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## Algebra 2

### Course: Algebra 2>Unit 2

Lesson 1: The imaginary unit i

# i as the principal root of -1

The video explores the intriguing concept of imaginary numbers, specifically the imaginary unit 'i'. 'i' is equal to the square root of -1, and every complex number can be expressed as a + bi, where a and b are real numbers. This video clarifies misconceptions about square roots of negative numbers. Created by Sal Khan.

## Want to join the conversation?

• what does principal square root mean? •   When taking square roots of a positive number you always get two answers. For example, the square root of 4 can be either 2 or -2. This is because the square of either one is 4: (2)(2) = 4 = (-2)(-2). So if you want to specify that you mean the positive answer then you would say the "principal square root". So...the principal square root of 4 is 2. The "other" one is -2.
• Wait. I'm lost. How can -1 be equal to 1?? •   It's not. What Sal is saying is that people try to prove that i = square root(-1) is wrong because they end up with an answer that 1 = -1 which obviously isn't true. But they're wrong because the square root multiplication rule doesn't apply when both numbers are negative.
Yes the absolute value of -1 = absolute value of 1
but -1 is not equal to one.
• So why exactly can't A and B both be negative? It sounds like they can't because of the "Because I said so" line of reasoning. If √a * √b = √a*b doesn't work when both numbers are negative then maybe something is wrong with the proof. •   Observe what happens.
1 = √1 = √(-1 * -1) = √-1 * √-1 = i * i = -1
Whoops, suddenly, 1 is equal to -1. We broke negative numbers :(

Assuming √(zw) = √z√w leads to contradictions such as above, that's why we ban it.
• What is the square root of i? • All the imaginary numbers can be written as a+ib where a is the real part and b the imaginary.
Let us assume that i^1/2 = a+ib for some a and b.
i=(a+ib)^2
i=a^2 + ib^2 + 2aib
ib^2 = -b^2
i=a^2-b^2+2aib
As the left hand side has only imaginary part, a^2 - b^2 =0
a=b (taking principal roots)
so, i = 2aib
2ab = 1
2a^2 = 1 (a=b)
a=1/√2
b=1/√2
i^1/2 = (1√2+i 1√2)
Taking 1/√2 common,
i^1/2 = 1/√2(1+i)
• At when it says the radical sign means the principle square root; what's the notation for either square root, or for the negative square root, in that case?

Also when talking about the third root, you wouldn't have to distinguish between principle and negative, right? Do you have to distinguish for all even roots, but not for odd roots? • The notation for both square roots is ±√x. If you want only the negative square root, then you would say -√x.

As for odd roots, you still have to distinguish between principal and non principal roots. The only difference is that a non principal odd root is not simply the negative of the principal root.

For instance, the cube root of -8 is -2, but it can also be 1±√3 i.
• How can we prove that both A and B cant be negative?(for any values) • Great question.
*Note that ⁺√ implies principal square root.*
We must prove that:
⁺√(ab) ≠ ⁺√a • ⁺√b
For
a, b < 0
If a and b are negative, then the square root of them must be imaginary:
⁺√a = xi
⁺√b = yi
x and y must be positive (and of course real), because we are dealing with the principal square roots.
⁺√a • ⁺√b = xi(yi) = -xy
-xy must be a negative real number because x and y are both positive real numbers.
On the other hand,
⁺√(ab) = √[(xi)²(yi)²] = (xyi²)² = (xy)²
Since ⁺√(ab) = (xy)² and ⁺√a • ⁺√b = -xy, our problem becomes to prove that:
(xy)² ≠ -xy
For
x, y > 0
Well this is easy! The left hand side is obviously positive and the right hand side is obviously negative, so they cannot be equal! Therefore, ⁺√(ab) ≠ ⁺√a • ⁺√b if a, b < 0. Q.E.D. Comment if you have any questions.
• In calculator, if -1, then press square root, why square root of -1 is error on calculator? • Because real numbers cannot be squared and equal a negative number. (I.e. -3 x -3 = 9). However, imaginary numbers (which are created outside of the normal and "real" numbers) make the square root of -1 possible, but that does not make it "real" or true. Try looking for an "i" sign on your calculator. It can be substituted for the square root of 1.
(1 vote)
• How to distinguish between a radical sign that indicates a principal square root and a radical sign that says the answer must be plus or minus the square root • Great question.
there is no radical sign that indicates both.
That is why when we solve equations like x^2=4 we list two solutions, the principal square root 2 and the negative square root -2. all positive real numbers have two distinct square roots that are opposites of each other.
When solving a problem, if you are looking for square roots, it is up to you to know when you need to consider the negative root. Sometimes it must be included, sometimes it makes no sense and can be discarded.
On the other hand, if you are reading a problem that has the square root symbol in it, it ALWAYS means the principal (positive) root. If the negative is intended, a minus sign will be in front of it. If both are intended they will either be listed separately or the +/- sign will be placed in front of the radical sign. The quadratic formula is usually written with a +/- since you need both the positive and negative roots to find both solutions of a quadratic equation.
• why is the principate square root used for negative numbers? can't it be used while multiplying or dividing fractions, however i'm am always asked to simplify the questions but how can i simplify 1/3 times the square root of -63 - the square root of -28? • I still do not understand...Why doesn't the square root property: √(A×B)=√(A)×√(B) not work when A and B are both negative?? In the video, Mr. Khan explained that A and B BOTH cannot be negative. • Consider sqrt(-a * -b).
sqrt(-a * -b)
= sqrt(-a) * sqrt(-b)
= sqrt(a) * i * sqrt(b) * i
= sqrt(a) * sqrt(b) * i^2
= -sqrt(a) * sqrt(b)
= -sqrt(a * b)

Now if we don't use the square root property.
sqrt(-a * -b)
= sqrt(a * b)
Which isn't equal.

Let sqrt(-a * -b) = c.
Notice how squaring the equation you get
(-a * -b) = c^2 = (-c)^2.
This causes ambiguous case, thus it "doesn't work".