The significant figures of a number are those digits that carry meaning contributing to its precision. This includes all digits except all leading zeros and trailing zeros when they are used to indicate the scale of the number.
The result of addition and subtraction should have as many decimal places as the term with less decimal places.
The result of multiplication and division should have as many significant figures as the least precise term.
If the first non-significant figure is a 5 not followed by any other digits or followed only by zeros, rounding requires a tie-breaking rule.
Round half up is the default rounding method used in many disciplines if the tie-breaking rule is not specified. Round half to even rounds to the nearest even number.
The power set P(S) is the set of all subsets of S, including the empty set and the set S itself. It has 2|S| elements.
Before proving the generalized version for all primes, I am going to prove that this is valid for a small prime. In this case, I’ve chosen 3. The easiest way I could find involved proving the contrapositive.
Let us prove that if n is not divisible by 3, n2 is also not divisible by 3.
If n is not divisible by 3, n may be written as either n = 3k + 1 or n = 3k + 2.
For the first case, n2 = 9k2 + 6k + 1 = 3(3k2 + 2k) + 1, which is not divisible by 3.
For the second case, n2 = 9k2 + 12k + 4 = 3(3k2 + 4k + 1) + 1, which is also not divisible by 3.
This concludes the proof that if n is not divisible by 3, n2 is also not divisible by 3, which also proves that if n2 (n ∈ ℕ) is divisible by 3, n is divisible by 3.
If n2 (n ∈ ℕ) is divisible by a prime p, n is divisible by p.
Again, I will prove the contrapositive. If n is not divisible by p, n2 is also not divisible by p.
If n = pk + q, for a positive q, q < p, then n2 = p2k2 + 2pkq + q2 Therefore n2 = p(pk2 + 2kq) + q2. As p is a prime, p ∤ q2 and therefore n2 is not divisible by p.
This proves that if n2 (n ∈ ℕ) is divisible by a prime p, n is divisible by p.
This is a quite short and simple proof. I’ve noticed that proofs for specific primes seem to be much more popular. However, the proofs are so similar that it does not even make sense to prove a specific case instead of the general one.
As one might expect, this is a proof by reduction to the impossible.
Suppose that the square root of p is a rational number. Therefore there exist two mutually prime numbers a and b such that the square root of p equals the ratio between a and b.
Therefore, p = a2 / b2. Then b2 p = a2, which implies that p divides a2, and, because p is a prime, p divides a. Therefore, b2 p = (k2 p2), which can be simplified to b2 = k2 p. This implies that p divides b2, and p divides b, which is absurd as a and b are mutually prime in our proof.
By reduction to the impossible, this proves that the square root of any prime number p is an irrational number.