Section 6.4 First consequences of the FTA
The impact of the FTA is so great, I cannot overstate its significance. This section collates a few examples, but you will see similar ones throughout the text, as well as in the next section, when we connect the theorem back to congruences. Most importantly, lots of theorems now have reasons, not just proofs. This distinction is an important point about mathematics! The difference boils down to the fact thatExample 6.4.1.
If
Since by the FTA
So if
Example 6.4.2.
Let's show that
To begin, let's write \(a=\prod p^e\text{.}\) Then
Similarly,
If these two numbers divide each other, then we can separate the product by each prime, so that for each \(p\text{,}\)
for some \(q\text{;}\) in fact we must have \(q=p\) for each such case 7 . But then \(p^{2f}=p^{2e}p^{(2f-2e)}\) and this can be viewed as \(2e\leq 2f\text{,}\) so \(e\leq f\) as well.
This is true for all the primes \(p\) dividing \(a\text{,}\) so \(p^e \mid p^f = q^f\) for all such \(p\text{;}\) multiplying these together shows that
as desired.
Definition 6.4.3.
Given two numbers
with an obvious extension to a min or max of a set consisting of more than two numbers.
Example 6.4.4.
Product formula:
Greatest common divisor formula:
Determining a quotient formula, assuming
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prime_divisors(693)
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factor(693)
Definition 6.4.5.
For
Definition 6.4.6.
We write
Example 6.4.7.
We can demonstrate these by saying
Example 6.4.8.
How many final zeros does twenty factorial have?
Either by hand or with help, we can see what the biggest powers of \(2\) and \(5\) in \(20!\) are.
Since \(2^{18} \parallel 20!\) and \(5^4\parallel 20!\text{,}\) we can conclude that \(20!\) ends with exactly 4 zeros merely from the prime factorization, which we could certainly get without multiplying it out (though in this case Sage does that first).
We can check this result: