Hi,
One method to find relative prime numbers for a given N is:
RP(55) = RP(5-1) * RP(11-1)
= 4 * 10
40.

Is the above a correct method, somebody please guide me.

Zulfi.

 

Huh???

Clearly 40 and 55 are NOT relatively prime (they are both divisible by 5).

 

Hi,
Does it say that there are 40 numbers relatively prime to 55? What does it tell us?

Zulfi.

 

Ah, it sounded like you were saying that you were using that method to find a number that was relatively prime to 55, but I see that you meant that you were finding how many numbers are relatively prime to 55 -- also known as Euler's totient function. Yes, what you showed is correct in the special case of a number that is a product of prime numbers, none of which are repeated.

 

Hi,
One method to find relative prime numbers for a given N is:
RP(55) = RP(5-1) * RP(11-1)
= 4 * 10
40.

Is the above a correct method, somebody please guide me.

Zulfi.

Assuming that the function RP( ) is the Totient function, the method shown doesn't seem to work:



RP(55) = RP(5-1) * RP(11-1) needs to be RP(55) = (5-1) * (11-1)

------------------------------------------------------
Furthermore, not all numbers of the class described by WBahn: a "number that is a product of prime numbers, none of which are repeated. "
work with the method the TS describes:

 

I didn't look closely enough. I was too focused on the steps and not what was actually written.

You are correct, it should be

RP(p1·p2·...·pn) = RP(p1)·RP(p2)·...·RP(pn)

if all of the p's are prime.

 

Assuming that the function RP( ) is the Totient function, the method shown doesn't seem to work:

View attachment 193613

RP(55) = RP(5-1) * RP(11-1) needs to be RP(55) = (5-1) * (11-1)

------------------------------------------------------
Furthermore, not all numbers of the class described by WBahn: a "number that is a product of prime numbers, none of which are repeated. "
work with the method the TS describes:

View attachment 193614

I'm not following how this is a counter example since 15 is not prime, and so you have

EulerPhi[165] = Euler[15]*EulerPhi[11] = EulerPhi[3]*EulerPhi[5]*EulerPhi[11] = (3-1)(5-1)(11-1) = 2·4·10 = 80

 

I'm not following how this is a counter example since 15 is not prime, and so you have

EulerPhi[165] = Euler[15]*EulerPhi[11] = EulerPhi[3]*EulerPhi[5]*EulerPhi[11] = (3-1)(5-1)(11-1) = 2·4·10 = 80

The TS showed a number (55) factored into two factors. He didn't say that the factors had to be individual prime factors of the starting number. I factored my example number (165) into two factors, one of which could be factored further, but nothing in post #1 said that the method being illustrated required it. It appeared that a procedure applied to the two factors would give the desired result. The method was not completely specified.

 

Hi,
phi(37) = 37-1 = 36
phi(21) is not a prime= phi (7) * phi(3) : Not both 7 & 3 are prime = 6 * 2 = 12

I think this is the example which I saw in the book. So 40 is correct.

Zulfi.

 

phi(21) is not a prime= phi (7) * phi(3) : Not both 7 & 3 are prime = 6 * 2 = 12

Both 7 & 3 are prime.

Did you mean to say: Now both 7 & 3 are prime = 6 * 2 = 12