If I have the product of two prime numbers, can any other two prime numbers multiplied equate to that same value?
I’m not entirely sure if that is applicable.
- They are dealing with all possible positive integers greater than 1 (as the output), where as I am limiting my output set to only the products of two prime numbers.
- They aren’t limiting the number of (input) primes needed to achieve those values.
Perhaps it is applicable, I just can’t follow their proofs well enough to tell. I just want to make sure that you’ve understood the specifics of my question before moving along.
No, Quartz’s post is applicable. The Fundamental Theorem of Arithmetic applies to all the natural numbers; since you’re only interested in a subset of the natural numbers, it carries over to that subset.
Of course, this all assumes you’re just interested in the natural numbers in the first place, rather than some non unique factorization domain, such as the set of all complex a+b*sqrt(-5), where a and b are integers.
Irrespective of what they’re dealing with, your conditions, as I understand it anyway, is that despite what number you wind up with, you are interested in:
1.) that the output value was achieved by the multiplication of two prime numbers, and
2.) that picking any 2 other primes and multiplying could yield the same answer.
The answer is no.
Now, if you change the number of inputs, then the question is different, which is to say that you want two cases:
1.) that the multiplication of any arbitrary number of primes gives some answer, and
2.) that choosing any 2 other primes could yield the same answer.
The questions are somewhat different, but if we choose a case with 3 primes, let’s say 2, 3 and 5, then we’ll get 30. There are no 2 prime number that when multiplied together can give the same answer.
It’s essentially dealing with the nature of factorizations: if the numbers aren’t factors of the product, then it’s no dice. In case with, let’s say, four prime numbers, say: 2, 3, 5, and 7, we’ll get 210. It’s no great surprise that factoring 210 will give us the factors that we originally had. Since no two of the factors of 210 can be multiplied together to give us aggregate we had to start with, then it’s not doable.
You can, of course, try this with as many combinations as you’d like, but you’ll find it’ll hold true*.
*If I’m understanding what it is that you’re asking.
My reading of the Theorem is that there is a subset of natural numbers which have multiple factor sets, not that every natural number has multiple factor sets, in which case, it would be theoretically possible to avoid those items.
Sage Rat writes:
> My reading of the Theorem is that there is a subset of natural numbers which
> have multiple factor sets, not that every natural number has multiple factor
> sets, in which case, it would be theoretically possible to avoid those items.
What does factor sets mean? What does multiple factor sets mean? Do you just mean the factors of a number? Are you asking if some numbers have several different factorizations? What do you mean by it being theoretically possible to avoid them? In any case, the answer is that each number has exactly one prime factorization. That’s what unique factorization means.
bolding added.
In which case the answer to my question would be “yes”, not “no”.
Let me give an example, the first 4 primes are:
2 3 5 7
We break that down into sets of two, and determine the products of the sets:
[2, 2] = 4
[2, 3] = 6
[3, 3] = 9
[2, 5] = 10
[2, 7] = 14
[3, 5] = 15
[3, 7] = 21
[5, 5] = 25
[5, 7] = 35
[7, 7] = 49
So I can say that within the first four prime numbers, there are no two sets which have the same product. So at least within that space, the answer to my question is yes; there are no two prime pairs that multiply out to the same value.
:smack: Sorry. Ignore me. For some reason I mistook the original “No” answer as saying that my theory was wrong.
My theory was confirmed, but the answer to the wording of my specific question was “no”. That got me confused, hence the follow ups.