What is the square root of minus 1.
When i do it in a scientific calculator, it says 'i' (imaginary unit) and i have no idea what that means

- anonymous

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- anonymous

\[\sqrt{-1} = x \] what is the value of x

- anonymous

please dont put links

- anonymous

Imaginary numbers result when you are taking the square root of any negative number:
For an example, \[\sqrt{-1}\] = i
\[\sqrt{-25} = 5i\]
The 'i' is kind of like the negative sign

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## More answers

- anonymous

so what is an imaginary unit

- anonymous

I know of a video you can watch if you want to

- anonymous

yes what is the vid @zaynahf

- hba

did you see the link @awn786
It has everything.

- anonymous

yes i did but what is an imaganary unit

- anonymous

1) http://www.youtube.com/watch?v=Iv_uzy4_gec
2)https://www.khanacademy.org/math/algebra/complex-numbers/complex_numbers/v/algebra-ii--imaginary-and-complex-numbers?v=C-2Ln0pK3kY

- anonymous

WOW OMG complicated @hba

- anonymous

The first link is really straight forward, try that

- hba

@awn786 Well,I tried to simplify and put everything :)

- anonymous

thanks for trying. @hba

- hba

You are welcome :)

- anonymous

listen and can u give a simple i word answer: \[\sqrt{-1}=x\] what is the value of x in numbers.

- anonymous

1 word answer*

- anonymous

not i word

- hba

I can help you understand what imaginary numbers are.
So,the base of imaginary numbers is actually laid upon some of the things like
\[\huge \ i=\sqrt{-1}\]
The other thing is
\[\huge\ \omega=\frac{ -1+i \sqrt{3} }{ 2 }\]

- anonymous

whats w

- hba

We call this omega,It's derivations is given in my link :)

- anonymous

whats w @hba

- hba

Now as per your question.
x=root{-1}
x=i
Here i is actually iota and it is an imaginary number.

- hba

That's omega @awn786

- anonymous

whats omega

- hba

@awn786 Please check the link i have provided.
It has everything
Please give it a read.

- anonymous

i said in numbers. your answer is basicly my question

- hba

Well,In numbers it is sqrt{-1}

- anonymous

it should be easy shouldent it.

- anonymous

i dont understand ur link

- anonymous

what is omega

- hba

I have already told you what omega is.

- anonymous

no u havent @hba

- hba

I said it is -1+i sqrt{3}/2

- anonymous

ok (i dont get it but ok) i' only in year 9 doing my gcse's

- anonymous

\[\sqrt{-1}\] doesn't exist as a the REAL number you are asking for, because there's no number that you can square (because of the SQUARE root) and gives you -1

- anonymous

you just need to know what a square root is

- anonymous

i know what a square root is ofc

- anonymous

that's why there's no sqrt of ANY negative numbers, because everthing squared is BIGGER than 0

- anonymous

(possitive)

- anonymous

oh i get it so its an imaganary number because it cant exist?

- anonymous

you have the best responce

- anonymous

EXACTLY, it CAN'T EXIST as a "normal number"

- anonymous

thank u so much.

- anonymous

^^^^^^ @GCR92

- anonymous

so mathematicians around XVII century had to figure out another type of numbers, called COMPLEX numbers, and the first number of complex numbers is \[\sqrt{-1}\]

- anonymous

they INVENTED it

- anonymous

you are welcome

- anonymous

ur a genius @GCR92

- anonymous

not really, but thanks!

- anonymous

ill draw you something

- anonymous

you can bring the most complex thing and make others understand. are u sure ur not related to prfsr.
brian cox

- anonymous

|dw:1358026644045:dw|

- anonymous

hahaha you amuse me, nothing is hard when you have an amazing teacher, and I have it right now.

- anonymous

you understand what I drew?

- anonymous

yeah u drew a number line. who is ur teacher

- anonymous

My uncle, who's been teaching maths since he was 15 and he's now 57, anyway, let's go back to our imagination

- anonymous

|dw:1358026974868:dw|

- anonymous

ALMOST

- anonymous

|dw:1358027044754:dw|

- anonymous

|dw:1358027175723:dw|

- anonymous

|dw:1358027034198:dw|

- anonymous

you put the axis pararell and it's perpendicular cutting in 0

- anonymous

|dw:1358027255837:dw|

- anonymous

that's why i said almost, because you figured out ANOTHER line, and that would make a 300 year old mathematician proud

- anonymous

yepeee

- anonymous

because they CAN be negative in their new axis, the imaginary axis

- anonymous

the imaginary line

- anonymous

it is not allowed to be a negative square root in the REAL line, but it is in the imaginary one

- anonymous

so in imaginative can they ever be 3i or always do they have to be -3i

- anonymous

so we just call it i, like we call 1 one

- anonymous

there are imaginary postive numbers and imaginary negative numbers, under 0

- anonymous

real negatives are at the left of 0, so why can't there be negative imaginary ones?

- anonymous

|dw:1358027454145:dw|

- anonymous

above 0 is positive?

- anonymous

if you delete the minus inside of the root, it would be a real number then!

- anonymous

Yes that's what i was writting

- anonymous

and under 0 negative, i forgot to draw it

- anonymous

just like the real line but perpendicular

- anonymous

above 0 goes posstive until inifinity and under 0 negative until infinity

- anonymous

and on the real line is it same?

- anonymous

|dw:1358027632303:dw|

- anonymous

they are two separated lines.

- anonymous

bye i need to go. if i see u tommorw here i will carry on the conversation.

- anonymous

okay, sure! bye! you are smart btw, you figured the other line very well.

- anonymous

figured out*

- anonymous

thanks

- anonymous

@GCR92 im back

- anonymous

complex numbers

- AravindG

i is actually an imaginary number used to give value to \[\sqrt{-1}\]

- anonymous

yes i get that.

- anonymous

Look above^

- ParthKohli

Wait, no!

- anonymous

what?

- ParthKohli

\(i\) is defined as the imaginary number such that \(i^2 = -1\), not \(i=
\sqrt{-1}\).

- anonymous

no ur wrong

- anonymous

wait

- ParthKohli

How so...?

- anonymous

watch

- anonymous

wait

- anonymous

http://imgur.com/kwevvqi

- anonymous

and also its just common knowlige

- anonymous

Knowledge*

- ParthKohli

Yes, \(i\) is one of the square roots of \(-1\). But \(i \ne \sqrt{-1}\).

- ParthKohli

I mean, \(\sqrt{-1} =i,-i\)

- ParthKohli

So you must define \(i^2 = -1\) if you don't want things to mess up

- anonymous

wow gcr92 was WAY better at explaining things

- ParthKohli

What would you like to learn?

- anonymous

apparantly his teacher is hgis uncle who started teaching him maths when he was 15 an now is 50

- anonymous

wait

- anonymous

carry on from:

- anonymous

|dw:1358591619083:dw|

- anonymous

look above^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

- ParthKohli

Re and Co, yes.

- anonymous

Here you go with the link for complete solution!
http://www.learnalberta.ca/content/memg/Division03/Square%20Root/index.html

- anonymous

ermmm it needs to be simple

- anonymous

nope its not

- anonymous

sorry

- anonymous

where is GCR92

- anonymous

Here. What's up?

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