Re: Can I have fries and a calculator with that?
- From: "Dave L. Renfro" <renfr1dl@xxxxxxxxx>
- Date: 9 Dec 2005 17:44:21 -0800
Dave L. Renfro wrote:
>> I came across the following while searching for
>> something else. It speaks for itself.
>>
>> ----------------------------
>>
>> http://www.vbforums.com/showthread.php?referrerid=43870&t=298609
>>
>> How do i rationalize the denominator in this?
>> 6/sqr(x) + sqr(3) I know for somthing like
>> 7/sqr(4) i can use a rationalizing factor of
>> sqr(4) to end up getting 7sqr(4)/4 but then
>> im not sure if this ends up being reduced to
>> 7sqr(1). So i guess im asking two quesitons.
>> Thanks.
>>
>> ----------------------------
>>
>> I'm almost tempted to think this was a troll post
>> given the "two quesitons" at the end, but I think
>> doing so requires too much textual interpretation
>> to be realistic.
Virgil wrote:
> If that denominator is (sqrt(x) + sqrt(3)), you can
> rationalize it by multiplying the numerator and denominator
> by (sqrt(x) - sqrt(3)) or by (-sqrt(x) + sqrt(3)).
>
> To rationalize a denomonator of sqrt(4), note that
> sqrt(4) = 2, which is already rational.
Uh ... I think we all know this, or at least I hope we do.
My point (see thread title) was that this person appears
to suffer from that malady some people get from calculator
overuse, where they aren't able to recognize sqrt(4) and
sqrt(1) as 2 and 1, and where they think sqrt(4) cancels
4 in sqrt(4)/4 to produce sqrt(1).
As for rationalizing denominators, here's a less trivial
example for you. We can rationalize the denominator of 1/b,
where
b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
by multiplying the numerator and denominator by f(b), where
f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
+ (1,125,983,744)*x^7 - (35,305,193,472)*x^5
+ (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
After multiplying, but before reducing, the denominator will be
- 525,242,269,696.
The numerator will be an integer-linear combination of 15
rationally independent square root terms.
Dave L. Renfro
.
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