|=-X AND |Y|=-Y
CASE 3
X>=0 AND Y<0...THAT IS |X|=X AND |Y|=-Y
CASE 4
X<0 AND Y>=0...THAT IS |X|=-X AND |Y|=Y
SO WE GET
==========================
CASE 1
X>=0 AND Y>=0...THAT IS |X|=X AND |Y|=Y......[TAKE X=Y=1 FORCLARITY IF NEEDED]
LHS=|X+Y|=X+Y
RHS=|X|+|Y|=X+Y
HENCE WE GET THE GIVEN PROPOSITION
|X+Y|<=|X|+|Y| AS CORRECT
-----------------------------------------
CASE 2
X<0 AND Y<0...THAT IS |X|=-X AND |Y|=-Y......[TAKE X=Y=-1 FORCLARITY IF NEEDED]
LHS=|X+Y|=-[X+Y]
RHS=|X|+|Y|=-X-Y=-[X+Y]
HENCE WE GET THE GIVEN PROPOSITION
|X+Y|<=|X|+|Y| AS CORRECT
------------------------------------------------------
CASE 3
X>=0 AND Y<0...THAT IS |X|=X AND |Y|=-Y......[TAKE X=1 ANDY=-2 FOR CLARITY IF NEEDED]
LHS=|X+Y|=POSITIVE VALUE OF DIFFERENCE BETWEEN X AND Y
RHS=|X|+|Y|=X-Y= SUM OF POSITIVE VALUES OF X AND Y .
HENCE WE GET THE GIVEN PROPOSITION
|X+Y|<=|X|+|Y| AS CORRECT
-------------------------------------------------
CASE 4
X<0 AND Y>=0...THAT IS |X|=-X AND |Y|=Y......[TAKE X=-2 ANDY=1 FOR CLARITY IF NEEDED]
LHS=|X+Y|=POSITIVE VALUE OF DIFFERENCE BETWEEN X AND Y
RHS=|X|+|Y|=-X+Y= SUM OF POSITIVE VALUES OF X AND Y .
HENCE WE GET THE GIVEN PROPOSITION
|X+Y|<=|X|+|Y| AS CORRECT

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