Are n1 and n2 Prime Factors of n?

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For n ≥ 2, where n is not prime, it can be expressed as a product of two factors n1 and n2, both greater than or equal to 2. The discussion centers on understanding how n being greater than both n1 and n2 implies that these factors must have prime factors themselves. The example of n = 6 illustrates that its prime factors, 2 and 3, validate the statement, although the smallest non-prime integer is actually 4. Clarification is sought on whether n1 and n2 are indeed prime factors of n. The conversation emphasizes the relationship between non-prime integers and their prime factors.
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Homework Statement



We have n ≥ 2, n not prime, n ∈ ℤ. Take the smallest such n. n is not prime and as such n is not irreducible and can be written as n = n1.n2; n1, n2 not units. We may take n1, n2 ≥ 2. However we have n > n1, n > n2 so n1, n2 have prime factors.

I'm not sure how n > n1, n > n2 implies that n1, n2 have prime factors.

Homework Equations


I'm not sure what's relevant here.

The Attempt at a Solution



From what I can see, the lowest possible n which meets the criteria is 6. 6 has the prime factors 2 and 3, which means that obviously what is stated is true. I'm just not sure how n > n1, n > n2 implies that its true.
 
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6 is not the smallest non-prime integer larger than or equal to 2, 4 is.

Anyway, if n is the smallest non-prime integer and n1 < n, what does this imply?
 
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Do you mean, "n1, n2 are prime factors"?
 

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