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Say I have a number, like 60. Which has a prime factorization of $$[2,2,3,5]$$ What function would take in 60 and remove redundant factors? In this case, it should return 30

Edit: I did some more research, and the term I'm wording in looking for is reducing a number to its distinct factors

2 Answers2

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Lets define the following function:

Given $x = \prod_{i=1}^{n}p_{i}^{a_{i}}$ i.e. the prime factorisation of $x$

$f(x):= \prod_{i=1}^{n}p_{i}$

Is this what you were looking for?

mrnovice
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  • Yes but I want it to reduce factors to plus or minus 1 multiplicity –  Mar 19 '17 at 01:16
  • I'm not sure I understand what you mean, doesn't this function achieve that? – mrnovice Mar 19 '17 at 01:19
  • possibly, I must admit I'm not totally mathematically literate and I don't quite understand your notation –  Mar 19 '17 at 01:20
  • It essentially means, that we are given $x$ in the form of the product of its prime factors, so then we define $f(x)$ as the the product of the prime factors of $x$, but each with exponent $1$ as required. The $\prod_{i=1}^{n}$ symbol means the product from $i=1$ to $n$ – mrnovice Mar 19 '17 at 01:21
  • I made an edit that has more meaning –  Mar 19 '17 at 01:23
  • as far as I can see, my function should still fit your purpose. – mrnovice Mar 19 '17 at 01:24
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In the context of ring theory, you can think of the number $n$ as the ideal $(n)$ in the ring $\mathbb{Z}$. The object you're looking for is called the radical of the ideal.

user263190
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