# test

• Sep 30th 2009, 04:39 PM
test
$\displaystyle \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac{ {\sqrt[n]{a}} {\sqrt[n]{b}} }\;\;\mbox{such that a and b}>\:0\:$
• Sep 30th 2009, 07:59 PM
Soroban

You have too many braces . . .

You typed:

[math ] \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac{ {\sqrt[n]{a}} {\sqrt[n]{b}} }\;\;\mbox{such that a and b}>\:0\:[/tex]

and produced this:

$\displaystyle \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac{ {\sqrt[n]{a}} {\sqrt[n]{b}} }\;\;\mbox{such that a and b}>\:0\:$

Eliminate the red braces above:

[math ] \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac {\sqrt[n]{a}} {\sqrt[n]{b}}\;\;\mbox{such that a and b}>\:0\:[/tex]

which produces:

$\displaystyle \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac {\sqrt[n]{a}} {\sqrt[n]{b}} \;\;\mbox{such that a and b}>\:0\:$

• Oct 8th 2009, 01:54 PM
just another test
[tex]j-i > 2^n \rightarrow A(i,j) \leq n[\math]

[tex]x^2\sqrt{x}[\math]
• Oct 9th 2009, 04:49 AM
Soroban

Quote:

[tex]j-i > 2^n \rightarrow A(i,j) \leq n[\math]

[tex]x^2\sqrt{x}[\math]

You must close LaTeX with: .[/math]

Then you will get:

$\displaystyle j-i > 2^n \rightarrow A(i,j) \leq n$

$\displaystyle x^2\sqrt{x}$

• Oct 9th 2009, 01:23 PM