You only forgot something. When you get a square root the answer will have a negative sign and a positive sign, so square root of 1 is -1 and 1. Those both numbers are posible answers. So i^2=-1 and 1, and there the -1 completes the term.
@oadg050693 i^2 always = -1. That's by definition, and that's how the complex number plane works. However, what you said about the square root of 1 is correct. The square root of 1 = 1 and -1.
From when the complex numbers first appear in your equasion, you cant use the multiplication as you generally do with real numbers, it has other rules in the C set.
You also made a false statement at the 5th step. -√(-1)*√(-1)=√(1)*√(1) (you forgot the - at the left side of the equasion). But -i=√(1)*√(1) is also false, so the solution lies in the fact that 1/i=-i.
@DarkNexarius yes, this is true. That is why i is an "imaginary" number, because it is not real. i is pretty much "no solution" in a pretty format. While this may sound like a useless waste of time, i can actually be used extensively in electrical engineering and pattern designing.
The problem is the fourth step: i / √(1) is NOT equal to √(1) / i. Besides, this breaks the fundamental rules of math. The linear function y = x states that whatever x is, y MUST be, and vice versa. A number cannot equal anything other than itself, otherwise its value is not its own and its false value doesn't have its own either. So, by saying that -1 = 1, you're pretty much saying that -1 is not -1. If -1 is not -1, then what is -1? It can't be 1 because you're implying that 1 is not 1.
@TheDBZisepic There is nothing wrong with the fourth step. You're dealing with complex numbers. For simplicity, I will use polar form: r cisθ i = 1cis(π/2) √1 = 1cis(0) or -1cis(0) (because √1 = ±1) so you have: 1cis(π/2) / 1cis(0) this = (1*1)cis(π/2 - 0) = 1cis(π/2) and -1cis(0) / 1cis(π/2) this = (-1*1)cis(0 - (π/2)) = -1cis(-π/2) = 1cis(π/2) They are the same.
@acnelson12 sorry, it deleted my line breaks. That makes it harder to read. There should be a break after "r cisθ" I think you can figure the rest out.
sqrt(x) is k and-k. So here is sqrt(1) -> 1 and -1. The definition for complex numbers is: i ² = -1 but it applies also (-i) ² =- 1 and it follows sqrt (-1) --> i and -i.
In this case: sqrt (-1) * sqrt (-1) = i *- i = 1 because in such case is -1 the same "x" and complex solutions are always conjugated.
@KaiForce we call it "i" or "j". It doesn't belong anywhere between -∞ and ∞ but you can still use it as long as you remember that taking the reciprocal is the same as negating it.
so what u need to do is to define the transformation f(g(x)) = f(h(x))*f(k(x)) for complex entries.
u can avoid this problem by using a symbolic function letter instead of the intuitive sqrt symbol which might lead to be used like other intuitive symbols like * and /.
i think using ± in front of the sqrt wouldnt make no difference, since the pos definition of the the sqrt function is used on both sides of the equal. but (-1)**(0.5) is not (-1)**(-0.5)
u need to be careful by replacing the function f(g(x)) by f(h(x))*f(k(x)) when f(x) means sqrt(x).
u cant transform from (-1/1)**(0.5) equal (1/-1)**(-0.5) to seperate sqrts, because u already have to pull out the i at that very point, to avoid errors of your pos definition of the sqrt, i think.
The argument of -1 is 180º.. right? So the square root of anything with 180º will have it's first root at 180º/2 and the other one 360º/2 further.. So I don't get why he says that the square root of -1 is 'i'..
The only thing this video does is reenforcing the stereotype about black people and their supposed inability to perform abstract cognitive processes like us normal people do..
@PatyTheImp I said, quoting, "..it's first root at 180º/2 and the other one 360º/2 further.." That would be 'i' and '-i'.. when did I say that those expressions are real numbers..?
Third equation is right, you have i at both sides. Fourth one is wrong: on the left, you have i, while on the right you have 1 / i which is equal to -i. And i =/= -i
The wrong part is that you did not square the left side (i^2). You cannot simply square only the right part (sqrt(1).sqrt(1)) and break the fundamental law of an equation ("do the same thing on both sides").
i like this proof of -1=1. it took me a couple of minutes but if you take the square root of a number, you get two solutions. a positive and a negitive solution. a calculator gives commonly the positive one out. but the square root delivers two [sqrt(1)=1 and -1] . so in this case, to have a mathematical correct solution, you have to take the negative solution of the square root.
taking the square root of a negative isn't wrong, it's just imaginary (he correctly replaced it with i). And when you square i, you get -1, just as he did. That part is correct.
What's incorrect is the right side of the equation- the sqrt(1)*sqrt(1). Both of those numbers are +/- 1, not just 1. (If you think about it, squaring -1 can also equal 1.) Replacing both sqrt(1)'s with 1 is extraneous; the other solution is -1 because -1*1=-1. so yep.
@DRKVideoProductions Taking the square root of a number is specifically defined as the PRINCIPLE square root... the real flaw is that sqrt(x/y) = sqrt(x)/sqrt(y) does not apply when x and y are both negative numbers.
he did not take the square root of -1. he replaced it with the letter i from the complex numbers. these allow you a to handle this square root easier in your equation.
@31093236 Well, that was my comment below yours. Many ppl missed the error.
i is as ubiquitous as Pi in math. 'i' is tightly coupled to harmonic oscillators. Eulers eqtn:
e^(i*x) = cos(x) + i*sin(x) [a beautiful eqtn]
is the reason behind this. Schrodingers eqtn also employs 'i' for the quantum mechanics (waves of probability). Study math and you will know the mind of God.
I don't see ANYONE who has the correct analysis here. I am an engineer who has taken a ton of math - trust me.
The 'error' occurs @ 1:10. You cannot arbitrarily split the radical of num & den when the combined divisional radicand (ie: 1/-1) is negative! That sign must remain in the numerator under the radical. Everything else follows from this error.
For example: This error immediately implies that ( i = 1/ i ) which is already wrong because ( 1/ i = - i)
@dwarduk2 Actually, no. He assumed if x=a then sqrt(x)=sqrt(a). Which is correct for real numbers, however in real numbers sqrt(-1) is not defined and hence there is a mistake in proof, and if you work in complex numbers then sqrt(1) is both -1 and 1 while sqrt(-1) is both i and -i so you actually have (1 or -1) = (1 or -1) which is correct.
@Rayqiorg x = a does indeed mean that sqrt(x) = sqrt(a). Taking your line of reasoning, the argument is broken because squaring isn't a 1 to 1 function. I don't have it to hand, but there is an extensive example in the Maths textbook I used last year on this.
As for my point, I was actually talking about the step of sqrt(-1)^2 = sqrt(1)^2, where x was sqrt(-1) and a was sqrt(1). One last thing - (-i)^2 = (i^3)^2 = i^6 = 1 * i^2 = -1, so it's -1 = (1 or -1), which was my whole point.
@dwarduk2 The mistake he actually made is the following: square root is defined only for NON-NEGATIVE numbers.
There is no square root of -1, but i^2=-1. That equation (i^2=-1) cannot be transformed by taking the square root of both sides, because you can do that only if the both sides are non-negative (obviously, in this case, the right side is negative).
Simply said: i is not the square root of -1. There is no square root of -1.
@SunTzutheWise It depends what stage of maths you are looking at, and the sets you are using. There is a square root function that applies to negative numbers, and also one that applies to complex numbers. They just aren't the same as the one for weakly positive reals. The complex/negative sqrt function is defined as the solution for z of z² = a with Im(z) >= 0. If you're using complex numbers at all, I'd say it's reasonable to assume you're using the complex extensions of functions.
There is no square root function for negative numbers. As you said yourself, there is a solution of the equation z^n=a (by the way, Im(z) can also be less than zero, not necessarily >=), which is called nth root of a (and there are n of them. I'm sorry if the terms I use here are incorrectly translated, as I'm learning math in Serbian, not in English language).
But the function itself is defined as: sqrt(a), a=>0, a e R, is a number that, when squared, equals a.
I forgot: sqrt(a), a=>0, a e R, is a NON-NEGATIVE number that, when squared, equals a. (Yes. There are two squares, opposite numbers, but the function gives only the non-negative one.)
(I've always been taught that j is an imaginary but j = i so doesn't matter) i.e. jj=-1 3rd line: √(-1/1) = √(1/-1) j / ± 1 = ± 1/ j said multiply by, √-1√1 = ± j carry through with +j: (j^2/± 1 = -1/±1) = (1/ j^2 = -1/± 1) or -j (-jj/± 1 = 1/±1) = (± 1/ -jj = ±1/1) both are correct and incorrect statements I think the problem with this is that the square-root of any positive real no. gives us two values (±) but with a negative value it only gives us one; +j .I may be wrong :(
the mistake is that a negative (excuse me if im use the wrong expression) number cant have a squareroot
the sqrt of -1 doest exist because the sqrt bringes you frome one number to the number that has to be multiplyed with itself and because 2 positive/negative numbers always become one positive its not possible ;)
@tubeyoukonto That is not the issue. In mathematics you can define "i" as an imaginary number, that does not exist but can be used in many ways in mathematics. The problem is that when you sqrt something, you have a positive and a negative result. ( Eg: sqrt(1) is 1 and is -1 )
What is a negative area? Everyone knows if you are on a graph and go 1 unit (up) and 1 unit (right) you make 1 unit of area. If you go -1 unit (down) and -1 unit (left) you also make one unit of area. So 1 and -1 are equivalent regarding squares, because they make the same area, that's all you show. The problem is in step 3 when you try to take a negative square root, and later call it "i". It's like me saying 1/0 is called x and using x as a regular algebra term. You can't just make letters up
Several people have given excellent responses to this. Look through the comments. But one explanation focuses on the fact that i squared is defined as -1 so i is defined as either positive square root of negative one or the negative square root of negative one. This problem only verifies that the positive square root of negative one is not the solution.
@themathmaster Not just that!!!When taking root of 1, you need to consider + or - 1 too... which you have not...so u need to consider +/-1 and +/-i. All solutions which gives such absurd answers are not solutions!!! :)
No guys, he added a negative sign when he split the radicals on the fractions. He put a - on the outside *and* the inside of the radical symbol. Thus, the sign of the left aide became negative.
@themathmaster ok, that makes more sense. Just after i posted the comment, i realised that it shouldn't be there anyways. That seriously looks like a negative sign though.
In that case, i believe the problem is that the property sqrt(a*b) = sqrt(a)*sqrt(b) cannot be used with complex numbers, which is pretty much what is being done in the 4th line.
Proof that -1 can't be equal to 1 : let's admit that -1=1 then 0=2. Let us have "a" as any real
yoh2292 1 day ago
i can't post my comment???
yoh2292 1 day ago
Step 5 is wrong. You doesn´t have sqrt-1 * sqrt-1. It is -(sqrt-1 * sqrt-1). So we have - (i * i). Than -(i²) an this is - (-1) and finally: 1=1.
Supremehandsup 2 days ago
squaring any # will never = a negative #. therefore I^2 does not = -1, it = +1
222056 5 days ago
@222056 i is an imaginary number, which by definition has -1 as its square.
Mitarohmar 2 days ago
@acnelson12 either i misread your post or you're suggesting that i = √1. i = √-1....
Plus, I haven't learned about using polar form yet, sadly, but I want to so bad! :'(
TheDBZisepic 5 days ago
non se po fa la radice di un numero negativo
lorenzogiusti 6 days ago
WRONG when proving something is equal to something else you have to work out one side at a time.
Noxtastic 6 days ago
You only forgot something. When you get a square root the answer will have a negative sign and a positive sign, so square root of 1 is -1 and 1. Those both numbers are posible answers. So i^2=-1 and 1, and there the -1 completes the term.
oadg050693 1 week ago 2
@oadg050693 Not too bad!
themathmaster 1 week ago
@oadg050693 i^2 always = -1. That's by definition, and that's how the complex number plane works. However, what you said about the square root of 1 is correct. The square root of 1 = 1 and -1.
acnelson12 5 days ago
very funny ....that's why they invented irational number :))))))...
pimpinel2210 1 week ago
hahahahahahahahahahaha "we know -1 = -1 of course... im going to make this into a fraction...." ahahahahahahahah
b0niK1337 1 week ago
[-1]=1
gityozene 1 week ago
Theoretically, 2nd step is already wrong.
WikiDota 1 week ago
From when the complex numbers first appear in your equasion, you cant use the multiplication as you generally do with real numbers, it has other rules in the C set.
mogyorospusztai 2 weeks ago
You also made a false statement at the 5th step. -√(-1)*√(-1)=√(1)*√(1) (you forgot the - at the left side of the equasion). But -i=√(1)*√(1) is also false, so the solution lies in the fact that 1/i=-i.
mogyorospusztai 2 weeks ago
My thumbs up for the LeVar Burton lookalike...
dispatcher7007 2 weeks ago
@DarkNexarius yes, this is true. That is why i is an "imaginary" number, because it is not real. i is pretty much "no solution" in a pretty format. While this may sound like a useless waste of time, i can actually be used extensively in electrical engineering and pattern designing.
TheDBZisepic 2 weeks ago
i think u cant have √-1 it has to be -√1 i didnt even watch the rest xD
cocyfadi 2 weeks ago
wait i heard in math lessons that we CANT have a negative number inside a √
DarkNexarius 2 weeks ago
This has been flagged as spam show
The "inside" the unsquare thing must be >0 or equal to 0. It can't be <0.
-1 is <0 so you are wrong.
iroNMike1995 2 weeks ago
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iroNMike1995 2 weeks ago
2:15 In conclusion, negitive 1 equals to negitive 1. Did you meant to say that? XD
YouLostAGame 2 weeks ago
The problem is the fourth step: i / √(1) is NOT equal to √(1) / i. Besides, this breaks the fundamental rules of math. The linear function y = x states that whatever x is, y MUST be, and vice versa. A number cannot equal anything other than itself, otherwise its value is not its own and its false value doesn't have its own either. So, by saying that -1 = 1, you're pretty much saying that -1 is not -1. If -1 is not -1, then what is -1? It can't be 1 because you're implying that 1 is not 1.
TheDBZisepic 3 weeks ago 6
@TheDBZisepic That's funny!!!
themathmaster 3 weeks ago 2
acnelson12 5 days ago
@acnelson12 sorry, it deleted my line breaks. That makes it harder to read. There should be a break after "r cisθ" I think you can figure the rest out.
acnelson12 5 days ago
this is only if you assume the positive square root of 1.... there is also the negative square root, so you proved -1=-1
theProgram24 3 weeks ago
Are you people for real? Clearly he knows it's wrong! He asked for people to tell why it was wrong, duh.
BER2ERKER 1 month ago
so your telling me this glass of milk equals to this glass of cow manure over there
sexking83 1 month ago
@sexking83 That's funny!
themathmaster 1 month ago
doesn't this just mean that -1 =/= 1. or am i losing it
onlyoneskiez 1 month ago
is dis niga 4 rel
oooo0oooo0oooo 2 months ago
sqrt(x/y)=sqrt(x)/sqrt(y) is only true when x and y are both positive.
tuba1423 2 months ago 3
@tuba1423 FINALLY someone who gets it right!
anticorncob6 2 months ago
This is just stupid.. i is defined as -1, then i^2 is -1*-1 .. Which is 1.
iasedu 2 months ago
@iasedu i = sqrt(-1)
i^2 = sqrt(-1)^2 = -1
GLiTcHmArInE 2 months ago
@GLiTcHmArInE My mistake, the simple fact is that sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1) is false.
iasedu 2 months ago
@iasedu i = √-1
BrothersFreedive 2 months ago
If you think this is a valid proof, you should go back to school.
kresimircindric 2 months ago
This has been flagged as spam show
For mathematically educated people
the correct calculation:
sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)
sqrt(-1)/1 =1/sqrt(-1)
i=1/(-i)
1=i*-i
1=1
hautster 2 months ago
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hautster 2 months ago
This has been flagged as spam show
The solution is quite simple ;-)
A square root always has two solutions.
sqrt(x) is k and-k. So here is sqrt(1) -> 1 and -1. The definition for complex numbers is: i ² = -1 but it applies also (-i) ² =- 1 and it follows sqrt (-1) --> i and -i.
In this case: sqrt (-1) * sqrt (-1) = i *- i = 1 because in such case is -1 the same "x" and complex solutions are always conjugated.
hautster 2 months ago
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hautster 2 months ago
cant take the square root of a negative number.
KaiForce 3 months ago
@KaiForce Hurr Durr, yes you can. Maybe finish primary school before you attempt big-boy questions...
blingadude 3 months ago
@KaiForce we call it "i" or "j". It doesn't belong anywhere between -∞ and ∞ but you can still use it as long as you remember that taking the reciprocal is the same as negating it.
guitardudeguy00 2 months ago
Square root of one is not one :) since you wrote in the square root symbol :)
It is plus or minus one. So the Second (or third) to last line should be:
i*i=1*-1
Which is:
-1=-1
TheElitebumbershoot 3 months ago
@gokcankrc67
4th is i = (1/i) (which does not hold)
now he multiplies both sides by i*1, i.e. i*i*1 = (1/i)*i*1...
5th is i*i = 1 (still does not hold)
GiorgioDeRa 3 months ago
why don't u just square both
churro3994 3 months ago
niggas dont know about shit maths
wworm0 3 months ago
-1 multiplya by -1 gives you 1, so 1=1, not -1=1.
antoshapink 3 months ago
WRONG! because radical -1 x radical -1 = |-1|=1
MetallicaHA 3 months ago
that's BS..!~
luznlun 3 months ago
at the end: "We can now say, that -1 = -1"
Manifest222 3 months ago
@Manifest222 read the description._.
iamahbudsnaguin 3 months ago
Are you Geordi LaForge?
Krator15 3 months ago
Comment removed
DrMadolite 3 months ago
2-1=1
ophidon9 3 months ago
at the end he said -1 = -1!
SuperXD25 3 months ago
@SuperXD25 read the description._.
iamahbudsnaguin 3 months ago
third step is wrong mate
marcisthabest 3 months ago
@marcisthabest Yeah I saw it now. ^.^
DrMadolite 3 months ago
This has been flagged as spam show
You flaw is that you split sqrt(a/b) into sqrt(a)/sqrt(b), that rule is valid only if both a and b are non-negative.
TheHarboe 3 months ago
Comment removed
TheHarboe 3 months ago
This has been flagged as spam show
so what u need to do is to define the transformation f(g(x)) = f(h(x))*f(k(x)) for complex entries.
u can avoid this problem by using a symbolic function letter instead of the intuitive sqrt symbol which might lead to be used like other intuitive symbols like * and /.
SonzOfGunz 3 months ago
This has been flagged as spam show
i think using ± in front of the sqrt wouldnt make no difference, since the pos definition of the the sqrt function is used on both sides of the equal. but (-1)**(0.5) is not (-1)**(-0.5)
u need to be careful by replacing the function f(g(x)) by f(h(x))*f(k(x)) when f(x) means sqrt(x).
u cant transform from (-1/1)**(0.5) equal (1/-1)**(-0.5) to seperate sqrts, because u already have to pull out the i at that very point, to avoid errors of your pos definition of the sqrt, i think.
SonzOfGunz 3 months ago
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SonzOfGunz 3 months ago
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SonzOfGunz 3 months ago
The argument of -1 is 180º.. right? So the square root of anything with 180º will have it's first root at 180º/2 and the other one 360º/2 further.. So I don't get why he says that the square root of -1 is 'i'..
The only thing this video does is reenforcing the stereotype about black people and their supposed inability to perform abstract cognitive processes like us normal people do..
UntakenNick 3 months ago
@UntakenNick a "normal person" would know that "i" is an imaginary number...
PatyTheImp 3 months ago
@PatyTheImp What does that have to do with my comment..? Even the person in the video seems to know that 'i' is the imaginary unit..
Or maybe you skipped the maths and just read the last line..
UntakenNick 3 months ago
@PatyTheImp I said, quoting, "..it's first root at 180º/2 and the other one 360º/2 further.." That would be 'i' and '-i'.. when did I say that those expressions are real numbers..?
UntakenNick 3 months ago
-1 X -1 = 1 X 1
same thing....
FitPeePee 3 months ago
the square root of one is - and + 1 so ±1=±1
danjoelabrenica 3 months ago
Third equation is right, you have i at both sides. Fourth one is wrong: on the left, you have i, while on the right you have 1 / i which is equal to -i. And i =/= -i
GiorgioDeRa 3 months ago 13
@GiorgioDeRa But, at 5th one he deletes that one. So, it doesn't matter.
gokcankrc67 3 months ago
@GiorgioDeRa Great, now what's the flaw?
anticorncob6 2 months ago
The square root of 1 is 1 or -1. The second last line is wrong.
NoodlemanNO1 3 months ago
the square root of a negative number should technically be both a negative number and a positive number at the same time
NoodlemanNO1 3 months ago
black man good at math
OrangutanNationz 3 months ago 2
erm... I think the sqrt need a '±' symbol so you should be getting:
±1=±1... I guess I'm wrong though lol
boilpoil 3 months ago
negativ x negative = positive
-1 x -1 = +1
happybirthday.....
ichTraylo 3 months ago
mistake is in the third action.
we must right +- befor ine of the squares.
eliote4 3 months ago
The wrong part is that you did not square the left side (i^2). You cannot simply square only the right part (sqrt(1).sqrt(1)) and break the fundamental law of an equation ("do the same thing on both sides").
kumostein 3 months ago
Really good one! :)
It all comes to:
(sqrt - square root, of course ;) and a^b - a to the power of b )
You either do:
[ sqrt (-1) ] ^ 2 = -1 /* the opposite operations
or:
sqrt [ (-1) ^ 2] = sqrt [ 1 ] = 1
which (at least for real numbers) should give the exact same result :D
RedRad1990 3 months ago
i like this proof of -1=1. it took me a couple of minutes but if you take the square root of a number, you get two solutions. a positive and a negitive solution. a calculator gives commonly the positive one out. but the square root delivers two [sqrt(1)=1 and -1] . so in this case, to have a mathematical correct solution, you have to take the negative solution of the square root.
addy0419 3 months ago
Blew my mind
forrest0no 3 months ago
@forrest0no Math is like that!
themathmaster 3 months ago
@themathmaster :D
forrest0no 3 months ago
@themathmaster YOU PROOF IS A FAIL.... the square root of -1 is i....
greywolf424 2 months ago
@greywolf424 Oh my god did I just find you on a random video again? Just like with EdwardCurrent's Building 7 video? OMG!!!
anticorncob6 2 months ago
WRONG! minus must be out of sqrt always: -1 = -sqrt(1) not sqrt(-1)
anonimanswer 3 months ago
guys... thumb this so everyone can see it...
taking the square root of a negative isn't wrong, it's just imaginary (he correctly replaced it with i). And when you square i, you get -1, just as he did. That part is correct.
What's incorrect is the right side of the equation- the sqrt(1)*sqrt(1). Both of those numbers are +/- 1, not just 1. (If you think about it, squaring -1 can also equal 1.) Replacing both sqrt(1)'s with 1 is extraneous; the other solution is -1 because -1*1=-1. so yep.
DRKVideoProductions 3 months ago 3
@DRKVideoProductions Taking the square root of a number is specifically defined as the PRINCIPLE square root... the real flaw is that sqrt(x/y) = sqrt(x)/sqrt(y) does not apply when x and y are both negative numbers.
anticorncob6 2 months ago 2
i hope you do realize you can't take the square root of a negative number ...
chimcharaapje 3 months ago 4
@chimcharaapje Oh ye of little faith!
themathmaster 3 months ago 2
@themathmaster i do agree that i = sqrt(-1), but you're still not allowed to write sqrt(-1)
chimcharaapje 3 months ago
@chimcharaapje It is traditional to replace the square root of -1 with the letter i.
It is same as the tradition as when simplifying 5x - 4x = 'x' rather than '1x'. So its not wrong, it's just not traditional.
themathmaster 3 months ago 7
he did not take the square root of -1. he replaced it with the letter i from the complex numbers. these allow you a to handle this square root easier in your equation.
addy0419 3 months ago
You just can't have sqrt of any negative number
RealYourHack 3 months ago
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mirandaq7g 3 months ago
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CrazyDuck1996 4 months ago
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CrazyDuck1996 4 months ago
@CrazyDuck1996 this does not work. you are left with 1=1 which is not proving that -1=1
Chriskon420 3 months ago
@Chriskon420 sorry i had fever that day
i didnt know what i was talkin about
CrazyDuck1996 3 months ago
sqrt(x/y)=sqrt(x)/sqrt(y) IF x and y are positive.
tuba1423 4 months ago
that's not fun....
the squareof -1 is a number called imaginary number,
i =√-1 (i is the word to represent √-1)
1/√-1 should be1/i ,
√-1/1 is i/1,
They are uneqaul`
31093236 4 months ago
@31093236 Well, that was my comment below yours. Many ppl missed the error.
i is as ubiquitous as Pi in math. 'i' is tightly coupled to harmonic oscillators. Eulers eqtn:
e^(i*x) = cos(x) + i*sin(x) [a beautiful eqtn]
is the reason behind this. Schrodingers eqtn also employs 'i' for the quantum mechanics (waves of probability). Study math and you will know the mind of God.
myrtlebox 4 months ago
I don't see ANYONE who has the correct analysis here. I am an engineer who has taken a ton of math - trust me.
The 'error' occurs @ 1:10. You cannot arbitrarily split the radical of num & den when the combined divisional radicand (ie: 1/-1) is negative! That sign must remain in the numerator under the radical. Everything else follows from this error.
For example: This error immediately implies that ( i = 1/ i ) which is already wrong because ( 1/ i = - i)
qed
myrtlebox 4 months ago
It is true to say that x² = a² => x = a or x = -a.
But this is not the reason because we don't have to go through this step to get -1=1.
as from √-1 * √-1 = √1*√1,
we have (√-1)^2 = (√1)^2, notice that (√x)^2 = x for any real number x,
so this gives us -1 = 1.
In fact,
-1/1=-1=1/-1 is true, -1=-1 is true, and -1/1=1/-1 is true,
but
-1=-1 does not imply -1/1=1/-1 though the inverse is true.
SkyEvolved 4 months ago
0:50 mistake! you can´t take the square root of a negative number
JakobRobert00 4 months ago
but in the part of -1 x -1 = 1?
because negative numbers in multiply = positive or im wrong?
mipiopo15 4 months ago
You cant have a negative number in tha square root!!!! We dont write that !!!
LaraReal 4 months ago
@LaraReal It's an imaginary number -.- Google that.
Gytax0 4 months ago
It is funny how people think they know what they are doing when they are actually pulling crap out of their ass...... :D
RandomBoardKat 4 months ago
I guess Reading Rainbow had to make some budget cuts.... :/
EatMyDuck84 4 months ago
Bet you cant do that with -2 and 2.
phoenix78240 4 months ago 11
@phoenix78240 Sure, just multiply both sides at the end by 2! LOLOL
themathmaster 4 months ago 16
@phoenix78240 I can. put ( and ) to left and right side, multiply by 2...
gokcankrc67 3 months ago
@phoenix78240 lol...like themathmaster said "just multiply both sides at the end by 2!"...wat a dumbass._.
iamahbudsnaguin 3 months ago
Complex numbers here don't matter. All he's done is the classic mistake:
assuming that x² = a² means x = a. It actually means x = a or x = -a, and in this case the negative solution is the correct one and you get -1 = -1.
dwarduk2 4 months ago 2
@dwarduk2 Thanks! Great point!
themathmaster 4 months ago
@dwarduk2 Actually, no. He assumed if x=a then sqrt(x)=sqrt(a). Which is correct for real numbers, however in real numbers sqrt(-1) is not defined and hence there is a mistake in proof, and if you work in complex numbers then sqrt(1) is both -1 and 1 while sqrt(-1) is both i and -i so you actually have (1 or -1) = (1 or -1) which is correct.
Rayqiorg 4 months ago
@Rayqiorg x = a does indeed mean that sqrt(x) = sqrt(a). Taking your line of reasoning, the argument is broken because squaring isn't a 1 to 1 function. I don't have it to hand, but there is an extensive example in the Maths textbook I used last year on this.
As for my point, I was actually talking about the step of sqrt(-1)^2 = sqrt(1)^2, where x was sqrt(-1) and a was sqrt(1). One last thing - (-i)^2 = (i^3)^2 = i^6 = 1 * i^2 = -1, so it's -1 = (1 or -1), which was my whole point.
dwarduk2 4 months ago
@dwarduk2 The mistake he actually made is the following: square root is defined only for NON-NEGATIVE numbers.
There is no square root of -1, but i^2=-1. That equation (i^2=-1) cannot be transformed by taking the square root of both sides, because you can do that only if the both sides are non-negative (obviously, in this case, the right side is negative).
Simply said: i is not the square root of -1. There is no square root of -1.
SunTzutheWise 4 months ago
@SunTzutheWise how is i^2= -1 if sqrt(-1) does not = i?
I thought imaginary i was defined as sqrt(-1) :o
TheKotassium 4 months ago
@TheKotassium
Simply: because square root is NOT defined for negative numbers. By the definition itself, there is no square root of -1.
And, as I said before - i is defined as a number which, when squared, equals -1. :)
SunTzutheWise 4 months ago
@SunTzutheWise It depends what stage of maths you are looking at, and the sets you are using. There is a square root function that applies to negative numbers, and also one that applies to complex numbers. They just aren't the same as the one for weakly positive reals. The complex/negative sqrt function is defined as the solution for z of z² = a with Im(z) >= 0. If you're using complex numbers at all, I'd say it's reasonable to assume you're using the complex extensions of functions.
dwarduk2 4 months ago
@dwarduk2 Of course, if both solutions have Im(z) = 0, it's the one with Re(z) >=0 because then z is real, not complex.
dwarduk2 4 months ago
@dwarduk2
There is no square root function for negative numbers. As you said yourself, there is a solution of the equation z^n=a (by the way, Im(z) can also be less than zero, not necessarily >=), which is called nth root of a (and there are n of them. I'm sorry if the terms I use here are incorrectly translated, as I'm learning math in Serbian, not in English language).
But the function itself is defined as: sqrt(a), a=>0, a e R, is a number that, when squared, equals a.
SunTzutheWise 4 months ago
@SunTzutheWise
I forgot: sqrt(a), a=>0, a e R, is a NON-NEGATIVE number that, when squared, equals a. (Yes. There are two squares, opposite numbers, but the function gives only the non-negative one.)
SunTzutheWise 4 months ago
@SunTzutheWise Imaginary number.
Gytax0 4 months ago
Comment removed
BlackSunSerenade 5 months ago
Bet u can devide by zero!!!!
wuzzuppp3452 5 months ago
-3 to the 2nd power is +9 :D
WICKEDspeeder17 5 months ago
@ themathmaster Well, the problem is (1)^0.5 = 1 and i ,both, so u did the classic trick of taking one root while ignoring the other. ... Funny proof
Dragogoel 5 months ago
mate you lost the "-" on the left hand side at 1:49, when u multiplied both sides. I do not miss arithmetic mistakes :P.
blueshift86 5 months ago
@blueshift86 there is no negative sign
bdiddy77777 4 months ago
This has been flagged as spam show
MrXenitha 5 months ago 2
Comment removed
MrXenitha 5 months ago
the mistake is that a negative (excuse me if im use the wrong expression) number cant have a squareroot
the sqrt of -1 doest exist because the sqrt bringes you frome one number to the number that has to be multiplyed with itself and because 2 positive/negative numbers always become one positive its not possible ;)
tubeyoukonto 5 months ago
@tubeyoukonto That is not the issue. In mathematics you can define "i" as an imaginary number, that does not exist but can be used in many ways in mathematics. The problem is that when you sqrt something, you have a positive and a negative result. ( Eg: sqrt(1) is 1 and is -1 )
SlipUp911 5 months ago
when sqrt both sides, need a +/- sign
jpcurley25 5 months ago
1 does not equal 1/-1 it eqauls 1/1
MrRafusa 5 months ago
LOLLLLLLLLLLL
acuario384 5 months ago
yeah every time you use square roots there is an extraneous solution.. because of plus or minus.
Charcharblaze 5 months ago
√1=±1, because -1*-1=1.
Or if you can't read those characters, sqrt(1) = + or - 1.
AlixFense 5 months ago
2:15 "and we have found the conclusion that negative 1 equals negative 1"
xXSabzzXx 5 months ago
What is a negative area? Everyone knows if you are on a graph and go 1 unit (up) and 1 unit (right) you make 1 unit of area. If you go -1 unit (down) and -1 unit (left) you also make one unit of area. So 1 and -1 are equivalent regarding squares, because they make the same area, that's all you show. The problem is in step 3 when you try to take a negative square root, and later call it "i". It's like me saying 1/0 is called x and using x as a regular algebra term. You can't just make letters up
RobBrooksMusic 5 months ago
@RobBrooksMusic
LOLOL...thanks for sharing Rob!
themathmaster 5 months ago
@themathmaster So what is the official correct answer to this? :)
RobBrooksMusic 5 months ago
@RobBrooksMusic
Several people have given excellent responses to this. Look through the comments. But one explanation focuses on the fact that i squared is defined as -1 so i is defined as either positive square root of negative one or the negative square root of negative one. This problem only verifies that the positive square root of negative one is not the solution.
themathmaster 5 months ago
@themathmaster Not just that!!!When taking root of 1, you need to consider + or - 1 too... which you have not...so u need to consider +/-1 and +/-i. All solutions which gives such absurd answers are not solutions!!! :)
Har1H 4 months ago
@RobBrooksMusic yes, after STEP 3 Uncertainty comes in and stops the whole proof thing.
kirstukas13 5 months ago
@RobBrooksMusic
hes not making up letters lol. its imaginary number -.-
MladiHudini 5 months ago
@MladiHudini I see his point though, it is like saying that 1/0=∞
2*0=3*0
∞(2*0)=∞(3*0)
2(0*∞)=3(0*∞)
2*1=3*1
2=3 which is impossible.
He's saying that, if we denote something that is impossible, we may get an impossible answer.
MrXenitha 5 months ago
@MrXenitha Isn't infinity a different class of number compared to integers so you can't combine the two in sums?
RobBrooksMusic 5 months ago
@MrXenitha never mind I just read this properly
RobBrooksMusic 5 months ago
Comment removed
MrXenitha 5 months ago
i like the conclusion ;) listen to it again ;)
voytecu 5 months ago
Square root of one = +or- 1
You get a correct answer when you have i^2 = (sqr rt 1)^2 as long as you make one of 1's roots positive and the other negative. As in -1*1 = -1
WolfYoghurt 5 months ago
1:05 he added the extra negative sign. (top left term)
GzusFreak1 5 months ago
No guys, he added a negative sign when he split the radicals on the fractions. He put a - on the outside *and* the inside of the radical symbol. Thus, the sign of the left aide became negative.
GzusFreak1 5 months ago
You missed a sign. Starting on the 4th line:
-i = 1/i
-i*i = 1/i*i
1=1
You didn't carry the -sign to the 5th line.
fedex12345678 6 months ago
This has been flagged as spam show
@fedex12345678 theres no such thing as -i
budsmokersonly2008 6 months ago
@fedex12345678
Oh, that was not a negative sign. It was a line that was apart of the root symbol.
themathmaster 6 months ago
@themathmaster ok, that makes more sense. Just after i posted the comment, i realised that it shouldn't be there anyways. That seriously looks like a negative sign though.
In that case, i believe the problem is that the property sqrt(a*b) = sqrt(a)*sqrt(b) cannot be used with complex numbers, which is pretty much what is being done in the 4th line.
fedex12345678 5 months ago
Wtf
thewillyagami123 6 months ago
The problem is taking the square root, as it's the step that creates false solutions, not squaring actually.
bigbig1604 6 months ago