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# nphw1 - xy x y 1 = xy y x 1 we may cancel xy on the left...

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Numbers and Polynomials Homework 1 0. Assume axioms AA , AID , AIV , MID and the following variant on the distributive axiom: D : if a, b, c R then a ( b + c ) = ba + ca . (a) Show that if z R then 0 z = 0. (b) Show that if x, y R then xy = yx . (c) Show that if x, y R then x + y = y + x . Extra credit : How far may we dispense with axiom MID ? Solutions : (a) Note first that 0 z + z MID = 0 z + 1 z D = z (0 + 1) AID = z 1 MID = z and then cancel z (add - z to both sides and invoke AA , AID , AIV as usual) to deduce 0 z = 0 as required. (b) Observe that the form of D effects a reversal of order in products. Take this as a hint: if c R is arbitrary then x ( y + c ) = yx + cx ; in particular, taking c to be 0 gives x ( y + 0) = yx + 0 x so that xy = yx on account of part (a) along with AID . (c) Notice that (b) converts D into the full distributive axiom D . Expand ( x + 1)( y + 1) in two ways using D : on the one hand, one obtains x ( y + 1) + 1( y + 1) = ( xy + x ) + ( y + 1) using also MID ; on the other hand, one obtains ( x + 1) y + ( x + 1)1 = ( xy + y ) + ( x + 1) again using MID . In the resulting equation (
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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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