# Please explain this calculus

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 ⇒ ??

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## Answers and Replies

Mentor
2022 Award
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.

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
You might want to consider the ##n## even and odd cases separately.

Homework Helper
2022 Award
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.