Unformatted text preview: xy + x ) + ( y + 1) = ( xy + y ) + ( x + 1) we may cancel xy on the left and 1 on the right, repeatedly using AA , AID and AIV to obtain x + y = y + x as desired. Extra credit : We may certainly dispense with MID in (a). Begin by noting that b D and AID yield 0 z + 0 z = z (0 + 0) = z 0. If we knew that z = z 0 then we could cancel an appearance on each side and be ﬁnished. Now b D and AID again yield 0 z = 0( z + 0) = z 0 + 00. If we knew that 00 = 0 then we would be ﬁnished by AID . Finally: 00 = 0(0 + 0) = 00 + 00 by further application of b D and AID , and cancellation settles things. Remark : Observe what happens when we replace b D by the requirement that if a, b, c ∈ R then a ( b + c ) = ca + ba (a further twist!): setting a = 1 at once shows that b + c = c + b ! 1...
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- Fall '08
- one hand, Axiom, distributive axiom, distributive axiom D.