# integral part of surd

$(\sqrt{a}+b)^n=N+f$ where $f \in (0,1)$

$(\sqrt{a}+b)^{n+2} =M+g$ where $g \in (0,1)$

Given that $0<\sqrt{a}-b<1$ and $(a,b)$ belongs to integers show that:

1. If $n$ is odd, $f>g$

2. If $n$ is even, $f<g$

-
I am new to StackExchange,could anyone edit my post to a readable form): –  Tom Lynd Sep 10 '13 at 16:50
Please see here for a guide to writing math with MathJax. Have I interpreted your question correctly with my edit? –  Zev Chonoles Sep 11 '13 at 4:59
yeah,thanks a million:) –  Tom Lynd Sep 11 '13 at 5:00
What is your exact definition of surd? Any restrictions on $a,b,n$ other than $0<\sqrt{a}-b<1$ and trivially $n>0,b\ne0$? Otherwise both parts are wrong (and your proof for part 1 must have a gap)! With obvious indexing of $f,g$ you get for $a=\frac{1}{2}$ and $b=\frac{1}{2}$ for odd $n$ $$f_1 \approx 0.2071 < g_1=f_3\approx0 .7589$$ and for even $n$ $$f_2\approx .4571 > g_2=f_4\approx .1231$$ –  gammatester Sep 11 '13 at 8:03
A third and IMO better alternative: You can answer your own question and show the proof(s). This gives reputation and will be informative to the other users, who up-voted your question or marked it as favourite. –  gammatester Sep 17 '13 at 8:46