# [Tutor] square root

Salim salim@nstp.com.my
Fri, 02 Feb 2001 15:38:23 +0800

```Please make a correction:
x**1/2==the square root of x

please note there is no bracket...

tq

At 09:21 AM 2/2/01 +0800, you wrote:
>There's no operator for squareroot (unless I've been misleading myself
>for this long).
>
>Use the power rule of x**(1/y). Here's the relavent rules:
>
>x**y == x to the power of y,
>x**(1/y) == x to the root of y,
>x**(z/y) == x to the root of y - all raised to the power of z,
>
>So x**(1/2) == the square root of x
>
>I don't think it was a dumb question, but then, I'm a newbie script
>kiddie! =)
>
>Andrew Wilkins
>
> > -----Original Message-----
> > W.W. van den Broek
> > Sent: Thursday, 1 February 2001 5:13
> > To: tutor@python.org
> > Subject: [Tutor] square root
> >
> >
> > How do you use the
> > square root as operator
> > in python?
> > Dumb question, but
> > thanks anyway,
> > walter
> > --
> > W.W. van den Broek
> > e-mail:
> > vandenbroek@psyd.azr.nl
> > AZR-Dijkzigt          fax:
> > 010-4633217
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> > tel:          010-4639222
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> > vdbroekw@wxs.nl (thuis)
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> > http://home.planet.nl/~vdbroekw
> >
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