Question #4182

If [i]a,b[/i]& are element of real numbers, show that [b]|a+b|=|a|+|b|[/b] & if [b]ab>=0[/b]

Expert's answer

from the definition of |a+b|:|a+b|=a+b if a+b>=0 and |a+b|=-(a+b) if a+b<=0

from the condition ab>=0 we have two opportunities 1) a,b>=0 or 2) a,b<=0

so let's consider the case |a+b|=a+b if a+b>=0

a+b>=0, and 1) a,b>=0 or 2) a,b<=0 =>a,b>=0 Hence |a+b|=a+b=|a|+|b|

now the second case |a+b|=-(a+b) if a+b<0

a+b<=0 and 1) a,b>=0 or 2) a,b<=0 =>a,b<=0

Hence |a+b|=-(a+b)=-a-b=|a|+|b|

so we showed that |a+b|=|a|+|b| if ab >=0.

from the condition ab>=0 we have two opportunities 1) a,b>=0 or 2) a,b<=0

so let's consider the case |a+b|=a+b if a+b>=0

a+b>=0, and 1) a,b>=0 or 2) a,b<=0 =>a,b>=0 Hence |a+b|=a+b=|a|+|b|

now the second case |a+b|=-(a+b) if a+b<0

a+b<=0 and 1) a,b>=0 or 2) a,b<=0 =>a,b<=0

Hence |a+b|=-(a+b)=-a-b=|a|+|b|

so we showed that |a+b|=|a|+|b| if ab >=0.

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