Given : Axiom 1.1. If m, n, and p are integers, then: 1.1. i. m+n=n+m 1.1. ii. (

Question

Given: Axiom 1.1. If m, n, and p are integers, then:
1.1. i. m+n=n+m
1.1. ii. (m+n)+p=m+(n+p)
1.1. iii. m*(n+p)=mn+mp
1.1. iv. mn=nm
1.1. v. (m*n)*p=m*(n*p)
Axiom 1.2. There exists an integer 0 such that for every integer, m, m+0=m
Axiom 1.3. There exists an integer 1 such that 1 is not 0 and whenever m is integer, m*1=m.
Axiom 1.4. For each integer, m, there exists an integer, denoted by -m, such that m+(-m)=0
Axiom 1.5. Let m, n, and p be integers. In m*n=m*p and m does not equal 0, then n=p.

Definition of Subtraction: m-n is defined to be m+(-n)

Explanation / Answer

The proof is actually quite easy, I will give you 2 proofs First: You have been given the well ordering principle(principle of induction is actually a consequence of the well ordering principle) Consider your set A we are given that A is a subset of Z, consider a new set B = {x-b+1|x belongs to A} This set is a subset of natural numbers, so by well ordering principle has a smallest element, you can easily see if y is the smallest element in B then y+b-1 is the smallest element in A Hence proved Second: Now a proof by induction on the number of elements in A P(1) is obviously true as A has only one element hence its the smallest element in A Hence proved Let P(k) be true, it any set of k element and a subset of Z has a smallest element. (Hypothesis) Lets prove it for P(k+1) Consider any element a in A Consider the set B = {x|x belong to A, not equal to a} this Set B has K elements, so by induction hypothesis it has one smallest element say y. y

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