# Number theory factorization proof

• dancergirlie
In summary: For n=20, k=0 and m=1, the statement holds.For n=21, k=1 and m=3, the statement holds.However, for n=22, k=2 and m=5, the statement does not hold.
dancergirlie

## Homework Statement

1. Homework Statement

If n is a nonzero integer, prove that n cam be written uniquely in the form n=(2^k)m, where It is in the primes and unique factorization chapter so maybe that every integer n (except 0 and 1) can be written as a product of primes

## Homework Equations

It is in the primes and unique factorization chapter so maybe that every integer n (except 0 and 1) can be written as a product of primes

## The Attempt at a Solution

I am not sure, but I tried to do a proof by induction on n

Let n=1
so, 1=(2^k)(m)
1=(1)(1)=1 where k=0 and m=1
so the statement holds for n=1

Now assume the statement holds for n=(r-1) for any integer (except 0 and 1)
so
(r-1)=(2^k)(m) where k is greater than or equal to zero and m is odd

Now let n=r
so r=(2^k)(m)

This is where I don't know what to do... that is why I'm thinking there is an easier way to do this besides induction. Any help would be appreciated =)

You left off the conditions in your problem statement.
If n is a nonzero integer, prove that n cam be written uniquely in the form n=(2^k)m, where
After this you tell us where the problem appeared in your text. What are the conditions on k and m?

Also, for your induction proof, I would start with 2, not 1.

ohh I'm sorry

k has to be greater than or equal to zero and m is odd

Let me repost the problem, cause I see some typos...

Let n be a nonzero integer, prove that n can be written uniquely in the form (2^k)(m) for any integer k greater than or equal to zero and for any odd integer m.

If i can avoid doing an induction proof, that would be great, but if this is the only way then i guess i can do it

This might not pass muster as a formal proof, but it will help you think about this problem.

There are two cases: n is even and n is odd.
1) If n is odd, it has no factors of 2, so n = 20*n, and you're done.
2) If n is even, there must be at least one factor of 2, so n = 21*m
a) If m is odd, you're done.
b) If m is even, apply step 1 again.

## What is number theory factorization proof?

Number theory factorization proof is a branch of mathematics that deals with the properties of integers and their relationships. It involves proving the factors of a given number and understanding the patterns and relationships between them.

## Why is number theory factorization proof important?

Number theory factorization proof is important because it helps us understand the fundamental building blocks of mathematics, the integers. It also has applications in cryptography, computer science, and other areas of mathematics.

## What are the different methods of number theory factorization proof?

There are several methods of number theory factorization proof, including trial division, Fermat's method, Pollard's rho method, quadratic sieve method, and number field sieve method. Each method has its own advantages and limitations.

## How do you prove the factors of a given number?

To prove the factors of a given number, you can use different factorization methods. These methods involve breaking down the number into smaller factors until you reach prime numbers, which cannot be further divided. The factors of a number are all the numbers that can divide it evenly without leaving a remainder.

## Are there any unsolved problems in number theory factorization proof?

Yes, there are still unsolved problems in number theory factorization proof, such as the Riemann Hypothesis and the Goldbach Conjecture. These problems have been studied by mathematicians for centuries and continue to be a topic of research and fascination.

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