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}


Answered Same DayDec 23, 2021

Answer To: 1. Recall that we discussed defining the real numbers as an axiomatic system with addition axioms...

David answered on Dec 23 2021
121 Votes
1. a) If we write the identity axiom for addition A4 for N it states that
There exists an element 0 € N s
uch 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
b) All the axioms are satisfied for Z
2. a) |b| <= a
b <= a , –b <=a
b <= 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
b) 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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