Multiplying Square Root of Negative Numbers

[ИльясМаметов-ио]

this is how government calculate your pension

[mr.d]

_The multiplicative property of square roots that √a*√b = √(a*b) only applies when a and b are more than or equal to 0._

[Schrodinger_]

There's a lot of confusion in the comments stemming from something that, unfortunately, they don't teach you until complex analysis class, which is that the √x operation is not the same as (x)^(1/2) operation, but rather, the former is a specific case of the latter specified on something called a "principal branch". Basically, if you take the equation y = x^(1/n), then there are n possible solutions of x on the complex plane, which makes x^(1/n) not strictly a function. If you want to define a function and call it the "nth root of x", you have to basically define it as "that solution to y = x^(1/n) that happens to lie in THIS region in the complex plane". The square root happens to be one of those functions, where it's basically "the square root of x is the solution to y = x^(1/2) that falls on the _right half_ of the complex plane."

The reason this means that the first method doesn't work and the second does is because you can show that, if you specify a branch, then a product of square roots is not generally equal to a square root of products. This can be proven using complex analysis, but here's an intuitive explanation. The principal branch square root requires an output in the right half of the complex plane. So, for a product of square roots, each factor must lie in the right half of the complex plane, but their product doesn't have to. However, by definition of principal square root, a square root of a product _must_ lie in the right half of the complex plane.

[samakolBanbol]

Imaginary numbers are easier to handle than hallucinary numbers

[Adi-fezg]

Man this guy roasts every regular middle school math student like crazy

[BRAINSPLATTER]

Im halfway through a bachelor's in computer engineering and this man is still HUMBLING me. 🤣🤣

[jasonarmstrong]

√16 is +4 ... Solving for n where n^2 = 16 is +-√16 so +- 4.. it is true to say when not zero a number has two square roots. but the √ symbol means principal square root ie positive so the video is correct. √-1 is i So √-2 is (√2)i so √2*√8*i^2 = √16 *-1 =-4.

[skit_inventor]

When extending the square root function to the complex plane (which cannot be avoided here), you _have_ to specify the branch that you're using, and also mention whether the extension to the negative reals was made through the upper or the lower half-plane, which would indicate whether sqrt(-1) is "i" or "-i". In fact, in the example from the video, the two roots don't necessarily have to be from the same branch. If you take different branches for each root, you get +4, and -4 otherwise.

P.S. no, the imaginary unit is _not_ defined as sqrt(-1) (because of this whole thing with multivalued functions that would make a lot of mess), rather, "i^2 = -1" is its property resulting from the multiplication rule in the complex plane (or, equivalently, it's defined as one of the roots of x^2 +1 = 0, doesn't matter which one, as long as it's fixed throughout the whole theory).

P.S. #2 Don't you find it strange that sqrt(-2)×sqrt(-8) cannot, according to the author, be written as sqrt((-2)×(-8)), but sqrt(-2) = sqrt((-1)×2) = sqrt(-1)×sqrt(2) no problem? -1 seems like a privileged individual that you can take in and out of the root, but not any other number 😅 (again, it all comes down to the specific branch and where you insert the branch cut)

[alllove]

Thank you for showing us the true and yet-to-be accepted way to show us how to write absolute value for the negative numbers ❤

[Psychobum]

I'm 32 now, haven't ever needed to use this information once in my life outside of high school but being given this crash course about negative square roots makes me remember that I used to LOVE this kind of stuff. It felt like puzzle solving, unlike a lot of other parts of math you learn in hs.

[atifavdovic]

More needs to be said on this matter. One should say something about "multi-value" functions of root and real and complex principal root. Without precisely determining the definition on what the "sqare root" is (since there are more different functions called the same name) the discussion on this topic can easily open the door of philosophy.

[kyleschutrum]

I changed the sqrt(-8) to sqrt(4) and sqrt(-2) then sqrt(-2) and sqrt(-2) just multiply to become -2 and then that’s multiplied by sqrt(4) which is 2. Final answer is -4

[ibrachaka]

I studied this, got a doctorate, forgot all of it 😂

[CHS]

I like these shorts of all these math problems I haven’t seen for decades, so I have forgotten a some of them, these are good reminders.

[kSevinVII]

That chalk hitting the board sounds so good

[rebase]

sqrt(a) as defined in high-school maths is ill-defined when 'a' is allowed to be negative.

In high school we define sqrt(a) as the nonnegative root of the polynomial x^2 - a. E.g. x^2 - 4 → x ∈ {2, -2} so we define sqrt(4) = 2.

However the roots of x^2 + 4 are {2i, -2i}. Neither of them are positive; they fall outside the real number line!

Ok, then define sqrt(-a) = sqrt(a)i (for positive real 'a'). You can do this, but the property sqrt(x)sqrt(y) = sqrt(xy) no longer holds! This is due to the fact that unlike the positive real numbers, the upper imaginary line is not closed under multiplication!

[adrianwoodruff]

Okay, so you have to take out the i before multiplying. Thanks for the free lesson.

[Rhah-]

Wow, I haven't done this stuff since high school. Always nice to refresh every now and again! Thanks guy!

[pilarleonor]

Very good explanation. Your voice clarifies it better.

[falkez]

it makes sense hear me out:
if some imaginary person gives you four apples
you somehow owe 4 apples to someone