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#1 2013-04-22 23:07:29

eunolwin
Member
Registered: 2013-04-22
Posts: 4

struggling with fractions in brackets with indices.

I'm struggling with   -4 to the power of 2 divided by 3
  (-4) to the power of 2 divided by (-4) to the power of 3
  -(-3) to the power of 3 divided by -(-4) to the power of 3
(-2) to the power of 0 divided by -4 to the power of 0 and others of the same ilk.

Is there a section I can study on your site?
Can you give me the gist of how to do these, please?
Many thanks:

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#2 2013-04-22 23:16:51

SteveB
Member
Registered: 2013-03-07
Posts: 547

Re: struggling with fractions in brackets with indices.

A number to the power of 2 means that you find the square of it.  (eg. 5 ^ 2 = 25)

Be careful when squaring a negative number because when you multiply a number by itself the signs cancel.
So for example if it were (-5) ^ 2 the (-5) x (-5) would be plus 25. (Obviously yours was (-4) ^ 2 )

I presume you know how to divide by 3 (?)

In general raising a number to the power of something (assuming it is a whole number) means that it is muliplied by itself
that many times. So (-5) ^ 3 = (-5) x (-5) x (-5)    [Notice that it is negative because there are 3 negatives here]

You could regard this as  (-5) ^ 3 = (1) x (-5) x (-5) x (-5) = -125  [EDIT: Sorry there was a typo there]
and that (-5) ^ 2 = (1) x (-5) x (-5) = 25

If you raise a number to the power of zero then this is always 1. 

So in my example of 5 :   

5 ^ 0 = 1

I have made up my own example there to make sure I am not giving away the answer.

5 ^ (1/2) = (square root of 5)
5 ^ (1/3) = (cube root of 5)
5 ^ (2/3) = (cube root of 5 squared)

Also note that if the power is negative then:

5 ^ (-1) =  (1/5)
5 ^ (-2) =  (1/25)
5 ^ (-3) =  (1/125)

When you say " (-4) to the power of 2 divided by 3 " do you mean  (-4) ^ (2/3)  or  (((-4)^2) / 3)  ? [These are not the same]

(EDIT: The first of these is very difficult and requires complex number theory, but the second is not too difficult.)
(EDIT2: Actually the real solution does work so the cube root of 4 squared is okay.)

Last edited by SteveB (2013-04-23 06:42:29)

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#3 2013-04-23 00:29:14

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,247

Re: struggling with fractions in brackets with indices.

hi eunolwin

Welcome to the forum.

Have a look at

http://www.mathsisfun.com/algebra/expon … ional.html

Do you know about complex numbers ?

Some power questions lead into complex numbers.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#4 2013-04-23 02:06:12

eunolwin
Member
Registered: 2013-04-22
Posts: 4

Re: struggling with fractions in brackets with indices.

My problem is the brackets. The calculations are fine, but it seems sometimes the brackets affect whether the answer is + or -. I know that 2 negatives multiplied become positive, but it seems that certain brackets change this.

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#5 2013-04-23 02:12:52

eunolwin
Member
Registered: 2013-04-22
Posts: 4

Re: struggling with fractions in brackets with indices.

Thanks Steve. I meant. - 4^2/3 . My answer for that would be-16/3
Whereas (-4)^2/(-4)^3 is -1/4?

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#6 2013-04-23 02:22:21

SteveB
Member
Registered: 2013-03-07
Posts: 547

Re: struggling with fractions in brackets with indices.

So let us first work out (-4) ^ 2

Well this is 16 because (-4) ^ 2 = (-4) x (-4) = -(-(16)) = 16

Then we get (16 / 3)

So  (((-4) ^ 2)/3) = (16/3)

On the other hand: -((4^2)/3) = -(16/3) 

Now how about ((-4) ^ 2) / ((-4) ^ 3) = 16/(-64) = 1/(-4) = -(1/4)

If there are an odd number of negative terms in a multiplication then the result is negative.

If there are an even number of negative terms in a multiplication then the result is positive.

Last edited by SteveB (2013-04-23 02:29:01)

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#7 2013-04-23 02:46:56

eunolwin
Member
Registered: 2013-04-22
Posts: 4

Re: struggling with fractions in brackets with indices.

Many thanks, Steve. I agree with that and got the same answer, but wasn't sure I was right. ALL the brackets confuse me.

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