Please show details Will Rate LIFE SAVER!!!!! For every pair of nonzero real num

Question

Please show details Will Rate LIFE SAVER!!!!!  

For every pair of nonzero real numbers a and b, (ab)-1 = b-1a-1. 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 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 . b) . c = a . (b . c). There exists a unique number 1 R such that a.1 = 1.a = a for every a R. For all nonzero a R, there exists a unique number a-1 R such that a .a-1 = a-1.a = 1. For all a, b R, a . b = b . a. For all a,b,c R, a.(b + c) = a.b + a.c. 1 0. For all a R, exactly one of the following statements is true: 0

Explanation / Answer

By M3, we know

aa-1 = 1

b-1b = 1

==> aa-1 · b-1b = 1

Also, (ab)(ab)-1 = 1

==> (ab)(ab)-1 = aa-1 · b-1b

By M1 and M4, (ab)(ab)-1 = (ab)(b-1a-1)

==> (ab)-1 = (b-1a-1)

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