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$p$ and $r$ are given real numbers, both are positive and p is greater than r ($p,r \in R$, $p>r>0$)

I need to prove or give a counter-example with the following conditions:

Exist $n,m \in Z$, where $n>0$ and $m>0$, and exist $q \in R$, where $r>q\ge0$ satisfying

$np=mr+q$

Additional exercise: in case of $m=2$, give valid $n,q$ or prove it is impossible ($np=2r+q$)

I think it is concerned somehow to the axiom of Arkhimedes, but I can't figure it out.

Any hint is welcome

User20141219
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1 Answers1

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I think I found the solution:

Let's choose an integer value for $n$

From Arkhimedes-axiom: exist $k \in Z$ satisfying $kr > np$

Let $k$ the minimal value which satisfies it.

Thus, $kr>np>(k-1)r$, $s=kr-np$

and $0<kr-np=s<r$

$np=kr-s=(k-1)r+(r-s)$

$0>-s$, thus $r>r-s$ and from $s<r$ implies $0<r-s$

So, $m=k-1$ and $q=r-s$ is good for us.

As, $n$ is arbitrary, $k$ could be greater than $1$, therefore $m>0$

User20141219
  • 35