He said there were no solutions in the real or complex numbers, and infinity is none of those, so we can't say it. If he said there were no solutions in the _extended_ real or complex numbers, that would be another matter.
Hey I just watched and liked one of your video, I heard you are offering computer major, can I ask you some questions as I am applying for a computer major, if you’d mind dropping any of your socials for me to reach you and then deleting it later after I reach you.
No computer scientist ever thinks that x=x+1 means "x is equal to x+1". It is just the assignment notation used by some languages (e.g. C and the like). Other languages have different symbols. For example, Pascal uses :=, which is indeed more obvious, R uses
@@9nikolai That's not what I mean. There's a difference between definitions explicitly designed to make some thing impossible, and undesired impossibilities arising „naturally“ from a given set of axioms. The insensibility of the first, the second, or the fifth equation here is an accidental consequence of „nice“ computational behaviour of numbers, but the square root function and the absolute value function are both outright defined to return only non-negative values.
@@xCorvus7x They are as much accidents as saying 1+1=3 is an accident. It's just an impossiblity based on the definition of mathematics. But even if something is impossible, we can always just make up something. Like how the square root of -1 is "i"? If it's impossible, it's by design or laziness.
3 and 4 are a matter of definition, we define √ as the positive square root and we define | | as the distance from the origin and, successively, we define any distance as a positive quantity. 1,2 and 5 give x an infinite magnitude which cannot be a well-defined number in any field.
@@9nikolai op is ahead of you here. 1, 2, and 5 work fine if you do not work over a field. taking the Alexandroff extension of the reals does this at the cost of some structure.
Nope limits don't give the exact results. Because limit of 1/x when x --> +inf is equal to 0.00000000......1 not exactly 0 but we just write that 0.00000....1 0 to make the study of a function easier. Hope that makes sense.
@@thecoolnewsguy That was a joke. We get zero on putting infinity, but we can't just put infinity. We have to take the limit. Thus, x has no numerical solution. We can get close to the solution using limits.
@@thecoolnewsguy No, limit x-> inf in 1/x is definitely 0, there is no 0.00001 and stuff. It is a limit. However, if we try tp divide 1/x without limits, where we take x as a large number, than we get 0.000... 1. If we use limits, we get exact 0.
it can be shown how √x cannot equal to -1 if √x=-1 then e^iθ/2 must equal to -1 which is e^iπ where θ is.in range [0,2π) but θ/2 is less than π so x^1/2 cannot equal to -1
Or you can say that by definition sqrt(x) lies in the nonnegative reals (or at least when you get a complex result the real part is nonnegative) Unless that's the formal way of saying that.
bbruh not nexessary, by definition a radical sign will always output a positive value, more specifically, it gives the principal value which is positive for the real domain
1):0/1 2) : 0/0 3):ua-cam.com/users/shorts5PGcA3LmASw?si=Mpy8mTrgCGcQ4Hf 4) :the next generation 5): the last generation "all thing have solution " Merkez nasir all of them have solution on complex and hypercomplex number
It’s kinda weird, because do the 3rd question, I immediately thought of i^4, but then I remembered that i^4 just equals one. Kinda weird tho how it works, very cool
well if you think about the complex square root sqrt(1)=1 and -1. So we kinda have the answer, 1. But we want sqrt(x) equal to only 1, not 1 and -1. idk
@@renangustavo3844 don't quote me but sqrt(36) = +6 as it's principle rather than plus/minus where it is used for as a function. Correct me if I'm wrong.
1) The only way for a fraction to evaluate to 0 is if the numerator becomes 0. That numerator will never be zero, no matter what x is, so there are no solutions. 2) Similarly, none of the factors of the second equations are 0, so the result can not be 0. 3) The positive square root of x can not be negative, even if x is imaginary. 4) Absolute value function always yields a positive number or 0; never negative. 5) Subtract x from both sides, shows 0 = 1, which is always false.
i love the last one x=x+1, or we called in programming increment by 1, in computer science. Absolutely I can relate on that. Yes I agree to all your claims in mathematics are true. 👍🤣👍❤️
Ridiculous... 1. Infinite 2. Negative infinite 3. 1 4. -1 (or any other point on the unit circle that falls on the left side of the complex plane) 5. Infinite
Actually contirary to what is taught in schools square roots only have one soloution as they are a function. They are defined as the inverse function of f(x) = x^2 with a range of [0,infinity)
anyone who didn't get the computer science joke it means that . . . you don't have basi concepts for programming . JK It means we are assigning the value of x + 1 to the variable x
For the 3rd one, if you sub in x = i⁴, would that work since √i⁴ = i^(4/2) = i² = -1? For the 4th one, could you do x = i²? How would that work with the absolute value?
@@The-Devils-Advocate I understand that. I was just curious as I never thought about what you could say about |i²|. It would be a colorful way of disproving the abs value by saying |-1| = |i²| = i² = -1 or something like that, but if you can go from -1 to i², you can also go from i² to -1 to disprove my original question. Math is always fun!
if u sub x = i^4 in 3, it gives you 1 not -1, as that symbol means "the positive square root of" (i.e. u should not consider -1 as a solution to positive square root of i^4)
That depends on what programming language you're using. I first learned programming with languages like BASIC and Pascal that don't have += or ++ (although in Pascal you'd write x := x + 1).
1. horizontal asymptote towards -inf and +inf, fair 2. vertical asymptote towards -inf, fair 3. essentially depends on using the principal root (or maybe it's by definition), as sqrt(1) could technically be +1 or -1, but we (at least normally) don't do this, so fair 4. by definition, fair 5. impossible outside of infinity or some kind of modular arithmetic (mod 1, i.e looking at decimals only, maybe?), so fair
No, because by definition, you finish evaluating the entire interior of the square root function, and then take the square root. sqrt(i^4) is the same thing as sqrt(1), which equals 1. Technically -1 is *a* square root of 1, but it isn't *the* square root of 1.
@@carultch Technically, neither 1 or -1 is the square root of 1. the square root of 1 is ±1 but, to solve √x=-1 all you need to do is to square both sides of the equation (√x)²=(-1)² now, the √ and ² cancel each other and we are left with x=(-1)² which is 1. easily solvable.
@@carultch my apologies for ignoring an irrelevant, and misleading, convention that only ignorants or lazy people use. the only reason to use only the positive square root is in cases where you can't use the ± and ∓ signs for some reason.
3 and 4 are kinda the same. The square root function is just defined to output the positive value so that the values can be distinguished by notation (+sqrt vs -sqrt)
You missed the solutions: 1) Infinity in the extended complex plane (it's a completely valid extension, through stereographic projection onto riemann sphere...!) 2) - Real Infinity + ib 3) x = 1 where we branch cut f(z) = z^0.5 along the positive imaginary axis, and use the branch pi/2 < argz < 5pi/2 so that square rooting halves the argument and produces a result satisfying pi/4 < arg(sqrt(z)) < 5*pi/4 - which much be the negative root when z = 1 4) Obviously this would work in a number system where -1 = 1 e.g. arithmetic modulo 1. I would like to find a more convoluted solution to this. 5) This is modular arithmetic (mod 1). But if you imagine a space of a unit square, where the left edge is glued to the right edge, and top edge glued to the bottom (e.g. in some space with torus-like curvature - e.g. in old computer games), adding one to the x or y coordinate will result in no change to the x or y coordinate.
In programming, we pre define the value of x and use that value to calculate and then define x again, but in maths we are taking x that has only one value for both sides.
Technically the answer to number 3 is 1, because any number rooted has a positive and negative answer, so the square roots of 1 are 1 and -1. And the answer for number 1 would be x= 1/0 which is undefined but is technically the answer
please correct me if im wrong, i might be making a stupid mistake in equation 3 √x = -1 square both sides x=1 lets put this in eq 3 √x= -1 √1= -1 +1 or -1= -1 so this is true right?
@@skullzs1983 it depends on the branch you take. On one branch, where √1=1, there's no solution. On the other branch, where √1=-1, there is a solution. So in total, there is a solution, but not along every branch.
This is an assignment operator, not a comparison operator, so x == (x+1) false. But (-x == x and x!=0) can be true if x = minimum integer. This is because the modulus for this number is greater than the maximum number.
I feel like #3 and #4 are essentially the same: #4 can be rewritten as (the square-root of x²) = -1. If there is no complex solution to #3 (i.e. there is no complex number whose square root is -1), then #4 would follow (there is no complex number which, when squared, would have a square root of -1) since the complex numbers are closed under multiplication (and thus squaring), and the set containing the result of squaring a complex number is a subset of the complex numbers.
x=x+1 is possible, but not with the limitations we've placed on algebra the limitation that prevents this equation from being true is the one that dictates that a variable can only be equal to 1 value its also possible in another way. if 1 represented 0. this would be true.
For x=x+1 => y=x+1 for y=0 x=-1 and for x=0 y=1 we graph and we see that is a line that come from -inf to the point(-1,0) to the point(0,1) to +inf ..so we need to exist a point such as that it meets this equation where its says (x,x) or (y,y) or as an equation y=x to find where it meets with equation y=x+1 ..now we have two equations with two unknowns and it is much easier to solve by equating them ...but wait if you both graph them ...they never meet because they are perfect parallel lines.....the end... so there is no solution..not that it is unsolvable...
For the 3rd one I think first we should take squares on both side to the square root would be removed from x and square of minus 1 equals 1 so x equals 1 Tell me whether I am wrong
I am a top scoring student and it's very simple that for the equation: (square root of)x≠(-1) that x = 1 because a negative multiplied by another negative is a positive, and 1 times 1 is 1, so negative 1 times negative 1 is 1. Simple!
@@skonaslp no, that entirely depends on your chosen definition. If I say we're working with the "negative" square root (the only branch other than the principal one), then √1=-1, √(-1)=-i, √2≈-1.41421, etc. and √:[0,∞)→R is a decreasing function.
Just invent/define a new value or imaginary number that can make 1/X = 0 and solve for x. g can be infinity as a number only. And the lying 8 can be infinity as a concept. Then you can solve the first 1. Then you have to go into philosphy of what infinity really means.
Now, allowing ordinals, Aleph_i satisfies the last equation, and maybe the first. IDK if ordinals get extended to signed ordinals the way naturals get extended to integers, but if so, 2^(-אᵢ) = 0, I think.
All those things are potential definitions if only you used a unique symbol instead of X. sqrt(:)) = 1 sqrt(:() = -1 would eliminate potential information loss and avoid the plus or minus answers. sqrt(9:() = -3
![Why you didn't learn tetration in school[Tetration]](http://i.ytimg.com/vi/Ea1_1aVfwl4/mqdefault.jpg)
![Why you didn't learn tetration in school[Tetration]](/img/tr.png)
I love the "I'm not talking about computer science".
That's why you use x++ instead XD
@@romerorobertoandres9761 you meant x+=1 ,right?
@@idiota-w-swiecie no he didn't
@@idiota-w-swiecie that would do the same thing regardless if its only increments of 1
@@idiota-w-swiecie In Java variable++; means increasing the value of that variable by one.
“You cannot say infinity”
Dammit
lim x-> infinity
If you could it would be:
1 ♾
2 -♾
5 ♾
@@peppy5121
2 ♾ and beyond
He said there were no solutions in the real or complex numbers, and infinity is none of those, so we can't say it. If he said there were no solutions in the _extended_ real or complex numbers, that would be another matter.
Dividing by infinity doesn't equal to 0 exactly because you're just going to divide by a very large number
As a computer science major I appreciate this guy for clarifying 😂😂
😂
Hey I just watched and liked one of your video, I heard you are offering computer major, can I ask you some questions as I am applying for a computer major, if you’d mind dropping any of your socials for me to reach you and then deleting it later after I reach you.
“You cannot say infinity”
*Negative Infinity*
😂
Infinity is a concept not a number
You still said infinity
x=x+1 for Computer Science: 🥳🤠😌🤪😜😘😝😛
x=x+1 for mathematicians: 😞😖😓😣🤬😠😡😩😫
Mathematicians do not necessarily hate x=x+1 because it can tell the mathematician that no solution exists
😂😂
No computer scientist ever thinks that x=x+1 means "x is equal to x+1". It is just the assignment notation used by some languages (e.g. C and the like). Other languages have different symbols. For example, Pascal uses :=, which is indeed more obvious, R uses
Java, JavaScript, C, *C++* devs: x++
Python devs: "++"?
@@Pacvalham
This is very True
Python: True
other languages: "True"?
0:01 "some solutions that have no equations" broooo the equations are right there!!
That was a misteak.
@@Aiden-xn6wo a misteak ...
@Ilya Sapronov yeah, but they don't have equations
i noticed that too lol
@@Aiden-xn6wo Is that when he intentionally says the wrong thing as a joke? Because I agree.
1 and 5 is approximately true for large x
And yes I am an engineer
You engineers do always think approximately means equals
@@lukerichmond1319 If you can't describe nature approximately, you'll never describe it precisely
@@kerim7158 Mathematics don't describe nature. Nature describes mathematics.
You could also say 2 is approx true for large but -ve X
@@wearygood1559 correct
why did I laugh so hard at “I’m not talking about computer science”
The sudden cut off made me laugh. You could here there was no period in this sentence.
I can't understand y😂
i have an equation that might not be solvable let's see if you can solve it
hold on I have to go get my math homework
😆
Don't believe him fast. He is trying to finish his hw with u lol 😂😂
Lol u replied love u r vids
Why is this relatable💀
Y@@danielintheworldoftheunive8240
This is a certifed black pen only moment
It should be a blue pen only
Frrrrr
3 and 4 feel kind of like cheating, since the respective functions were defined to only give positive answers
That's the point of unsolvable though. If it's not unsolvable by definition, then it's solvable.
@@9nikolai That's not what I mean.
There's a difference between definitions explicitly designed to make some thing impossible, and undesired impossibilities arising „naturally“ from a given set of axioms.
The insensibility of the first, the second, or the fifth equation here is an accidental consequence of „nice“ computational behaviour of numbers, but the square root function and the absolute value function are both outright defined to return only non-negative values.
@@xCorvus7x They are as much accidents as saying 1+1=3 is an accident. It's just an impossiblity based on the definition of mathematics.
But even if something is impossible, we can always just make up something. Like how the square root of -1 is "i"?
If it's impossible, it's by design or laziness.
@@9nikolai ?
Don't you mean they are accidents as much as saying 1+1=2 is an accident?
@@xCorvus7x No, because 1+1=2 is actually true, while 1/x=0 isn't.
"I'm not talking about computer science."
Literally the first thing that came to mind when he wrote that. He really is on top of things!
3 and 4 are a matter of definition, we define √ as the positive square root and we define | | as the distance from the origin and, successively, we define any distance as a positive quantity.
1,2 and 5 give x an infinite magnitude which cannot be a well-defined number in any field.
1, 2 and 5 do not give x and infinite, but an indefinite. Even with 1 divided by infinity can only _approach_ 0.
@@9nikolai op is ahead of you here. 1, 2, and 5 work fine if you do not work over a field. taking the Alexandroff extension of the reals does this at the cost of some structure.
@@joeg579what the hell did you just say?
Calculus for the first two be like: Hold my limits.
For last two: imma head right out :D
Nope limits don't give the exact results. Because limit of 1/x when x --> +inf is equal to 0.00000000......1 not exactly 0 but we just write that 0.00000....1 0 to make the study of a function easier. Hope that makes sense.
@@thecoolnewsguy That was a joke. We get zero on putting infinity, but we can't just put infinity. We have to take the limit. Thus, x has no numerical solution. We can get close to the solution using limits.
@@thecoolnewsguy No, limit x-> inf in 1/x is definitely 0, there is no 0.00001 and stuff. It is a limit. However, if we try tp divide 1/x without limits, where we take x as a large number, than we get 0.000... 1. If we use limits, we get exact 0.
@@ilyakrikun6528 Ah you're right I reversed 😅
@@ilyakrikun6528limit x approaching negative infinity in 1/x is also 0 😔
it can be shown how √x cannot equal to -1
if √x=-1 then e^iθ/2 must equal to -1 which is e^iπ where θ is.in range [0,2π)
but θ/2 is less than π so x^1/2 cannot equal to -1
Or you can say that by definition sqrt(x) lies in the nonnegative reals (or at least when you get a complex result the real part is nonnegative)
Unless that's the formal way of saying that.
@@nanamacapagal8342 nonnegative reals*
@@nanamacapagal8342 but that's not true because sqrt(i) =e^(πi/4) = (1+i)/sqrt(2) which is not real
sqrt(1)=±1
bbruh not nexessary, by definition a radical sign will always output a positive value, more specifically, it gives the principal value which is positive for the real domain
“Do not say infinity.” Lol, love that. And you’re absolutely right.
i'm not talking about computer science, u make my day.
"Don't say infinity okay?" Had me fucking rolling 🤣 💀
*Don't* majorly curse in the forum, because it is rude, ignorant, and needless.
Stop your major cursing! It is ignorant and needless.
🤓🤓⬆️
That '' I'm not talking about computer science '' got me😭
For sqrt(x), you can get -1 in the complex world if you consider both branches of sqrt. Of course, I'm talking about x=1.
sqrt function will only return non-negative values, what im saying is that there exists no "both branches of sqrt"
@@oinkityoink So what will sqrt(i) return? Which is the "non-negative value"? Again, as I said, this only applies in the *complex world*.
@@andreimiga8101 it returns the principle branch
@@toniokettner4821 in the complex world, sqrt is a multivalued function
@@andreimiga8101 NO. do you know what a function is?
As the days pass by, he gets closer to becoming a Shaolin Math Monk
😂😂
1):0/1
2) : 0/0
3):ua-cam.com/users/shorts5PGcA3LmASw?si=Mpy8mTrgCGcQ4Hf
4) :the next generation
5): the last generation
"all thing have solution " Merkez nasir
all of them have solution on complex and hypercomplex number
First: Ur right. Second: -infinity, Third: i, 4: Ur right. 5: programmers would say the answer is adding one to X but for math, Ur right.
The third one must have an answer... What about complex numbers? Ah my brain explode.
It’s kinda weird, because do the 3rd question, I immediately thought of i^4, but then I remembered that i^4 just equals one. Kinda weird tho how it works, very cool
I mean it IS one, but square roots can’t have a negative output
this equation does have a solution, if x=1 the solutions are ±1 which shows that the square root of x can be equal to -1
well if you think about the complex square root sqrt(1)=1 and -1. So we kinda have the answer, 1. But we want sqrt(x) equal to only 1, not 1 and -1. idk
1/(1/0)
*accidentally breaks math*
= 1*(0/1)
= 0
me: “uhh guys, I ended up in an illegal situation”
I don’t get it, you divided 1 into an undefined number of pieces, then just said 0/1
@@destdrom He substituted division with multiplication by the reciprocal (doesn't work for zero ofc but that's what he tried)
It doesn't work because 1/0 is not a real or complex number
In 3 )x can be equal to 1 , so we will get 2 solutions ie. +1and -1
Another equation
X^n + Y^n = Z^n where n is a while number greater than two
Since we're allowed to use the complex numbers to try finding the solutions, wouldn't √x = -1 result in x = i⁴?
Like √(i⁴) = √((i²)²)= i² = -1 ?
But i⁴ = i².i² = (-1).(-1) = 1, so √1 = -1, what's technically strange , but ok, since √36 = ±6, so should be √1 = ±1
Hope he noticed this
@@renangustavo3844 don't quote me but sqrt(36) = +6 as it's principle rather than plus/minus where it is used for as a function. Correct me if I'm wrong.
@@notjayk8057 You're not wrong.
I think you are right :)
In complex numbers square root is not unique and sqrt(1) = { 1, -1}.
At least I think so 🤣
1) The only way for a fraction to evaluate to 0 is if the numerator becomes 0. That numerator will never be zero, no matter what x is, so there are no solutions.
2) Similarly, none of the factors of the second equations are 0, so the result can not be 0.
3) The positive square root of x can not be negative, even if x is imaginary.
4) Absolute value function always yields a positive number or 0; never negative.
5) Subtract x from both sides, shows 0 = 1, which is always false.
0 ring
i love the last one x=x+1, or we called in programming increment by 1, in computer science. Absolutely I can relate on that. Yes I agree to all your claims in mathematics are true. 👍🤣👍❤️
Ridiculous...
1. Infinite
2. Negative infinite
3. 1
4. -1 (or any other point on the unit circle that falls on the left side of the complex plane)
5. Infinite
Actually contirary to what is taught in schools square roots only have one soloution as they are a function. They are defined as the inverse function of f(x) = x^2 with a range of [0,infinity)
Him: x = x + 1
Me (an engineer): I know this one!
Also him: I am not talking about computer science
🙄🙄🙄🙄
I don't know about the other equations but for the 1st one we can take the value of 'x' as 1/0?
i was thinking that
1/0 is more than 0 thus not equal
@@noahb885 wdym
Except 1/0 has no value.
I'm calling those the lalaland world. We got the real world, we got the imaginary world, now we got lalaland world. There we go, we'll solve for that
anyone who didn't get the computer science joke it means that
.
.
.
you don't have basi concepts for programming
.
JK
It means we are assigning the value of x + 1 to the variable x
There's one more: X! = 0
Sgn(x) = n, n€ R-{-1,0,1}
For the 3rd one, if you sub in x = i⁴, would that work since √i⁴ = i^(4/2) = i² = -1?
For the 4th one, could you do x = i²? How would that work with the absolute value?
You can’t have a negative distance from zero, so there are no solutions for the fourth one
@@The-Devils-Advocate I understand that. I was just curious as I never thought about what you could say about |i²|. It would be a colorful way of disproving the abs value by saying |-1| = |i²| = i² = -1 or something like that, but if you can go from -1 to i², you can also go from i² to -1 to disprove my original question.
Math is always fun!
@@OganySupreme |i²|=|-1|=1
If x=i⁴, x just equals 1, and the square root of 1 is defined to be 1 even though (-1)² also equals 1.
if u sub x = i^4 in 3, it gives you 1 not -1, as that symbol means "the positive square root of" (i.e. u should not consider -1 as a solution to positive square root of i^4)
Ok now who the f even does x=x+1, you'd always simplify it to x+=1 and even then, the +=1 can just be simplified to x++
That depends on what programming language you're using. I first learned programming with languages like BASIC and Pascal that don't have += or ++ (although in Pascal you'd write
x := x + 1).
In the third equation, X can be 1
Since the square root of 1 is +1/-1
-1*-1= 1
Edit: correct me if Im wrong
nope, the square root function only ever returns the positive solution.
x^2 = 1 has both 1,-1 as solutions
but sqrt(1) = 1 always
"Some solutions that have absolutely no equations"
Or tan(x)=i
tan(x) for your input.
@@gamingmusicandjokesandabit1240 I-
Oh yea! Lol
Basically, tan^-1(i) will be solution to e^ix = 0. Since this is like 2, it will not have any solution
Of course his name is gonna be Albert Einstein, what else should I expect
1. horizontal asymptote towards -inf and +inf, fair
2. vertical asymptote towards -inf, fair
3. essentially depends on using the principal root (or maybe it's by definition), as sqrt(1) could technically be +1 or -1, but we (at least normally) don't do this, so fair
4. by definition, fair
5. impossible outside of infinity or some kind of modular arithmetic (mod 1, i.e looking at decimals only, maybe?), so fair
I came upon the last one 23 times in my math hw when i simplified and just stared at for an hour straight 😂. Well now i know its unsolvable.
For 3, wouldn’t i^4 work?
No, because by definition, you finish evaluating the entire interior of the square root function, and then take the square root.
sqrt(i^4) is the same thing as sqrt(1), which equals 1.
Technically -1 is *a* square root of 1, but it isn't *the* square root of 1.
@@carultch Technically, neither 1 or -1 is the square root of 1. the square root of 1 is ±1
but, to solve √x=-1 all you need to do is to square both sides of the equation (√x)²=(-1)²
now, the √ and ² cancel each other and we are left with x=(-1)² which is 1.
easily solvable.
@@WilliamWizer By convention, the square root function refers to the positive square root, unless context specifies otherwise.
@@carultch my apologies for ignoring an irrelevant, and misleading, convention that only ignorants or lazy people use.
the only reason to use only the positive square root is in cases where you can't use the ± and ∓ signs for some reason.
Hey bprp, obviously log_2(0) is the solution to the 2nd, just letting you know!
Kidding
Well isn't that also equal to minus infinity?
So isn't there a solution?
@@wtfisthisnewfeaturebruv reread the entire comment
@@thewitchking2556 you mean by he said "don't say infinity"
Is that the point or am i missing something
Note: I am aware infinity is not a number
@@wtfisthisnewfeaturebruv it says kidding if you click read more on my comment
3 and 4 are kinda the same. The square root function is just defined to output the positive value so that the values can be distinguished by notation (+sqrt vs -sqrt)
1/Infinity=0
2^(-infinity)=0
√j=-1(i create the new type of imaginary number)
|j|=-1
basically for f: S -> R we have equation f(x)=y, where y is not in Im(S)
yeah... and I guess you mean Im(f)
Why the hell is no one noticing that he said at the end of the video "this is not computer science".
Just because you can write it ,it doesn't have to be correct.
Wise man...
First one x=tan 90°
Second one x=-tan 90°
Third one x=i^4
Fourth one x=dont know
Fifth one x=tan 90°
You missed the solutions:
1) Infinity in the extended complex plane (it's a completely valid extension, through stereographic projection onto riemann sphere...!)
2) - Real Infinity + ib
3) x = 1 where we branch cut f(z) = z^0.5 along the positive imaginary axis, and use the branch pi/2 < argz < 5pi/2 so that square rooting halves the argument and produces a result satisfying pi/4 < arg(sqrt(z)) < 5*pi/4 - which much be the negative root when z = 1
4) Obviously this would work in a number system where -1 = 1 e.g. arithmetic modulo 1. I would like to find a more convoluted solution to this.
5) This is modular arithmetic (mod 1). But if you imagine a space of a unit square, where the left edge is glued to the right edge, and top edge glued to the bottom (e.g. in some space with torus-like curvature - e.g. in old computer games), adding one to the x or y coordinate will result in no change to the x or y coordinate.
He missed nothing.
We can do field extension but it wouldn't have archimedian property anymore
as a 6th grader learning python, i appreciate the clarification on the last one.
If he didn't, you would have got him good.
Lets be honest, no real programmer would actually write "x=x+1"
They'd instead use "x++" or "x+=1"
In programming, we pre define the value of x and use that value to calculate and then define x again, but in maths we are taking x that has only one value for both sides.
"do not say infinity ohkay?"
he is so good in math that he takes a t(h)ree root of his head
Technically the answer to number 3 is 1, because any number rooted has a positive and negative answer, so the square roots of 1 are 1 and -1. And the answer for number 1 would be x= 1/0 which is undefined but is technically the answer
nope. the square root function only returns the positive solution.
x^2 = 1 has x = 1 and -1 as solutions
but
sqrt(1) = 1 always
For the third one
√x = -1
Squaring on both sides
(√x)² = (-1)²
Will give us
x = 1
So a solution exists right?
sqrt (1) = 1
For third you can take complex number or square both sides ❤
nope
please correct me if im wrong, i might be making a stupid mistake
in equation 3
√x = -1
square both sides
x=1
lets put this in eq 3
√x= -1
√1= -1
+1 or -1= -1
so this is true right?
But then u have 2 answers, positive and negative. X can be both, which means it would be half correct i guess
@@skullzs1983 it depends on the branch you take. On one branch, where √1=1, there's no solution. On the other branch, where √1=-1, there is a solution. So in total, there is a solution, but not along every branch.
1. Unsolvable
2. X≈(-infinity), since 2 to the power of bigger negative numbers gets closer to 0
3.x=1 imaginary unit
4. Unsolvable
5. Unsolvable
Waiting for mathematicians to make "Impossible numbers" now
This is an assignment operator, not a comparison operator, so x == (x+1) false.
But (-x == x and x!=0) can be true if x = minimum integer. This is because the modulus for this number is greater than the maximum number.
I feel like #3 and #4 are essentially the same:
#4 can be rewritten as (the square-root of x²) = -1. If there is no complex solution to #3 (i.e. there is no complex number whose square root is -1), then #4 would follow (there is no complex number which, when squared, would have a square root of -1) since the complex numbers are closed under multiplication (and thus squaring), and the set containing the result of squaring a complex number is a subset of the complex numbers.
There is no real or complex number whose square root is negative anything.
3 may have a solution depending on your choice of branch cut
The answer to the 2nd question is 2 to the power of -1545
2) as x approaches negative infinity the answer approaches 0 so if -∞ was a number it would be the root, and 1)+∞
4 is possible if you take it to mean the determinant of a matrix 👀
Ans 1
Multiplying both side by 0
1/x × 0 = 0 × 0
=> 0 = 0
Rhs = lhs
Hence proved
Ans 5
Putting x = 1
Then we'll get
+1, -1 ✓
In 1 you didn't find any solution for x, you just multiplied by 0
In 5, plugging 1 for x
1 = 1 + 1
1 = 2
So no
What? Are you tricking me? I dont know about complex numbers but the 3rd one definitely has 1 as its solution!!!!!!
Plug it in and you get 1=-1
Bro giving inequalities to confuse us with his Brilliant(!!) Moves, (no factorial).
the last one and 1st one is basically same try adding 1 on both sides of equation 1 you will understand
x=x+1 is possible, but not with the limitations we've placed on algebra
the limitation that prevents this equation from being true is the one that dictates that a variable can only be equal to 1 value
its also possible in another way.
if 1 represented 0.
this would be true.
In a C1 group，X = X + 1.
In a Cn group，X = X + n.
For x=x+1 => y=x+1 for y=0 x=-1 and for x=0 y=1 we graph and we see that is a line that come from -inf to the point(-1,0) to the point(0,1) to +inf ..so we need to exist a point such as that it meets this equation where its says (x,x) or (y,y) or as an equation y=x to find where it meets with equation y=x+1 ..now we have two equations with two unknowns and it is much easier to solve by equating them ...but wait if you both graph them ...they never meet because they are perfect parallel lines.....the end... so there is no solution..not that it is unsolvable...
Sun Tzu once said that "At the end of the day it's the end of the day"
1/(-1)! = 1(0)
2^-(-1)! = 2^(-2)! = 2^-infinity -> might equal some collection of zeroes greater than 1(0)
"Do not say infinite, ok?"❤😅
Is it just me or is the third equation actually solvable
I mean just square both the sides and you get x = 1
Sure, but then check your work by taking the square root of x, and it's not -1. Bzzzzzt! You can't claim it's a solution if it gives the wrong answer!
3rd can have answer as
√i²=-1
Maths teacher when you say 1/♾️ is 0 : 🤬🤬$$***#**
a whole new branch of mathematics will be invented for the third and fourth equations
In third equation what if we take natural log of both sides and write -1 as e^(2n+1 pi) ?
Equation 1 and 5 are the same!
1/x=0 if we add 1 in both sides we have 1/x + 1 = 1 then we multiply by x and get x+1 = x
having a phd in computer science I like this guys clarification 😂
infinity IS NOT a number, it's a limit
For the 3rd one I think first we should take squares on both side to the square root would be removed from x and square of minus 1 equals 1 so x equals 1
Tell me whether I am wrong
imaginary numbers have entered the chat
how can you solve these in the complex plane?
Complex arithmetic changes nothing here.
The moment the third one came up I thought i then was like "wait no"
Time for a new imaginary number, make a more complex 3d space
\sqrt{x} = -1 is surprising! \sqrt{x} = e^{i(2n+1)\pi }, x = e^{(2n+1)2i \pi} should work, right?
1. a
2. b
3. c
4. d
5. e
Since mathematicians literally made up an answer to x²=-1, i can make up my own answers
1. X= +or- infinity
2. X = i² I know this ain't possible
Sure it's possible i² = -1. It's just not the right answer.
I am a top scoring student and it's very simple that for the equation:
(square root of)x≠(-1)
that x = 1 because a negative multiplied by another negative is a positive, and 1 times 1 is 1, so negative 1 times negative 1 is 1. Simple!
Try it, plug x=1 into √x, so √1;
√1 is 1, as it is only the principal square root (without ± sign)
Unless there is ± in front, √x cannot be negative
@@skonaslp no, that entirely depends on your chosen definition. If I say we're working with the "negative" square root (the only branch other than the principal one), then √1=-1, √(-1)=-i, √2≈-1.41421, etc. and √:[0,∞)→R is a decreasing function.
Just invent/define a new value or imaginary number that can make 1/X = 0 and solve for x. g can be infinity as a number only. And the lying 8 can be infinity as a concept. Then you can solve the first 1. Then you have to go into philosphy of what infinity really means.
Now, allowing ordinals, Aleph_i satisfies the last equation, and maybe the first. IDK if ordinals get extended to signed ordinals the way naturals get extended to integers, but if so, 2^(-אᵢ) = 0, I think.
All those things are potential definitions if only you used a unique symbol instead of X. sqrt(:)) = 1 sqrt(:() = -1 would eliminate potential information loss and avoid the plus or minus answers. sqrt(9:() = -3
In the third equation If we square both side we will have x=1 right?
In 3rd equation we can solve : √x =-1 =x = (-1) whole square = x=1