How to Solve Calculus Problems with Recurrence: A Step-by-Step Explanation

[avata4]

Homework Statement

∀n ℕ-{0}, ∃ (p,q) ℕ², n=2^p(2q+1)

Relevant Equations

∀n ℕ-{0}, ∃ (p,q) ℕ², n=2^p(2q+1)

i think solution with récurrence

for n=1 then 1=2¨¨^0(2x0 +1) true
suppose that n=2¨^p(2q+1) is true shows that n+1=2^p( 2q +1)?
n+1=2¨^p(2q+1) +1 ⇒ ??

 

[fresh_42]

Homework Statement:: ∀n ℕ-{0}, ∃ (p,q) ℕ², n=2^p(2q+1)
Relevant Equations:: ∀n ℕ-{0}, ∃ (p,q) ℕ², n=2^p(2q+1)

i think solution with récurrence

for n=1 then 1=2¨¨^0(2x0 +1) true
suppose that n=2¨^p(2q+1) is true shows that n+1=2^p( 2q +1)?
n+1=2¨^p(2q+1) +1 ⇒ ??

You could use an induction of the length of the decomposition into primes, or what is also common when it comes to natural numbers, assume a smallest number $n_0$ for which it is not the case and deduce a contradiction.

 

[vela]

You might want to consider the $n$ even and odd cases separately.

 

[pasmith]

Homework Statement:: ∀n ℕ-{0}, ∃ (p,q) ℕ², n=2^p(2q+1)
Relevant Equations:: ∀n ℕ-{0}, ∃ (p,q) ℕ², n=2^p(2q+1)

i think solution with récurrence

for n=1 then 1=2¨¨^0(2x0 +1) true
suppose that n=2¨^p(2q+1) is true shows that n+1=2^p( 2q +1)?
n+1=2¨^p(2q+1) +1 ⇒ ??

The statement is asserting that every strictly positive natural number is a power of two times an odd number.

This follows from the unique decomposition of $n$ as a product of powers of primes.