0 has no multiplication inverse. That is, there exists no real number a such tha
ID: 2943614 • Letter: 0
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0 has no multiplication inverse. That is, there exists no real number a such that a0 = 1. Suggestion: Suppose instead that such a real number a does exist. Now consider Proposition 4 and Axiom NT1. For all a, b, c R, (a + b) + c = a + (b + c). There exists a unique number 0 R such that a + 0 = 0 + a = a for every a R. For all a epsilon R, there exists a unique number -a R such that a + (-a) = (-a) + a = 0. For all a, b R, a + b = b + a. For all a, b, c R, (a middot b) middot c = a middot (b middot c). There exists a unique number 1 epsilon R such that a middot 1 = 1 middot a = a for every a R. For all nonzero a R. there exists a unique number a-1 R such that a middot a-l = a-1 middot a = 1. For all a, b, c R, a middot b = b middot a. For all a, b, c R, a middot (b + c) = a middot b + a middot c. NT1. 1 0. For all c R, exactly one of the following statements is true: 0Explanation / Answer
Assume that there exists a number z such that z is the reciprocal of zero. In symbols, z = 0^-1 = 1/0. The inverse property of multiplication (M3) states that any number x has an inverse 1/x, x not equal to 0, such that x*1/x = x/x = 1. So, applying the rule, we have z * 0 = 1. This statement doesn't make sense. If there exists a number z in the real numbers, then z * 0 = 0. Since (NT1) 1 does not equal 0
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