Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

MATH 301 Homework 1 Solutions, Exercises of Discrete Mathematics

Homework 1 with questions and answers

Typology: Exercises

2018/2019

Uploaded on 09/27/2024

john-brown-ost
john-brown-ost 🇺🇸

1 document

1 / 1

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 301 Homework 1 Charlie Lane McMenomy
Proposition 1.8. If m is an integer, then (m) + m= 0.
Proof. Let mZ. Then by axiom 1.4, m+ (m) = 0. Applying axiom 1.1(i) gives
(m) + m= 0.
Proposition 1.10. Let m, x1, x2Z. If m, x1, x2satisfy the equations m+x1= 0 and
m+x2= 0, then x1=x2.
Proof. Let m, x1, x2Z, and let m+x1= 0 and m+x2= 0. Since = is both symmetric
and transitive, m+x1=m+x2. Adding (m) to each side of the equation gives
(m)+(m+x1)=(m)+(m+x2)
By axiom 1.1.(ii),
((m) + m) + x1= ((m) + m) + x2
Then Proposition 1.8 can be applied to both sides of the equation to get 0 + x1= 0 + x2.
Applying axiom 1.1.(i) gives x1+ 0 = x2+ 0. It then remains to use axiom 1.2 to get
x1=x2.
Revised
Proposition 1.14. For all mZ,m·0=0=0·m.
Proof. Let mZ. By axiom 1.2, 0 Zand 0 + 0 = 0. By replacement, m·0 = m·(0 + 0).
Using axiom 1.1.(iii), m·(0 + 0) = m·0 + m·0. Returning to axiom 1.2, m·0 + 0 = m·0.
Since = is transitive, m·0 + 0 = m·0 + m·0. Then by proposition 1.9, 0 = m·0. Applying
axiom 1.1.(iv) gives m·0 = 0. Therefore, m·0 = 0 = m·0.
1

Partial preview of the text

Download MATH 301 Homework 1 Solutions and more Exercises Discrete Mathematics in PDF only on Docsity!

Math 301 Homework 1 Charlie Lane McMenomy

Proposition 1.8. If m is an integer, then (−m) + m = 0. Proof. Let m ∈ Z. Then by axiom 1.4, m + (−m) = 0. Applying axiom 1.1(i) gives (−m) + m = 0. 

Proposition 1.10. Let m, x 1 , x 2 ∈ Z. If m, x 1 , x 2 satisfy the equations m + x 1 = 0 and m + x 2 = 0, then x 1 = x 2. Proof. Let m, x 1 , x 2 ∈ Z, and let m + x 1 = 0 and m + x 2 = 0. Since = is both symmetric and transitive, m + x 1 = m + x 2. Adding (−m) to each side of the equation gives (−m) + (m + x 1 ) = (−m) + (m + x 2 ) By axiom 1.1.(ii), ((−m) + m) + x 1 = ((−m) + m) + x 2 Then Proposition 1.8 can be applied to both sides of the equation to get 0 + x 1 = 0 + x 2. Applying axiom 1.1.(i) gives x 1 + 0 = x 2 + 0. It then remains to use axiom 1.2 to get x 1 = x 2. 

Revised Proposition 1.14. For all m ∈ Z, m · 0 = 0 = 0 · m. Proof. Let m ∈ Z. By axiom 1.2, 0 ∈ Z and 0 + 0 = 0. By replacement, m · 0 = m · (0 + 0). Using axiom 1.1.(iii), m · (0 + 0) = m · 0 + m · 0. Returning to axiom 1.2, m · 0 + 0 = m · 0. Since = is transitive, m · 0 + 0 = m · 0 + m · 0. Then by proposition 1.9, 0 = m · 0. Applying axiom 1.1.(iv) gives m · 0 = 0. Therefore, m · 0 = 0 = m · 0. 

1