Gavin Does Math

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@bobbyheffley4955 2 күн бұрын

The error is in line 5: if both numbers under the square root sign are negative, you must change the sign of the product of the individual square roots.

@juliem3667 2 күн бұрын

This person is very smart

@diamondaforce 4 күн бұрын

Great explanation. I dont see much stuff about Lambert W but this is a cool idea taking the derivative. Makes me wonder what application doing so could have.

@goldenplucker 5 күн бұрын

js plug it into desmos

@realcaesium 5 күн бұрын

awesome video, gavin! you have such a gift for making complex math concepts both clear and exciting. the way you explain things really shows your passion and creativity, thanks for inspiring us to see the beauty in math! good stuff mate!

@innermoon6194 5 күн бұрын

since when is 2^(-1/2) = 1?? 2:40

@gavindoesmath 5 күн бұрын

That was multiplying everything by -1/2 might have looked like a superscript

@melongl5697 6 күн бұрын

W Lambert!

@NoamMasterLearn-e7t 2 ай бұрын

Hey! I teach math too! :D

@gavindoesmath 2 ай бұрын

@@NoamMasterLearn-e7t 🙂

@ChristianBrickerD 4 ай бұрын

It would be great if you did more SAT-style problems. This was a great one. You just earned a new subscriber.

@franciscoduque7703 7 ай бұрын

LIKE.

@saburousaitoh 8 ай бұрын

Please look: viXra:1905.0008 submitted on 2019-05-01 20:40:00, An Interpretation of the Identity $ 0.999999...... =1$

@thetaomegatheta 8 ай бұрын

That work is kind of garbage. A lot of hand-waving, no proofs, all to approach a question the answer to which is obvious to anybody who understands the relevant definitions. Also, the author outright states that a series is equal to its sum, which is nonsense.

@mia.serable 10 ай бұрын

now I understand why your channel infuriates the people of the twenty first century

@thinking_stardust 10 ай бұрын

What would happen in the arctan if the real part (a) was equal to 0?

@gavindoesmath 10 ай бұрын

Arctan=0 means theta is 0, so it is on the positive real axis

@thinking_stardust 10 ай бұрын

@@gavindoesmath noo i dont mean if the arctan itself was equal to 0, i mean a=0 in arctan=b/a

@gavindoesmath 10 ай бұрын

@@thinking_stardust arctan is not defined but that means it only has an imaginary part so it is a quadrangle

@A2Moonclaw 10 ай бұрын

Wrong, because the square root of 1 results in two possible solutions, 1 and -1. You only used the negative roots and thus this proof is wrong.

@HamzaAlmusawi20 10 ай бұрын

You can also prove it with a geometric series! By writing out .9999… as fractions adding infinitely, you’ll get 9/10 / 1 - 1-10, which is just 1!

@zacklee5787 10 ай бұрын

That's not really a "proof" since you're assuming that you can move the decimal point to the right in the same way you can with finite decimal numbers. This happens to be true but it essentially amounts to to same thing as just assuming 0.999... = 1. The only reason this is true is because that's how we defined it when we came up with the real number system.

@gavindoesmath 10 ай бұрын

This is intuitive and easy to prove using sum of infinite converging series

@zacklee5787 10 ай бұрын

@@gavindoesmath Yes, but when we start from the definition of infinite decimals as converging series we can simply calculate 0.999... directly as a converging series. That's why your proof is essentially begging the question.

@ashtar3876 10 ай бұрын

So uhhh is this a real thing or is there an error somewhere

@zkuppy9885 10 ай бұрын

it’s correct

@joeymikaelson9845 10 ай бұрын

It is real

@TheRenaSystem 10 ай бұрын

Haven't watched yet, but just factor out 5 to the x, divide both sides by 4, take the natural log of both sides, done. X = ln(250)/ln(5) Edit: Whyd you add so many extra steps lmao

@lutis5692 10 ай бұрын

One other proof for 0.999... = 1: We know from logical point of view in math that if there are 2 numbers and one of them is smaller then the other one then there is still FOR SURE infinitely many other numbers that are between these two numbers. And if there's not it means the numbers are equal. So... can you find at least one number that is between 0.999.... and 1?

@zacklee5787 10 ай бұрын

That's not a proof though. You would have to actually prove that no such number exists, which you could do by showing every number less than 1 is also less than 0.999...

@lutis5692 10 ай бұрын

@@zacklee5787 I mean... yeah. Its not quite a "legit" proof but I hope you've got the point :)

@lutis5692 10 ай бұрын

I don't really remember if that proof of yours is the one I saw in college but for people who don't know 0.999... is indeed equal to 1

@puppergump4117 10 ай бұрын

The reason we have this in the first place is because of base 10 rounding errors. Also, there's a better proof. 1/9 = 0.111... 2/9 = 0.222... 3.9 = 0.333... ... 9/9 = 1 You can't get to 0.999... through any fractional expression.

@reedoken6143 10 ай бұрын

this isn't actually a rigorous proof. assuming 1/9 = .11111... is making the same assumption as .999.... = 9/9. they're both true, but you can't use one to prove the other because they're mathematically equivalent statements

@puppergump4117 10 ай бұрын

@@reedoken6143 Transitive property guarantees that since a = b and b = c, a = c. Because 9/9 = 0.999... in this pattern, but 9/9 = 1 normally, 0.999... = 1. It's the same way of proving x^0 = 1

@reedoken6143 10 ай бұрын

@@puppergump4117 yes, which is why it is trivial to say that because 1/3 = .333... then 1 = .9999. there is no proof there.

@puppergump4117 10 ай бұрын

@@reedoken6143 Sorry, what I mean is that because of base 10 rounding errors, there really is no such thing as 0.999... without being rounded to 1. The proof is that 9/9 converges that rounding error to 1.

@reedoken6143 10 ай бұрын

@@puppergump4117 ahhhh, I see, repeating decimals like .999... only exist as a consequence of base 10. I misunderstood you, sorry. thanks for explaining

@kpika911 10 ай бұрын

Gavin should stop doing math

@calengo454 10 ай бұрын

@-YELDAH 10 ай бұрын

This infinitesimal slander will not go unpunished!

@thatcooldudeisawesome876 10 ай бұрын

I am fine with letting a equal that number, but because it is infinite in length there is no certain way that multiplying it by 10 will produce the number we assume.

@ladripper47874 10 ай бұрын

But it does because all multiplying by 10 does is shift the decimal point

@BODYBUILDERS_AGAINST_FEMINISM 10 ай бұрын

Logical error you have infinitely many 9's or 3's which invalidates the argument

@igniteshadow2902 10 ай бұрын

Can you explain how that's a logical error? I didn't get it.

@BODYBUILDERS_AGAINST_FEMINISM 10 ай бұрын

@@igniteshadow2902 something like 1/3 equals 0.333333 forever which is undefined. There's roughly two hundred trillion videos on KZbin with the proof give it a gander

@BODYBUILDERS_AGAINST_FEMINISM 10 ай бұрын

@@igniteshadow2902 1/3 does NOT equal 0.33333333. 1/3 is irrational therefore to say it equals anything is unverifiable. There's many KZbin videos on it.

@giists5197 10 ай бұрын

@@BODYBUILDERS_AGAINST_FEMINISM 1/3 is irrational... now ive seen everything

@BODYBUILDERS_AGAINST_FEMINISM 10 ай бұрын

@@giists5197it's true.

@melongl5697 10 ай бұрын

Why did you change to natural log at the end

@RubyPiec 10 ай бұрын

log base x of n = log(x)/log(n) or ln(x)/ln(n)

@melongl5697 10 ай бұрын

@@RubyPiec how does that help solve the equation though

@RubyPiec 10 ай бұрын

@@melongl5697 good question. i assume it's because if you had to solve this with a calculator, more people know log(5) and log(2) than log_2(5)

@melongl5697 10 ай бұрын

Thanks gavin!!!

@Alberto6800 10 ай бұрын

Wouldn't it have been easier to go the 3/3=1 1/3=0.333 *3=0.999 ?

@aidan-mx2fy 10 ай бұрын

that’s not how it works

@BODYBUILDERS_AGAINST_FEMINISM 10 ай бұрын

1/3 doesn't equal 0.333 logical error unfortunately

@thetaomegatheta 8 ай бұрын

@@aidan-mx2fy That is how it works. (Assuming they meant '0.333...' and '0.999...' and not '0.333' and '0.999'.

@Triznels 10 ай бұрын

Gavin so the beef still going

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