# What are the elements of the set A? A = {x is in Z: x=(6y-7)/(2y+1), y is in N}

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### 2 Answers

In order to determine the elements of the set A, we'll consider the given constraint of the set, that is that all the elements are integer numbers (x integer).

For x to be integer, the denominator (2y+1) has to be the divisor of the numerator (6y-7).

We'll write the division with reminder:

(6y-7)=3(2y+1)-10

We'll divide both sides by (2y+1) and we'l get:

(6y-7)/(2y+1) = 3-10/(2y+1)

But x = (6y-7)/(2y+1) and x integer, so, in order to obtain x integer, (2y+1) has to be a divisor of 10.

We'll write 10 divisors:

D10=+/-2;+/-5;+/-1;+/-10

(2y+1)=1

2y=0,

y=0

**x=(6*0-7)/(2*0+1)**

**x = -7/1**

**x = -7**

(2y+1)=-1

2y=-2

y=-1

**x=(6*(-1)-7)/(2*(-1)+1)**

**x = 13**

(2y+1)=2

2y=1

y=1/2 rejected (not integer)

(2y+1)=-2

(2y+1)=5

2y=4

y=2

**x=(6*(2)-7)/(2*(2)+1)**

**x=5/5**

**x = 1**

(2y+1)=-5

2y=-6

y=-3

**x=(6*(-3)-7)/(2*(-3)+1)**

**x = 5**

(2y+1)=10

2y=9

y=9/2 rejected (not integer)

(2y+1)=-10

2y=-11

y=-11/2 rejected (not integer)

**A={-7,1,5,13}**

A ={x is in Z : x = (6y-7)/(2y+1) , y is in N}

We know N is the set of all natural numbers: 1,2,3,.....

We know Z is the set of all integers (positve and negative): .....4,3,2,1,0, 1,2, 3..... equally saced on the number line.

x = (6y-7)/(2y+1) .

Since y is in N, y = 1,2,3,....

x1 = (6*1-7)/(2*1+1) = -1/3.

x2 = (6*2-7)/(2*2+1) = 5/5 = 1. So x2 belongs to Z.

x3 = (6*3-7)/(2*3+1) = 11/7.

x4 = (6*4-7)/(2*4+1) = 17/9. So x4 < 2

x5 = (6*5-7)/(2*5+1) = 23/11. X5 > 2. So Xn does not take the value 2.

xn = (6n-7)/(2n+1).

Lt n--> infinity xn = (6n-7/(2n+1) = (6-1/n)/(2+1/n) = 3.

We see that for y > = 2 and for x in n, the nth term Xn steadily increase from 1 to 3.

So 1 < xn < 3.

Therefore the only integer value xn takes is x2 = (6*2-7)/(2*2+1) = 1 when y = 2.

Therefore A = A = {x is in Z: x=(6y-7)/(2y+1), y is in N} = {x: x= 1} .