Question

# Prove the Fundamental Theorem of Arithmetic. Every integer than 1 is a prime or a product of primes. This product is unique, exept for the order in which the factors appear.

Polynomial arithmetic
Prove the Fundamental Theorem of Arithmetic. Every integer than 1 is a prime or a product of primes. This product is unique, exept for the order in which the factors appear.

The objective is to prove the Fundamental Theorem of Arithmetic, that states: “Every number greater than 1 is a prime or a product of primes. This product is unique, except for the order in which the factors appear.” To prove the existence, use induction: $$2>1$$ is a prime number itself so it satisfy the statement. Now let all the integers from 1 to k, i.e. $$1k=ab$$ where $$\displaystyle{1}{<}{a},{\quad\text{and}\quad}\ {b}{<}{k}$$ By our asumption since both a and b are between 1 and k they both can be written as product of primes So, $$k = ab$$ can be written as product of primes. Thus, by induction all integers gratee than 1 are either prime or can be written as product of primes. Now, let’s prove uniqueness of the product for that let a integer n can be expressed as a product of primes in two ways: $$\displaystyle{n}={p}_{{1}}\cdot{p}_{{2}}\cdot\ldots.\cdot{p}_{{k}},{\quad\text{and}\quad}{n}={s}_{{1}}\cdot{s}_{{2}}\cdot\ldots\cdot{s}+{m}$$
Where, $$\displaystyle{p}_{{1}},{p}_{{2}},\ldots.,{p}_{{k}}{\quad\text{and}\quad}{s}_{{1}},{s}_{{2}},\ldots,{s}_{{m}}$$ are primes.
$$\displaystyle\Rightarrow{n}={p}_{{1}}\cdot{p}_{{2}}\cdot\ldots.\cdot{p}_{{k}}={s}_{{1}}\cdot{s}_{{2}}\cdot\ldots\cdot{s}_{{m}}\Rightarrow{p}_{{i}}{\mid}{s}_{{1}}\cdot{s}_{{2}}\cdot\ldots\cdot{s}_{{m}}{f}{\quad\text{or}\quad}{a}{n}{y}{1}\leq{i}{<}{k}$$
Now, by Euclid’s Lemma $$\displaystyle\exists{s}_{{j}}$$ for $$\displaystyle{1}\leq{j}{<}{m}$$ such that $$\displaystyle{p}_{{i}}{\mid}{s}_{{j}}$$ And since both $$\displaystyle{p}_{{i}},$$ and $$\displaystyle{s}_{{j}}$$ are primes $$\displaystyle\Rightarrow{p}_{{i}}={s}_{{j}}$$ Divide these two common factors from $$\displaystyle{p}_{{1}}\cdot{p}_{{2}}\cdot\ldots\cdot{p}_{{k}}={s}_{{1}}\cdot{s}_{{2}}\cdot\ldots\cdot{s}_{{m}}$$ and repeat the process untill all common factors have been divided out. Also note that $$k = m$$ unless then after repeating above process we will be left with $$\displaystyle{p}_{{{i}_{{1}}}}\cdot{p}_{{{i}_{{2}}}}\cdot\ldots\cdot{p}_{{{i}_{{r}}}}={1}{\left({\quad\text{if}\quad}{k}{>}{m}\right)}{\quad\text{or}\quad}{s}_{{{m}_{{1}}}}\cdot{s}_{{{m}_{{2}}}}\cdot\ldots\cdot{s}_{{{m}_{{n}}}}={1}{\left({\quad\text{if}\quad}{k}{<}{m}\right)}$$ And, all primes are greater than 1 so both of above cases are not possible, thus, $$\displaystyle{k}={m}.$$ And after repeating the process we have $$\displaystyle{p}_{{i}}={s}_{{r}}\forall{1}\leq{i}\leq{k},{\quad\text{and}\quad}{1}\leq{r}\leq{m}$$ Thus, product of prime factors of a number is unique.