anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sqrt{-1} = x \] what is the value of x
anonymous
  • anonymous
please dont put links
anonymous
  • 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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anonymous
  • anonymous
so what is an imaginary unit
anonymous
  • anonymous
I know of a video you can watch if you want to
anonymous
  • anonymous
yes what is the vid @zaynahf
hba
  • hba
did you see the link @awn786 It has everything.
anonymous
  • anonymous
yes i did but what is an imaganary unit
anonymous
  • 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
  • anonymous
WOW OMG complicated @hba
anonymous
  • anonymous
The first link is really straight forward, try that
hba
  • hba
@awn786 Well,I tried to simplify and put everything :)
anonymous
  • anonymous
thanks for trying. @hba
hba
  • hba
You are welcome :)
anonymous
  • anonymous
listen and can u give a simple i word answer: \[\sqrt{-1}=x\] what is the value of x in numbers.
anonymous
  • anonymous
1 word answer*
anonymous
  • anonymous
not i word
hba
  • 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
  • anonymous
whats w
hba
  • hba
We call this omega,It's derivations is given in my link :)
anonymous
  • anonymous
whats w @hba
hba
  • hba
Now as per your question. x=root{-1} x=i Here i is actually iota and it is an imaginary number.
hba
  • hba
That's omega @awn786
anonymous
  • anonymous
whats omega
hba
  • hba
@awn786 Please check the link i have provided. It has everything Please give it a read.
anonymous
  • anonymous
i said in numbers. your answer is basicly my question
hba
  • hba
Well,In numbers it is sqrt{-1}
anonymous
  • anonymous
it should be easy shouldent it.
anonymous
  • anonymous
i dont understand ur link
anonymous
  • anonymous
what is omega
hba
  • hba
I have already told you what omega is.
anonymous
  • anonymous
no u havent @hba
hba
  • hba
I said it is -1+i sqrt{3}/2
anonymous
  • anonymous
ok (i dont get it but ok) i' only in year 9 doing my gcse's
anonymous
  • 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
  • anonymous
you just need to know what a square root is
anonymous
  • anonymous
i know what a square root is ofc
anonymous
  • anonymous
that's why there's no sqrt of ANY negative numbers, because everthing squared is BIGGER than 0
anonymous
  • anonymous
(possitive)
anonymous
  • anonymous
oh i get it so its an imaganary number because it cant exist?
anonymous
  • anonymous
you have the best responce
anonymous
  • anonymous
EXACTLY, it CAN'T EXIST as a "normal number"
anonymous
  • anonymous
thank u so much.
anonymous
  • anonymous
^^^^^^ @GCR92
anonymous
  • 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
  • anonymous
they INVENTED it
anonymous
  • anonymous
you are welcome
anonymous
  • anonymous
ur a genius @GCR92
anonymous
  • anonymous
not really, but thanks!
anonymous
  • anonymous
ill draw you something
anonymous
  • anonymous
you can bring the most complex thing and make others understand. are u sure ur not related to prfsr. brian cox
anonymous
  • anonymous
|dw:1358026644045:dw|
anonymous
  • anonymous
hahaha you amuse me, nothing is hard when you have an amazing teacher, and I have it right now.
anonymous
  • anonymous
you understand what I drew?
anonymous
  • anonymous
yeah u drew a number line. who is ur teacher
anonymous
  • 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
  • anonymous
|dw:1358026974868:dw|
anonymous
  • anonymous
ALMOST
anonymous
  • anonymous
|dw:1358027044754:dw|
anonymous
  • anonymous
|dw:1358027175723:dw|
anonymous
  • anonymous
|dw:1358027034198:dw|
anonymous
  • anonymous
you put the axis pararell and it's perpendicular cutting in 0
anonymous
  • anonymous
|dw:1358027255837:dw|
anonymous
  • anonymous
that's why i said almost, because you figured out ANOTHER line, and that would make a 300 year old mathematician proud
anonymous
  • anonymous
yepeee
anonymous
  • anonymous
because they CAN be negative in their new axis, the imaginary axis
anonymous
  • anonymous
the imaginary line
anonymous
  • anonymous
it is not allowed to be a negative square root in the REAL line, but it is in the imaginary one
anonymous
  • anonymous
so in imaginative can they ever be 3i or always do they have to be -3i
anonymous
  • anonymous
so we just call it i, like we call 1 one
anonymous
  • anonymous
there are imaginary postive numbers and imaginary negative numbers, under 0
anonymous
  • anonymous
real negatives are at the left of 0, so why can't there be negative imaginary ones?
anonymous
  • anonymous
|dw:1358027454145:dw|
anonymous
  • anonymous
above 0 is positive?
anonymous
  • anonymous
if you delete the minus inside of the root, it would be a real number then!
anonymous
  • anonymous
Yes that's what i was writting
anonymous
  • anonymous
and under 0 negative, i forgot to draw it
anonymous
  • anonymous
just like the real line but perpendicular
anonymous
  • anonymous
above 0 goes posstive until inifinity and under 0 negative until infinity
anonymous
  • anonymous
and on the real line is it same?
anonymous
  • anonymous
|dw:1358027632303:dw|
anonymous
  • anonymous
they are two separated lines.
anonymous
  • anonymous
bye i need to go. if i see u tommorw here i will carry on the conversation.
anonymous
  • anonymous
okay, sure! bye! you are smart btw, you figured the other line very well.
anonymous
  • anonymous
figured out*
anonymous
  • anonymous
thanks
anonymous
  • anonymous
@GCR92 im back
anonymous
  • anonymous
complex numbers
AravindG
  • AravindG
i is actually an imaginary number used to give value to \[\sqrt{-1}\]
anonymous
  • anonymous
yes i get that.
anonymous
  • anonymous
Look above^
ParthKohli
  • ParthKohli
Wait, no!
anonymous
  • anonymous
what?
ParthKohli
  • ParthKohli
\(i\) is defined as the imaginary number such that \(i^2 = -1\), not \(i= \sqrt{-1}\).
anonymous
  • anonymous
no ur wrong
anonymous
  • anonymous
wait
ParthKohli
  • ParthKohli
How so...?
anonymous
  • anonymous
watch
anonymous
  • anonymous
wait
anonymous
  • anonymous
http://imgur.com/kwevvqi
anonymous
  • anonymous
and also its just common knowlige
anonymous
  • anonymous
Knowledge*
ParthKohli
  • ParthKohli
Yes, \(i\) is one of the square roots of \(-1\). But \(i \ne \sqrt{-1}\).
ParthKohli
  • ParthKohli
I mean, \(\sqrt{-1} =i,-i\)
ParthKohli
  • ParthKohli
So you must define \(i^2 = -1\) if you don't want things to mess up
anonymous
  • anonymous
wow gcr92 was WAY better at explaining things
ParthKohli
  • ParthKohli
What would you like to learn?
anonymous
  • anonymous
apparantly his teacher is hgis uncle who started teaching him maths when he was 15 an now is 50
anonymous
  • anonymous
wait
anonymous
  • anonymous
carry on from:
anonymous
  • anonymous
|dw:1358591619083:dw|
anonymous
  • anonymous
look above^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
ParthKohli
  • ParthKohli
Re and Co, yes.
anonymous
  • anonymous
Here you go with the link for complete solution! http://www.learnalberta.ca/content/memg/Division03/Square%20Root/index.html
anonymous
  • anonymous
ermmm it needs to be simple
anonymous
  • anonymous
nope its not
anonymous
  • anonymous
sorry
anonymous
  • anonymous
where is GCR92
anonymous
  • anonymous
Here. What's up?

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