Bertrand`s postulate
... Bertrand’s postulate Bertrand’s postulate states that for each integer n ≥ 2 there is a prime number p with n < p < 2n. The following proof is due to Erdős. This account is based on my reading of Hardy and Wright, Introduction to the Theory of Numbers and Rose, A Course in Number Theory (both Oxfor ...
... Bertrand’s postulate Bertrand’s postulate states that for each integer n ≥ 2 there is a prime number p with n < p < 2n. The following proof is due to Erdős. This account is based on my reading of Hardy and Wright, Introduction to the Theory of Numbers and Rose, A Course in Number Theory (both Oxfor ...
Square values of Euler`s function
... The number Vϕ(x) of ϕ-values in [1, x] is O(x/(log x)c), where c = 1e log 2 = 0.254 . . . . Pillai’s idea: There are not many values ϕ(n) when n has few prime factors, and if n has more than a few prime factors, then ϕ(n) is divisible by a high power of 2. Since ϕ(p) = p − 1, we have Vϕ(x) ≥ π(x + 1 ...
... The number Vϕ(x) of ϕ-values in [1, x] is O(x/(log x)c), where c = 1e log 2 = 0.254 . . . . Pillai’s idea: There are not many values ϕ(n) when n has few prime factors, and if n has more than a few prime factors, then ϕ(n) is divisible by a high power of 2. Since ϕ(p) = p − 1, we have Vϕ(x) ≥ π(x + 1 ...
Proof that 2+2=4
... for i 6= 2, because f2 − f1 = f0 and f0 = 0, but f1 = 1, and the Fibonacci sequence is nondecreasing. Hence (5 − 2) − 2 > 0. Now, suppose that ∃b ∈ Z such that b > 0 and (k − 1) − 2 = 2 + b. We need to prove that, for some b0 > 0, k − 2 = 2 + b0 . Our inductive hypothesis is equivalent to: k−1−2=2+b ...
... for i 6= 2, because f2 − f1 = f0 and f0 = 0, but f1 = 1, and the Fibonacci sequence is nondecreasing. Hence (5 − 2) − 2 > 0. Now, suppose that ∃b ∈ Z such that b > 0 and (k − 1) − 2 = 2 + b. We need to prove that, for some b0 > 0, k − 2 = 2 + b0 . Our inductive hypothesis is equivalent to: k−1−2=2+b ...
