Number of roots of a number

Identity · 2009-07-31 00:57:34

This issue came to my attention while doing a practice exam paper. The question is to find

. The solution goes:

So I tried extending this to real roots:

Suppose

for some number a.

I always learnt in school that

implies the positive root though! What is correct??

Also, is the story the same for

?

Ricky · 2009-07-31 03:32:32

Remember that once you write:

You are no longer talking about an equation. Rather, you are now talking about a set of equations.

Now try the rest and you'll see it all makes sense.

Identity · 2009-07-31 04:08:02

I don't see how it makes a difference? If you multiply the right equation by -1 you still get

,

Ricky · 2009-07-31 05:28:16

You seem to be thinking that a must be positive. Also remember that in the two different equations, they are not the same 'a'.

1a2b3c2212 · 2009-07-31 15:29:46

√2 implies the positive root unless the paper gives other wise

Last edited by 1a2b3c2212 (2009-07-31 15:30:06)

Identity · 2009-07-31 23:59:10

So does

not ?

Last edited by Identity (2009-08-01 00:26:32)

1a2b3c2212 · 2009-08-01 00:21:15

√-1 should be i. but if u have i²=-1, then √-1 should either be i or -i.

Ricky · 2009-08-01 03:36:43

In the most general terms, √a can be any roots of the equation:

x^2 = a

However, standard practice is to make it the "positive" root.

Identity · 2009-08-01 03:44:30

But who's to say

is 'more' positive than ?

After all, aren't they off the real-number line?

riad zaidan · 2009-08-01 03:58:45

Hi
By definition i is root of -1 which satisfies the equation i^2=-1 . It lookslike that we have root of 4 equals 2, which does not mean the number that if it is multiplied by itself gives 4, but it means the +ve real number that if it is multiplied by itself gives 4.
w.b.w
Riad Zaidan

Ricky · 2009-08-01 07:09:43

riad zaidan wrote:

By definition i is root of -1 which satisfies the equation i^2=-1 .

The issue is that there are two numbers you can then take for the value of i, either what we currently call i or what we currently call -i. Both of these are square roots of -1.

Identity wrote:

But who's to say
is 'more' positive than
?
After all, aren't they off the real-number line?

Yes, you can't algebraically order the complex numbers so that there is no sense of positive or negative. You can however lexicographically order them, in which case the natural way is that -i < i. However, if you went with -i instead, everything would work out precisely the same, modulo a few negative signs.

Identity · 2009-08-01 08:28:45

Ah ok, thanks ricky

Math Is Fun Forum

#1 2009-07-31 00:57:34

Number of roots of a number

#2 2009-07-31 03:32:32

Re: Number of roots of a number

#3 2009-07-31 04:08:02

Re: Number of roots of a number

#4 2009-07-31 05:28:16

Re: Number of roots of a number

#5 2009-07-31 15:29:46

Re: Number of roots of a number

#6 2009-07-31 23:59:10

Re: Number of roots of a number

#7 2009-08-01 00:21:15

Re: Number of roots of a number

#8 2009-08-01 03:36:43

Re: Number of roots of a number

#9 2009-08-01 03:44:30

Re: Number of roots of a number

#10 2009-08-01 03:58:45

Re: Number of roots of a number

#11 2009-08-01 07:09:43

Re: Number of roots of a number

#12 2009-08-01 08:28:45

Re: Number of roots of a number

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