high school proof question, absolute value?

Question

for any two numbers a and b proove abs(ab) = abs(a)*abs(b)

Answer 1

Recall the definition of absolute value:

|x| = {x if x≥0, -x if x<0}

Note that if x≤0, then either x<0, so |x| = -x, or x = 0, thus |x| = |0| = 0 = -0 = -x, so in either case x≤0 ⇒ |x| = -x. This simplifies the proof slightly. Now, we divide into four cases:

Case 1: a≥0, b≥0
In this case, ab≥0, so |ab| = ab = |a|·|b|

Case 2: a<0, b≥0
In this case, ab≤0, so |ab| = -ab = (-a)b = |a|·|b|

Case 3: a≥0, b<0
In this case, ab≤0, so |ab| = -ab = a(-b) = |a|·|b|

Case 4: a<0, b<0
In this case, ab>0, so |ab| = ab = (-a)(-b) = |a|·|b|

So in all of these cases, |ab| = |a|·|b|. Since these cases are collectively exhaustive, it follows that |ab| = |a|·|b| is an identity, and we are done.

Answer 2

Absolute Value Proof

Answer 3

i think u mean absolute value when u say abs? well, next time just us to Ls, (l l) to show absolute value.
l ab l= l a l * l b l
well, yes, because despite wat the actual values of a and b are, l a l and l b l and l ab l will all be positive. just input values:
a=2, b=-3
l (2)(-3) l = l 2 l * l -3 l
l-6l=2*3
6=6

Answer 4

Just work through all four possible cases:

a > 0, b > 0:
|ab| = ab = |a| |b|

a > 0, b < 0:
|ab| = -ab = a(-b) = |a| |b|

a < 0, b > 0:
|ab| = -ab = (-a)b = |a| |b|

a < 0, b < 0:
|ab| = ab = (-1)(-1)ab = (-a)(-b) = |a| |b|

Answer 5

Well, I don't really know how to right a proof, but it makes sense if you think about it -- The absolute value of a number will always be positive, so the "positive form", or whatever you want to call it, of a * b would be equal to the "positive form" of a times the "positive form" of b