a← 1 + 0j1×1+⎕ct b← (1+⎕ct) + 0j1×1+⎕ct×2 a = b ⍝ therefore (|a-b) ≤ ⎕ct×(|a)⌈|b 1 (|a) = |b ⍝ therefore (|(|a)-|b) > ⎕ct×(|a)⌈|b 0 ⍝ therefore (|(|a)-|b) > |a-b