Aldfrith Aern he may just say “Excellent point. The thing is, it was digits, the plural for digit. 1 and -1 already have so many other famous properties.” Or something along those lines.
@@jamesclark6864 but ultimately the talk about "favorite number" surely having this property and many other famous properties would make it even more of a favorite. I for one love Wau which has similar properties to the number 1.
to everyone saying that they were bad audience, keep in mind that they were British school children where it's taught interrupting during presentations is rude.
Found this myself by writing program (in Ruby) which looks for such numbers: i = 0 loop do digits = i.to_s.chars.map(&:to_i) j = digits.map{|x| x**x }.sum puts i if i == j i += 1 end
It makes me really sad that when I watch Matt's live talks, most of the time he doesn't get good reactions to jokes I laugh my ass off at. WHY DON'T THESE PEOPLE APPRECIATE MATHS JOKES??
one reason is the mic placement, its more important to capture the speaker than the audience. But also, a lot of times in science humor the jokes only land if you understand what the joke is riffing on. Ergo, in teaching something comedically, a joke based in the thing you are currently educating the audience in has a hard time finding the mark as the listener might not have made the connection the joke relies on for a successful lampoon. And its not like it can be explained and get a big laugh. Because dissecting a joke is like dissecting a frog.....no body laughs and the frog dies ;)
Probably not, since this video would make that number extremely unsafe for him. =P He may use some tricky version of it though, like maybe the first four numbers in (3+4+3+5)^(3+4+3+5) backwards or something like that.
+ Laurelindo = 15¹⁵ = 437,893,890,380,859,375 Or the first (or last; or middle) 4 digits of 3435³⁴³⁵. [It has 12146 digits, so there actually *are* a middle 4. If *those* are Matt's PIN, then his secret is *definitely* safe with me!]
to all of you mentioning 1. at 2:44 he said "this is the only number for which if you raise the digit-S [...]" since 1 doesnt have more than 1 digit, it doesnt evalidate his statement check and mate
+Ben E No it does not, because you cannot have plural with just one thing. Your formalisation is correct, but it has one problem, namely that mathematics doesn't make any distinction between different numbers of things. If you were right, then we should be able to say that 1 has a digits in it, and that's simply nonsensical. One thing is never in plural, unlike his quote had the word digit in plural.
Ben E This collision of maths and languages is also part of the reason why zero factorial seems so confusing to some people, they are thinking with language, not with mathematical logic.
Ben E And I just managed to delete my original reply, how marvelous! But my point was mainly that you can't describe some mathematical ideas verbally, and that's why we can actually argue for and against. I'm into both maths and languages, but enjoy linguistics a bit more. And where as English is unable to make an unambiguous sentence to cover all integers with a positive number of digits, I believe, though I'm not entirely sure, some Sino-Tibetan languages might be able to. At least Mandarin Chinese doesn't make any distinction between singular and plural, unless it is by use of particles and numerals.
Ben E Yep, and the whole thing is about Munchhausen numbers, and 1 definitely is a Munchhausen number. Actually the only one besides 3435. But that's the thing, while you can write down a formula which covers all numbers with any number of digits, you can't really say it aloud in English. One of the more fascinating aspects of maths and logic.
Yeah. I have a bit of a pet peeve for people talking about properties of a representation of a number as if it has something to do with the number itself, which is why I don't find the fact that 3301 and 1033 are both prime compelling at all.
"My lucky number's 4 billion, that doesn't come in real handy when I'm gambling. "C'mon 4 billion; Fuck, 7. Not even close, I need more dice"" --Mitch Hedberg
Yes, Matt, you're right that 0⁰ ≠ 0; rather, it is undefined. Mainly because n² = n³/n n¹ = n²/n n⁰ = n¹/n = n/n = 1 all of which works as long as n ≠ 0. Because with n = 0, you're getting 0⁰ = 0/0, a very well-known undefined expression. If you introduce the multiplicative identity, 1, into it, then you could say: n² = 1·n·n n¹ = 1·n = n n⁰ = 1 and you could insist on that as defining even 0⁰ as 1. But if instead, you come at it from the powers-of-0 angle, then you could just as well insist that any power of 0 is 0. In truth, you have to consider nᵐ in which both *m* and *n* are going to 0. If you take the limit as n→x ≥ 0 and m→y, then as long as x and y aren't *both* 0, it won't matter how you arrange that double limit: lim ( lim nᵐ) = lim ( lim nᵐ) = xʸ n→x⁺ m→y m→y n→x⁺ But as soon as you let x=y=0, there's trouble. lim ( lim nᵐ) = lim 1 = 1 n→0⁺ m→0 . . but: lim ( lim nᵐ) = lim 0 = 0 m→0 n→0⁺ And there are infinitely many m:n dependences you can dream up to boil this down to a single limit, that will produce various results. But the simplest of these is m = n, which will yield: lim mᵐ = 1 m→0⁺ So personally, I would say that if you *have* to assign it a value, that value should be 1. But in the end, you really can't escape the fact that 0⁰ is undefined.
ffggddss I personally prefer to think of every sigma or pi notation as a for loop. \Sigma_{i=a}^{b}{f(i)} = sum of f(i) for i = a to b. (pardon the tex notation). As a for loop this would be: res = 0; b your result is zero, the identity of sigma sums for(i = a; i b --> res = 1 is returned. I suppose it's how you looked at it, but with some computer science thrown in there.
+Anthony Northrup Interesting perspective there, I've never actually thought about sigma notation that way, despite it being basically the same concept.
Yeah but I'd think that for this kind of recreational math that's kind of a trivial solution. 3435 is much more impressive in that it's not a unit element.
Are there any cycles of such numbers? (For simplicity of typing, I'll use p(n) to mean n after it's been split into digits and each digit exponentiated and added). For example, any cases where p(n) = m and p(m)=n ? Or p(n)=m; p(m)=q; p(q)=n ? What about other number bases? Or cycles across multiple bases?
Trying to understand the calculation behind 0^0 breaks my mind. And I make jokes about quantum mechanic: _Why did the particle cross the event horizon?_ _It was never actually on the other side._
You could argue that since 0^0 is undefined it could equal anything you want it to Which would mean any number that has a zero could have this property since you just make the zero equal whatever number you need. This would also include every number if you count lead zeroes
My favorite number is 17 it is a prime number equal to 2^3+3^2, but my second favorite number is 89 because it is a prime whose digits add up to 17, my favorite number
Okay, first of all, who figures this stuff out? What sort of mind looks at a number and goes "You know, if you were to do this obscure set of operations this one would have a unique property that pretty much only it has". However, talking about the 0^0 thing... it's complicated, isn't it? Generally X^0=1, the reason being that we're talking about repeated multiplication and the multiplicative identity is 1. Among other reasons is X^0 was 0, then it would cause a lot of products to just disappear and wipe out a lot of mathematician's hard work. There are other better reasons than that. But, 0^0 is generally considered to be 1, mostly so it will play nice polynomial identities, but as many people throughout the years have pointed out that doesn't really make sense because how does 0, the concept of nothing, become something like this? And so they argue that 0^0 should be 0, except that breaks polynomial identities (and power series and a whole bunch of things) and so they settle on "Fine, 0^0 is undefined unless you need it to be something that won't break everything, so then you can call it 1, but I'm not going to be happy about it" and most of mathematics is happy to just ignore those folks. Though, I will point out that in the case of 438579088, with 0^0 being 1 in there, it's only off by 1, resulting in a sum of 438579089. Meaning the result is adjacent to 3435, so we can call 438579088 3435's derpy little brother or something. OR you could just ignore the ^0, because that's got a lot of baggage, and you get that nice circular identity that Matt's so exited about. By the way, if you type 3435 into the video while it's playing you get Matt saying "Only but only thousand" which doesn't mean a darn thing.
My all time favourite number is 288 and so it has been for over 10 years (I'm now 17), the strange thing is that I have no clue why it's my favourite number
You've picked a reasonably interesting one. 1^1 + 2^2 + 3^3 + 4^4 = 1 + 4 + 27 + 256 = 288 1! x 2! x 3! x 4! = 1 x 2 x 6 x 24 = 288 In base 12 it's a nice round 200 (referred to as two gross) Average surface temperature of Earth using Kelvin as the unit of measurement.
It makes perfect sense. He said if you add up the digits raised to their own power. But since 1 only has one digit, there is no adding to be done. He seemed to very specifically avoid saying Munchausen Numbers, so he wouldn't have to include 1 or -1. And it's also why he can say that 0^0 != 0, since, for Munchausen Numbers, they define it that way. That's why all those extra zeroes at the beginning or end still work. When doing addition, 0^0 is often defined as 0, just like it's often defined as 1 for multiplication.
Kalum Batsch I'm not drunk...I just understand the concept of "plural" and you don't. If someone is discussing adding the "digits" of a number...clearly single digit numbers can't qualify.
So... how do we know it is the only one? Is there an underlying rule to this or by going up through series of numbers raised to themselves, might we eventually discover another, bigger one? (It would be less impressive, because I usually think the most impressive unique-ish numbers are the smaller ones that you may end up actually using, but still.)
There's an upper bound on numbers that could work, since each new digit can only add up to 9^9 to the sum, but increases the number by at least 10^n (where n is the number of digits), so the sum will be too small.
0^0 is often assumed to be 1. Most calculators would agre. However 1 in any base certainly has those properties so I have at least infinite solutions with 1.
yes. it would be correct to say "take the paper from the desk" not papers. if you said that to someone they would probably respond something like "wheres the other papers?" or "wtf there's only one!". Calling something many of something and it only being one is just wrong.
It's called the Munchausen number. You get -1 and 1 in every base. (And 0 if you allow 0^0 = 0, as most do). To get the rest, I wrote a quick JavaScript program. These are the Munchausen numbers under one million in bases 2 through 36. (base 3): 12; 22 (base 4): 131; 313 (base 6): 22352; 23452 (base 7): 13454 (base 9): 31; 156262; 1656457 (base 10): 3435 (base 13): 33661 (base 14): 23 (base 17): 33 (base 20): 6534 (base 25): 13; 513 JavaScript does not natively support bases higher than 36.
See the other videos on the channel, but basically Matt hosted a bunch of shows containing a variety of other speakers and entertainers over a huge range of disciplines
I would say 1, but not -1. If -1 counted then -3435 would also count. "-1" is not a digit, so the (-) would have to be left to the outside of the "raise each digit to itself" operation.
Actually, it has been the topic of debate whether 0 to the power of 0 is one or zero. On the one hand, you would usually get n to the power of 0 as n to the kth power divided by itself, which in fact gives all zeroth powers the value 1, but this does not apply to 0, because that would be division by 0, which is not allowed. This is why many people claim that 0 to the power of 0 equals all other powers of 0, much like how all real numbered powers of 1 equal 1. During actual calculations, this barely ever comes forth, so it is not something that needs to be dealt with immediately, but it is definitely a not yet solved issue. Until the zeroth power is more closely defined, I will have to disagree with you about 438579088.
But it is more closely defined. It's been two years, so you've probably forgotten you've left this comment and have been corrected and thus making this comment pointless, or I may just come off as rude, but I've seen 3 proofs, 2 of which are in this comment section: The limit of x^x as x approached 0 appears to be 1 if you were to graph it. But that's forgetting complex numbers, who don't follow this rule. (Credit to a Numberphile video I found once) y = 2x y/x = 2 y * x^(-1) = 2 We know that when x=0, y=0, so it is possible for x=y to be true. Let's take the limit as y tends to x: x * x^(-1) = 2 x^1 * x^(-1) = 2 x^(1-1) = 2 x^0 = 2 Now note that the only way in which y can tend to x in the equation y=2x is if x tends to 0 (and y tends to 0). So let's now take the limit as x tends to 0: 0^0 = 2. (Credit to Smathlax) x^0 is x/x (hence why x^-y = 1/x^y.) Saying 0^0 is saying 0/0, which is undefined. [This one I'm not so sure is accurate.] (Credit to TheMoonRover) Sorry if you already learned one of these proofs in the two years since you commented this, but I had to say it.
0^0 doesn't equal 0? Isn't it indeterminant? So it technically can equal anything you want? It would still make that second number not quite work though.
One way to think about it is that there are two conflicting rules being applied. 0 times anything is always 0, and anything to the 0 power is always 1, so what does it mean when you apply both of those at once?
There are two conflicting rules, but those aren't the ones as there is no multiplying by zero occurring. Raising a number to the power 0 can be defined as dividing it by itself: 1/1=1, 874/874=1, (-0.3)/(-0.3)=1, e/e=1, 0/0 is undefined. Raising a number to the power _n_ can be defined as taking 1 and multiplying it by the number _n_ times, which for n=0 results in 1 no matter what number you choose. Contradiction.
0^0 = anything because 0^0 = 0^(x - x) = (0^x)/(0^x) = 0/0 = anything since 0/0 = anything because if you inverse it you get anything times zero equals zero which is correct.
