1. ## Equivalence Classes

Hi, I need some help with this question if anyone can help me.

For part a and b my answer is:

There are infinitely many E-equivalence classes, and each E-equivalence class is infinite because the set of Natural numbers is infinite and there are infinitely many primes.

Is this correct? And how do I do part c and d?

Thanks

2. Originally Posted by Nguyen
Hi, I need some help with this question if anyone can help me.

For part a and b my answer is:

There are infinitely many E-equivalence classes, and each E-equivalence class is infinite because the set of Natural numbers is infinite and there are infinitely many primes.

Is this correct? And how do I do part c and d?

Thanks

For (a) I'd rather say there are infinite E-eq. classes since there are infinite primes and we can form a

natural number with any arbitrary number of different primes

For (b) the answer is not since the E-eq. class of {1} has ony two elements (which natural is the other one?)

For (c) check that for example [101] = [103] (why?), [102]=[105] (why?), etc.

For (d) check how many different primes divide 120, and then form the 6 lowest possible numbers with those divisors

Tonio

3. Originally Posted by tonio
For (a) I'd rather say there are infinite E-eq. classes since there are infinite primes and we can form a

natural number with any arbitrary number of different primes

For (b) the answer is not since the E-eq. class of {1} has ony two elements (which natural is the other one?)

For (c) check that for example [101] = [103] (why?), [102]=[105] (why?), etc.

For (d) check how many different primes divide 120, and then form the 6 lowest possible numbers with those divisors

Tonio

Thanks Tonio.

For b) the E-eq. class {1} has only one element which is 1 from what I understand now. What is the second?
And to check my understanding, what are the elements of the E-eq. class {4}?

For c) so [101] = [103] because 101 and 103 are prime numbers?
and [102]=[105] because 102 and 105 can be written as the product of 3 primes?

So I get [101]=[103]=[107]=[109] , [102]=[105]=[110] and [104]=[106]=[108] ?
So there is a total of 3 different E-eq. classes.

for d) 120=2^3 x 3 x 5, so what do I do from here?

Thanks heaps!

4. Originally Posted by Nguyen
for d) 120=2^3 x 3 x 5, so what do I do from here?
Here is the smallest $2\cdot 3\cdot 5=30.$
Then $2\cdot 3\cdot 7=42$ and $2^2\cdot 3\cdot 5=60$.
Can you finish?

5. Originally Posted by Plato
Here is the smallest $2\cdot 3\cdot 5=30.$
Then $2\cdot 3\cdot 7=42$ and $2^2\cdot 3\cdot 5=60$.
Can you finish?
oh yeah, thanks Plato. I fully understand now.

6. Originally Posted by Nguyen
Thanks Tonio.

For b) the E-eq. class {1} has only one element which is 1 from what I understand now. What is the second?

My bad: I thought the equivalence is defined on the integers and not on the integers.

Tonio

And to check my understanding, what are the elements of the E-eq. class {4}?

For c) so [101] = [103] because 101 and 103 are prime numbers?
and [102]=[105] because 102 and 105 can be written as the product of 3 primes?

So I get [101]=[103]=[107]=[109] , [102]=[105]=[110] and [104]=[106]=[108] ?
So there is a total of 3 different E-eq. classes.

for d) 120=2^3 x 3 x 5, so what do I do from here?

Thanks heaps!
.