upper bounds have analogous versions for greatest lower bounds. The Axiom of Com
ID: 1890839 • Letter: U
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upper bounds have analogous versions for greatest lower bounds. The Axiom of Completeness does not explicitly assert that, a nonempty set bounded below has an infimum, but this is because we do not need to assume this fact as part of the axiom. Using the Axiom of Completeness, there are several ways to prove that greatest lower bounds exist for bounded sets. One such proof is explored in Exercise 1.3.3. Let Z5 = {0, 1, 2, 3, 4} and define addition and multiplication modulo 5, In other words, compute the integer remainder when a + b and ab are divided by 5, and use this as the value for the sum and product, respectively. Show that, given any element z Z5, there exists an element y such that z + y = 0. The element y is called the additive inverse of z. Show that, given any z 0 in Z5, there exists an element x such that zz = 1. The element x is called the multiplicative inverse of z. The existence of additive and multiplicative inverses is part of the definition of a field. Investigate the set Z4 = {0, 1, 2, 3} (where addition and multiplication are denned modulo 4) for the existence of additive and multiplicative inverses. Make a conjecture about the values of n for which additive inverses exist in Zn, and then form another conjecture about the existence of multiplicative inverses. Write a formal definition in the style of Definition 1.3.2 for the infimum or greatest lower bound of a set. Now, state and prove a version of Lemma 1.3.7 for greatest lower bounds. Let A be bounded below, and define B = {b R :Explanation / Answer
A) Since there are only 5 element we might as well just do each individually 0+0 = 0 (mod 5) 1+4 = 0 (mod 5) 2+3 = 0 (mod 5) 3+2 = 0 (mod 5) 4+1 = 0 (mod 5) Thus every element in Z5 has an additive inverse in Z5 B) Again we can just show it for each one: 1*1 = 1 (mod 5) 2*3 = 1 (mod 5) 3*2 = 1 (mod 5) 4*4 = 1 (mod 5) C) For Z4 Additive Inverses 0+0 = 0 (mod 4) 1+3 = 0 (mod 4) 2+2 = 0 (mod 4) 3+1 = 0 (mod 4) Multiplicative Inverses 1*1 = 1 (mod 4) 2*1 = 2 (mod 4) 2*2 = 0 (mod 4) 2*3 = 2(mod 4) Thus 2 has no inverse 3*3 = 1 (mod 4) My conjecture (biased by having previously taken this class) is that for all n in the natural numbers elements will have additive inverses, but in order for each element to have a multiplicative inverse, n must be prime.
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