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1. Recall that we discussed defining the real numbers as an axiomatic system with addition axioms A1-A4, multiplication axioms M1-M4, a distributive law (DL), and order axioms 01-05. (a) Which of the...

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1. Recall that we discussed defining the real numbers as an axiomatic system with addition axioms A1-A4, multiplication axioms M1-M4, a distributive law (DL), and order axioms 01-05.
(a) Which of the properties Al-A4, M1-M4, DL, 01-05 fail for N? In each of those cases, demonstrate that the property fails through an example. (b) Repeat part (a) for Z instead of N. 2. (a) For a, b E R, prove that Ibl
3. (a) Use the triangle inequality to prove that la + b + ci
dal+ a2 + • • + an.1 5_ lad ± lath + • • lard for all al, a2, , E H, where n is an integer with n > 2. 4. Let a and ,3 be Dedekind cuts. (a) Show that the set a + ,3 defined as a + = {r+slrect,sE 3}
is a Dedekind cut. (b) Explain why the following definition for "multiplication" of Dedekind sets is a poor definition: a • 13 = {r•sir a,s 3}
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Answered Same Day Dec 23, 2021

Solution

David answered on Dec 23 2021
123 Votes
1. a) If we write the identity axiom for addition A4 for N it states that
There exists an element 0 € N such that a+0 = 0+a = a
The identity axiom for addition A4 fails for N since 0 does not belong to N
Rest of all the axioms are satisfied for N
) All the axioms are satisfied for Z
2. a) |b| <= a
<= a , –b <=a
<= a , b >= -a
(multiplying the second inequality with -1 on both sides the inequality changes as it is
multiplied by a negative number)
So, from the above two inequalities we can write
-a <= b <= a
) Triangle inequality states that |a+b| <= |a| + |b|
Without loss of generality, assume that |a| ≥ |b|. (If |b| ≥ |a|, then we could just swap the role of a and b in the inequality.) So it suffices to show that .
 However, we can write a as a-b+b, ie think of it as a = (a-b) + b.
This means that |a| = | a-b+b|. But,
|a-b+b| <= |a-b| + |b| by the Triangle Inequality, and so we have that
|a| <= |a-b| + |b|. Now, subtract |b| from both sides. This gives us:
          |a|-|b| <= |a-b|.
From this we can write ||a| - |b|| <=...
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