# Consider the p-adic field $\ \mathbb{Q}_p \$ . Define $\operatorname{ord}_p(x) \$ to be the p-adic valuation of $\ x \$

Consider the p-adic field $$\ \mathbb{Q}_p \$$ . Define $$\operatorname{ord}_p(x) \$$ to be the $$p$$-adic valuation of $$\ x \$$ by $$\operatorname{ord}_p(x)=\max \{r : \ p^r \ \text{ divides } \ x \} \$$.

The inequality $$1 \leq p^{\operatorname{ord}_p(n)} \leq n$$ holds good for every $$\ n \$$.

My question is -

Does the inequality $$\ 1 \leq p^{\operatorname{ord}_p(\large n^3)} \leq n \$$ holds?

That is , we are replacing $$\ n^3 \$$ by $$\ n \$$.

Let $$\ \ n=p^r \cdot \frac{a}{b} , \ b \neq 0 , \ p \ \text{ does not divide}\ a,b \$$

Then

$$\operatorname{ord}_p(n)=r \$$

So,

$$n^3=(p^r \cdot \frac{a}{b}) (p^r \cdot \frac{a}{b} ) (p^r \cdot \frac{a}{b}) =p^{3r} \left( \frac{a}{b} \right)^3 \$$

Thus,

$$ord_p(n^3)=3r \$$

So we have,

$$1 \leq p^{\operatorname{ord}_p(n)} \leq p^{\operatorname{ord}_p(n^3)} \leq n \$$

Am I right?

Please kindly check my work and correct it if necessary . Thanks.

• In your example with $n=p^r \cdot \frac{a}{b},\,b\ne 0$, it should be $\text{ord}_p(n)=r$, not $\text{ord}_p(n)={\large{\frac{1}{p^r}}}$. Commented Aug 23, 2018 at 13:28
• Oh yes. thanks. But can you conclude about the question?
– MAS
Commented Aug 23, 2018 at 13:40
• I don't understand something here. Wouldn't increasing $b$ decrease $n$ but leave $ord_p(n)$ unchanged? So how could the inequality be true? Commented Aug 23, 2018 at 13:46
• the first inequality is true. My question is whether the 2nd inequality is true or not.
– MAS
Commented Aug 23, 2018 at 14:05

Presumably, for the question at hand, $n$ is required to be a positive integer.

But even with that restriction, the inequality $p^{\text{ord}_p(n^3)}\le n$ is not always true.

For example, if $p$ is prime, and $n=p$, we get $$p^{\text{ord}_p(n^3)}=p^{\text{ord}_p(p^3)}=p^3 > p = n$$

• so what would be the correct inequality? Can you help little or can you predict an inequality which i written?
– MAS
Commented Aug 23, 2018 at 14:07
• Can you give any similar inequality for the case $\ ord_p(n^3) \$ ?
– MAS
Commented Aug 23, 2018 at 14:08
• For $p$ prime, and $n$ a positive integer, you can say $$p^{\text{ord}_p(n^3)}\le n^3$$ but you already knew that. Commented Aug 23, 2018 at 14:09