Prove or disprove: for any $n \in \mathbb{N}$, for any $r_1,r_2, \ldots, r_n \in R$ then there exists $a_j, 1 \leq j \leq n$ such that $r_j > 1$.

Question

I have to prove or disprove:

For any $n \in \mathbb{N}$, for any $r_1,r_2,\ldots, r_n \in R$ such that $\forall i, 1 \leq i \leq n, r_i > 0$ and such that $r_1 r_2 \cdots r_n = 1$, if there is an $i$, $1 \leq i \leq n$ such that $r_i < 1$, then there exists a $j$, $1 \leq j \leq n$ such that $r_j > 1$.

Hi if i can get some help for this one here please I'm new to discrete math sorry for my lack of knowledge thank you.

The product o fall others is $=\frac1{r_i}$ – Hagen von Eitzen Feb 16 '19 at 05:11 — Hagen von Eitzen, Feb 16 '19 at 05:11

score 4 · Answer 1 · 2019-02-16T05:30:29.627

4

if we suppose that for each $j$, $1 ≤ j ≤ n$ with $j≠i$, then we have $r_j ≤ 1$ . given that $\forall j$ ${r_j>0}$
we can conclude that $r_1 × r_2 × ... × r_n ≤ r_i <1$ . contradiction.

edited Feb 16 '19 at 05:30

answered Feb 16 '19 at 05:20

Way to go, *+1, endorsed!* Now bring it home with MathJax and $\LaTeX$ify the thing! Cheers! – Robert Lewis Feb 16 '19 at 05:27
Very well done, wish I could upvote again! ;) – Robert Lewis Feb 16 '19 at 06:00
Thank you very much for your help – NaiMomo Feb 16 '19 at 13:08

Prove or disprove: for any $n \in \mathbb{N}$, for any $r_1,r_2, \ldots, r_n \in R$ then there exists $a_j, 1 \leq j \leq n$ such that $r_j > 1$.

1 Answers1