   Chapter 2.1, Problem 13E

Chapter
Section
Textbook Problem

In Exercises 13 − 24 , prove the statements concerning the relation < on the set Z of all integers.If x < y , then x + z < y + z .

To determine

To prove: The statement, ‘If x<y, then x+z<y+z’ on the set of integers.

Explanation

Formula Used:

Subtraction:

Subtraction is defined by xy=x+(y) for arbitrary x and y in Z.

The order relation less than:

For integers x and y, x<yifandonlyifyxZ+,

Where yx=y+(x).

x+(y+z)=(x+y)+z for arbitrary elements x,y,zZ.

The distributive law:

x(y+z)=xy+xz holds for all elements x,y,zZ.

x+y=y+x for arbitrary x,yZ.

For each xZ, there is an additive inverse of x in Z, denoted by x, such that x+(x)=0=(x)+x.

Proof:

Consider the given information.

x<yyxZ+bydefinitionoforderrelationlessthan

Consider x,y,zZ

(y+z)(x+z)=(y+z)+((x+z))bydefinitionofsubtraction=y+z+(1)(x+z)byassociativity

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