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

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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
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\[\sqrt{-1} = x \] what is the value of x
please dont put links
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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Other answers:

so what is an imaginary unit
I know of a video you can watch if you want to
yes what is the vid @zaynahf
  • hba
did you see the link @awn786 It has everything.
yes i did but what is an imaganary unit
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
WOW OMG complicated @hba
The first link is really straight forward, try that
  • hba
@awn786 Well,I tried to simplify and put everything :)
thanks for trying. @hba
  • hba
You are welcome :)
listen and can u give a simple i word answer: \[\sqrt{-1}=x\] what is the value of x in numbers.
1 word answer*
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 \[\huge\ \omega=\frac{ -1+i \sqrt{3} }{ 2 }\]
whats w
  • hba
We call this omega,It's derivations is given in my link :)
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
whats omega
  • hba
@awn786 Please check the link i have provided. It has everything Please give it a read.
i said in numbers. your answer is basicly my question
  • hba
Well,In numbers it is sqrt{-1}
it should be easy shouldent it.
i dont understand ur link
what is omega
  • hba
I have already told you what omega is.
no u havent @hba
  • hba
I said it is -1+i sqrt{3}/2
ok (i dont get it but ok) i' only in year 9 doing my gcse's
\[\sqrt{-1}\] 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
you just need to know what a square root is
i know what a square root is ofc
that's why there's no sqrt of ANY negative numbers, because everthing squared is BIGGER than 0
(possitive)
oh i get it so its an imaganary number because it cant exist?
you have the best responce
EXACTLY, it CAN'T EXIST as a "normal number"
thank u so much.
^^^^^^ @GCR92
so mathematicians around XVII century had to figure out another type of numbers, called COMPLEX numbers, and the first number of complex numbers is \[\sqrt{-1}\]
they INVENTED it
you are welcome
ur a genius @GCR92
not really, but thanks!
ill draw you something
you can bring the most complex thing and make others understand. are u sure ur not related to prfsr. brian cox
|dw:1358026644045:dw|
hahaha you amuse me, nothing is hard when you have an amazing teacher, and I have it right now.
you understand what I drew?
yeah u drew a number line. who is ur teacher
My uncle, who's been teaching maths since he was 15 and he's now 57, anyway, let's go back to our imagination
|dw:1358026974868:dw|
ALMOST
|dw:1358027044754:dw|
|dw:1358027175723:dw|
|dw:1358027034198:dw|
you put the axis pararell and it's perpendicular cutting in 0
|dw:1358027255837:dw|
that's why i said almost, because you figured out ANOTHER line, and that would make a 300 year old mathematician proud
yepeee
because they CAN be negative in their new axis, the imaginary axis
the imaginary line
it is not allowed to be a negative square root in the REAL line, but it is in the imaginary one
so in imaginative can they ever be 3i or always do they have to be -3i
so we just call it i, like we call 1 one
there are imaginary postive numbers and imaginary negative numbers, under 0
real negatives are at the left of 0, so why can't there be negative imaginary ones?
|dw:1358027454145:dw|
above 0 is positive?
if you delete the minus inside of the root, it would be a real number then!
Yes that's what i was writting
and under 0 negative, i forgot to draw it
just like the real line but perpendicular
above 0 goes posstive until inifinity and under 0 negative until infinity
and on the real line is it same?
|dw:1358027632303:dw|
they are two separated lines.
bye i need to go. if i see u tommorw here i will carry on the conversation.
okay, sure! bye! you are smart btw, you figured the other line very well.
figured out*
thanks
@GCR92 im back
complex numbers
i is actually an imaginary number used to give value to \[\sqrt{-1}\]
yes i get that.
Look above^
Wait, no!
what?
\(i\) is defined as the imaginary number such that \(i^2 = -1\), not \(i= \sqrt{-1}\).
no ur wrong
wait
How so...?
watch
wait
http://imgur.com/kwevvqi
and also its just common knowlige
Knowledge*
Yes, \(i\) is one of the square roots of \(-1\). But \(i \ne \sqrt{-1}\).
I mean, \(\sqrt{-1} =i,-i\)
So you must define \(i^2 = -1\) if you don't want things to mess up
wow gcr92 was WAY better at explaining things
What would you like to learn?
apparantly his teacher is hgis uncle who started teaching him maths when he was 15 an now is 50
wait
carry on from:
|dw:1358591619083:dw|
look above^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Re and Co, yes.
Here you go with the link for complete solution! http://www.learnalberta.ca/content/memg/Division03/Square%20Root/index.html
ermmm it needs to be simple
nope its not
sorry
where is GCR92
Here. What's up?

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