Exercise 1. For every r N we define rZ := {r · n : n Z}. E.g. 2Z = {..., 4, 2, 0

Question

Exercise 1. For every r N we define rZ := {r · n : n Z}. E.g. 2Z = {..., 4, 2, 0, 2, 4, ...}.

1. Give all elements of the sets 3Z, 5Z and 7Z in the interval [20, 20].

2. Let r N fixed. Investigate the functions fr : Z Z, x 7 r · x gr : Z Z, x 7 j x r k . Prove or disprove that the functions injective or surjective. Caution: it might depend on the choice of r; distinct different cases if necessary..

3. Show that you can change the domain or codomain of one of the functions fr, gr such that it becomes a bijection (for all r). Use this fact to conclude the cardinality of rZ.

Explanation / Answer

1. rZ gives all multiples of r

Hence, 3Z gives multiples of 3. Multiples of 3 in [-20,20] are:

{-18,-15,-12,-9,-6,-3,0,3,6,9,12,15,18}

Hence, 5Z gives multiples of 5. Multiples of 5 in [-20,20] are:

{-20,-15,-10,-5,0,5,10,15,20}

Hence, 7Z gives multiples of 7. Multiples of 7 in [-20,20] are:

{-14,-7,0,7,14}

2.

First we look at the function: f_r

f_r(x)=rx

rZ gives all multiples of r

Case 1: r=1,-1

So f_r is surjective only for r=1 and also injective

Case 2:|r|>1

Let x and y be to distinct integers.

rx-ry=r(x-y) which is not equal to 0

Hence, f_r is injective. for |r|>1

Now we consider : g_r

Consider the interval:I_m [mr,(m+1)r)

Let x belong to I_m

Hence g_r(x)=floor(x/r)=m

So all integers in the interval: [mr,(m+1)r) are mapped to m under the g_r

Case 1 |r|=1

The interval [m,m+1) has only one integer =m. Hence g_r is injective and surjective

Case 2 |r|>1

The interval [mr,(m+1)r) has more than one integer. Hence g_r is not injective but it is surjective.

3.

For g_r we can change the domain to rZ

Any integer, x in rZ is of the form: rm where m is in Z

Hence, g_r(x)=floor(rm/r)=m

Let x and y be distinct integers in rZ

so, x=rm,y=rn for some distinct m,n in Z

g_r(x)-g_r(y)=m-n which is non zero

Hence, g_r is injective and g_r is surjective as for any m in Z, g_r(rm)=m

Hence, g_r is bijection with domain being rZ and codomain being Z

Hence cardinality of rZ =cardinality of Z

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