Proof i A,B are real number and |A| < inf then |B||A|>=|B|-|A|

Question

measure and intergration

Answer 1

Given: A,B∈ℝ, |A|<∞.

|B||A|≥0 because |B| and |A| are both positive by definition of the absolute function (||).

If |B|<|A|, then |B|-|A|<0, hence |B||A|≥|B|-|A|, because all positive quantities are greater than negative quantities.

If |B|=|A| then |B|-|A|=0, so |B||A|≥|B|-|A| because |B||A|≥|B|-|A|=0.

If |B|>|A| then |B|=|A|+k where k>0; |B||A|=(|A|+k)|A|=A²+k|A|; |B|-|A|=k.

|B||A|-(|B|-|A|)=A²+k|A|-k=A²+k(|A|-1). If |A|<1 then A²<|A| so k(|A|-1)<0 and A²+k(|A|-1) can be negative.

For example, let |A|=½, then A²=¼, so we have ¼-½k<0 when ¼<½k, that is, when k>½.

This means that if, for example, |A|=½ and |B|=2, then |B||A|-(|B|-|A|)=1-1½<0, so  |B||A|<|B|-|A|.

Therefore the proposed statement is false.