Hi,

I want to prove fundamental theorem of arithmetic by using strong induction and proof by smallest counterexample.

Can someone by any chance help me? (with great detail please)?

I would appreciate people's help.

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- Nov 15th 2012, 04:04 AMGpinkFundamental theory of arithmetic
Hi,

I want to prove fundamental theorem of arithmetic by using strong induction and proof by smallest counterexample.

Can someone by any chance help me? (with great detail please)?

I would appreciate people's help. - Nov 15th 2012, 06:51 AMGpinkRe: Fundamental theory of arithmetic
Can anybody please help me please?

- Nov 15th 2012, 11:39 AMemakarovRe: Fundamental theory of arithmetic
I am not sure about smallest counterexamples, but for the inductive proof of the fundamental theorem of arithmetic see Wikipedia.

- Nov 15th 2012, 02:43 PMDevenoRe: Fundamental theory of arithmetic
if the theorem is not true, there exists a minimal counter-example, a.

by supposition, there exists two distinct prime factorizations of a:

$\displaystyle a = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r} = q_1^{m_1}q_2^{m_2}\cdots q_s^{m_s}$ with $\displaystyle p_1 < p_2 < \cdots < p_r, q_1 < q_2 < \cdots < q_s$.

now $\displaystyle p_1|a$ so $\displaystyle p_1|q_i^{m_i}$ for some i = 1,2,...,s.

this, in turn, implies $\displaystyle p_1|q_i$. since $\displaystyle q_i$ is prime, and $\displaystyle p_1 \neq 1$, we conclude $\displaystyle p_1 = q_i$.

thus since$\displaystyle \frac{a}{p_1^{k_1}} < a$,

we conclude that $\displaystyle p_2^{k_2}\cdots p_r^{k_r} = \frac{q_1^{m_1}q_2^{m_2}\cdots q_s^{m_s}}{q_i^{m_i}}$ are, in fact, the same factorization.

hence since we have r-1 factors on the left, and s-1 factors on the right:

r - 1 = s - 1, so r = s.

we now have two possibilities:

i = 1 (on the RHS),

i ≠ 1.

if i ≠ 1, then q_{1}= p_{2}, which implies p_{2}= q_{1}< q_{i}= p_{1}, contradicting our original factorization of a.

thus we must have p_{1}= q_{1}, and since the factorization of $\displaystyle \frac{a}{p_1^{k_1}}$ IS unique,

p_{i}= q_{i}for all i > 1, and k_{i}= m_{i}, for all i > 1.

thus the only difference between:

$\displaystyle a = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r} = p_1^{m_1}p_2^{k_2}\cdots p_r^{k_r}$

is the power of p_{1}.

but $\displaystyle p_1^{k_1} = \frac{a}{p_2^{k_2}\cdots p_r^{k_r}} < a$, so we must have k_{1}= m_{1}, and a does not exist, which proves the theorem.