0

I'm now thinking about a question that:

If $0<p_1,~p_2,~\ldots,~p_k<1$(further assume these $p_i$ are rational numbers) and $p_1+\cdots+p_k=1$, then there exists $m\in\Bbb Z$ and $n_1,~\ldots,~n_k\in\Bbb Z$ such that $p_1=n_1/m,~p_2=n_2/m,~\ldots,~p_k=n_k/m$?

Is it true or false? How to prove this?

Michael Hardy
  • 1
Eric
  • 4,517
  • 1
  • 18
  • 36
  • 1
    You are assuming that the $p_i$ are rational? Are you in effect asking whether a finite number of rationals can be put over a common denominator? – Angina Seng Jun 24 '17 at 03:47
  • @LordSharktheUnknown Yes, $p_i$ are rational. Yes, you're right. – Eric Jun 24 '17 at 03:48

2 Answers2

2

If $0<p_1,~p_2,~\cdots,~p_k<1$ (...) and $p_1+...+p_k=1$

Neither of these conditions is necessary.

assume these $p_i$ are rational numbers (...) then there exists $m\in\Bbb Z$ and $n_1,~\cdots,~n_k\in\Bbb Z$ such that $p_1=n_1/m,~p_2=n_2/m,~\cdots,~p_k=n_k/m$?

Yes. Let $\,p_i = a_i / b_i\,$ in irreducible form, then take for example $\,m=\operatorname{lcm} \,\{b_i \mid i=1,2,\cdots,k\}\,$ and $\,n_i = a_i m / b_i\,$, where $\,n_i \in \mathbb{Z}\,$ because $\,b_i \mid m\,$. The representation is not unique unless you add more conditions, such as $\,\gcd \,\{n_i \mid i=1,2,\cdots,k\}=1\,$.

dxiv
  • 76,497
0

Let $p_i = \frac{n_i}{m_i}$ for $i = 1 \dots k$, multiply by $\frac{\prod_{j \neq i,j=1}^k m_j}{\prod_{j \neq i,j=1}^k m_j} = \frac{m_1 \dots m_{i-1} m_{i+1} \dots m_k}{m_1 \dots m_{i-1} m_{i+1} \dots m_k}$ so that $$p_i = n_i \frac{\prod_{j \neq i,j=1}^k m_j}{\prod_{j=1}^k m_j} = 1 - \sum_{j \neq i}^k p_j = \frac{r_i}{m}$$ where $$m = \prod_{j=1}^k m_j = m_1 \dots m_k, \qquad r_i = \prod_{j=1}^k m_j - \sum_{j \neq i}^k n_j \prod_{l \neq j,l=1}^k m_l$$

Weaam
  • 2,849