Prove that for each positive integer $r$, there exists a rational number $c$, s.t. $c > r.$ Could we let $c$ = $(r+1)$$/1$ and then conclude $r+1 > r$ so $c > r$?
Asked
Active
Viewed 98 times
0
-
1$c = (r+1)/1$? Of course this works because natural numbers are rational. But if you need a rational that's not an integer, how about just taking the midpoint of $r$ and $r+1$? – aduh Sep 09 '16 at 01:26
-
so let $c = (r+r+1)/2 = r+1/2$? Isn't that the same principle though? – Jay3 Sep 09 '16 at 01:31
-
For these very elementary results, it's useful to know how things have been defined. What's a rational number? How are they ordered? Have you already shown that all integers are rational? – aduh Sep 09 '16 at 01:39
-
No I haven't. The class I'm in is Real Analysis so we've defined what rationals are in that it's a integer divided by a nonzero integer. – Jay3 Sep 09 '16 at 01:44
-
In that case, just apply the definitions and the problem should be straightforward. I'll let you make an attempt before writing anything more myself. – aduh Sep 09 '16 at 01:46
-
Alright so we are given r. Let c $\in$ Q. Then $c= m/n$ for some $m,n$ $\in$ Z, $m \nequal$ 0. Since $r$ is an integer,$ r+1$ is also an integer. Let $m$ = $r+1$ Since 1 is non-zero integer, let $n$= 1. Then $c$= $(r+1)/1$. $(r+1)/1 = 1$ so $c$=$r+1$. Since $r+1 > r$, $c > r$. Would that be sufficient? – Jay3 Sep 09 '16 at 01:57
-
Let us continue this discussion in chat. – aduh Sep 09 '16 at 02:01