Prove that for every natural number $a$ there are integers $t \geq 0$ and $r$ such that $a = 3^tr$ and $3 \not | r$.
Would I use the well ordering principle for this? There’s so many variables so I’m a bit lost.
Prove that for every natural number $a$ there are integers $t \geq 0$ and $r$ such that $a = 3^tr$ and $3 \not | r$.
Would I use the well ordering principle for this? There’s so many variables so I’m a bit lost.
Let $a=2^{r_1}3^{r_2}\cdot\cdot$ $\cdot% $ $ p^{r_n}$. If $a$ is not a multiple of $3$, take $t=0$ and $r=a$. If $a$ is a multiple of $3$, take $t=r_2$ and so the product of the rest of the factors of $a$ is a product of primes which does not contain a $3$. So, take $r=2^{r_1} \cdot \cdot \cdot p^{r_n}$. Hence every natural $a$ can we written in this way.
Use the fundamental theorem of arithmetic to obtain a prime decomposition of a given number: $t$ is gonna be the exponent of $3$, and $r$ is simply the product of all other primes appearing in it, $r$ obviously not divisible by $3$
For a natural number, it's prime factors can consist of many other prime numbers.
For '$a$' it's prime factors can be any prime natural number including 3.
But for '$r$' its prime factors can be any integer excluding 3.
If we look at $3^t r$ as a whole (not whole number) then it can be any natural number as it's prime factors have included every prime number.