- #1
xax
- 26
- 0
I need to prove this for any n natural, n>= 5, n not prime.
When we say that "(n-1) is divisible by n", it means that the remainder when (n-1) is divided by n is equal to 0. In other words, n is a factor of (n-1).
To prove that "(n-1) is divisible by n", you can use the division algorithm. Divide (n-1) by n and if the remainder is equal to 0, then n is a factor of (n-1) and therefore, (n-1) is divisible by n.
Yes, for example, if n=5, then (n-1) is 4 and when we divide 4 by 5, the remainder is 0. Therefore, 5 is a factor of (n-1), and (n-1) is divisible by n.
No, it is not always true. In fact, it is only true for specific numbers. For example, if n=3, then (n-1) is 2 and when we divide 2 by 3, the remainder is not 0. Therefore, 3 is not a factor of (n-1) and (n-1) is not divisible by n.
The concept of "(n-1) is divisible by n" is useful in various mathematical concepts such as number theory, algebra, and even in proving certain theorems. It is also used in solving equations and in finding the factors of a number.