a discovery of irrationality from 470 BC
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The offices at the Pentagon have been real quiet since this dropped
Of course they are real. You can visit them 🙄
@@christosvoskresye Well, things can be complex even without imaginary parts. Real numbers-rational and irrational alike-are complex. Anything natural is real, and therefore, complex as well. Even zero is complex. So saying that something is complex doesn't mean what I think you want me to think it means. 😜
BTW, I don't think the military contains imaginary elements. Adding imaginary elements to real ones creates a complex field, which unfortunately cannot be ordered. And we know how important following orders is within the military!
It could lead to dissolution of NATO!
Long Live BREXIT! Traiasca ROEXIT!
It's worth to mention that according to the legend Hippasus's colleagues weren't very fond of the proof (and the existence of irrational numbers in general) and literally thrown poor guy overboard.
“Hail science!”
This reminds me of one of the scenes in "Donald in Mathemagicland"
Super cool proof! Super elegant too
Edit: golden ratio jumpscare
when I first saw the title I thought "people have been irrational long before 470BC" 😁
I would love to see more historical results like this proven. Bonus points if they are represented in a similar manner as they were back then.
You can simplify the proof slightly by starting with the assumption that you have selected a measurement scale such that lengths AB and BN are relatively prime, i.e. that gcd(AB, BN) = 1. Then you proceed as in the video and get to the point where you note that BC/ CR = AB/BN, but that BC and CR are both smaller than AB and BN respectively. But that’s impossible since AB and BN are relatively prime, if BC and CR have the same ratio but are each smaller then they have a common scaling factor with AB and BN.
So you don’t actually need the infinite descent part of the argument so long as you start with the initial integers being relatively prime.
Nice! Not bad for 470 BC.
It's also a short step from that conclusion to the fact that √5 is irrational.
Fred
13:16
People: OMG it’s the satanic pentagram 😱
Me: Dallas Cowboys logo 🗿
When I was at school we learnt this demonstration and we were taught that all the segments (taken with some criteria) were all congruent by construction
And a last comment... we know that for real numbers x/x ≡ 1 (ambiguities about zero dividers and infinities ignored for the time being)
So it corresponds well with result of two functions in quotient form illustrating 0/0 ≡ 1 along with the results at signed and mixed sign infinities giving results +1 or -1 accordingly.
After all a constant is a constant for all real numbers.
This also suggests that conclusions depend upon the question being asked.
Can we call this infinite descent?
Fermat's method of infinite descent, 2000 years earlier.
@@MichaelRothwell1 thanks
Let's hope Michael hasn't just summoned something 😶🌫️
He did summon the golden ratio out of this air and thin chalk 🌚
😂😂😂😂😂😂
What a plot twist! But CZcams spoiled it for me by recommending video right below named "The Golden Ratio" 😁
Ah well - I was hoping for a question or observation about post below . Here is something I was going to put.
The quotient format Q(x) given earlier may also be represented as Q(x,y) or Q(y, x) where these represent x/y or y/x respectively.
I think this may be called a projection of x on to y or y on to x but I'll have to read up on that a bit more.
In either case it originally suited me to use y/x rather than x/y because I may have had a bit more explaining to do about zero dividers.
Some observations:
obvious Q(x) as originally given (see below) is a tied two degrees of freedom relation as x=y whether in denominator or numerator position whereas both x/y and y/x have strictly free two degrees of freedom. The obvious knot being set y=x=y in the alternative representations.
Note that x and y are names and so WLOG x=y=x or even a=z=a are equally valid.
So does the original Q(x) operate in 2-space with a free association on x,y get tied to a limited association where y takes its values directly from x ie y:=x
Or does Q(x) live in 1-space since x:=x?
Implications(?) on Q(x) as given there are different behaviors in 1-space and 2-space?
Fermat's method of infinite descent, 2000 years earlier.
music notes coming from a pentagram segments lengths is just magical
Off topic but interesting (! or ?) You decide?
Take a couple of functions and put them in quotient form. Use f(x)=x and the functions in quotient format r/s is f(x)/f(x) and that is exactly x/x
Let Q(x) denote f(x)/f(x)
Evaluate Q(x) at zero ie Q(0) Uh-ho zero divider warning!
Now consider changing the signs + or - on numerator/denominator and we have potential +/+ -/- +/- and -/+ , Say signed[Q(+x or -x)] as a hand waving descriptor
For example (-x)/(+x) evaluate this at zero.Uh-ho other zero divider warnings.
When signs are the same we see well behaved behaviors with Q(x) = 1 and when signs differ we also see good behaviors with Q(x) = -1.
It seems reasonable to expect Q(x) to be infinite, infinitesimal, +1 or -1.
But l'Ho yields funny result when signs on numerator or denominator differ for example (+1)/(-1) or (-1)/(+1) and what can we deduce from that? 🙂
And strangely enough as x tends to infinity in either direction Q(x) remains exactly 1
EDIT Interim conclusion by taking limits when x tends to 0 from left or right and when x tend to signed infinities +/-∞ :
0/0 = 1
(-0)/0 = -1
0/(-0) = -1
(-0)/(-0) = 1
∞/∞ = 1
(-∞)/∞ = -1
∞/(-∞) = -1
(-∞)/(-∞) = 1
and these results are obvious.
Good choice of tshirt, goes with the topic.
I'd start from showing geometrically that AB/BN = BM/MN (yay golden ratio (reciprocal)!), and since BN = BM+MN and MN=AB is natural in our choice of measurement, therefore BM must also be natural (if a whole and a part are both natural, then so must be the remaining part). But this means that the ratio AB/BN cannot be expressed as an irreducible fraction, because BM/MN would then be even more reduced. That's contradiction enough?
This proves, without a shadow of a doubt, that Mathematics is better than sex!
Long Live BREXIT! Traiasca ROEXIT!
Fun, thanks.
Very clever.
Brilliant! Numbers are just numbers, ... or are they? 🙂
EDIT @ 7:46 where AB/BN is assumed as an element in rationals and accepting that it is indeed a real number would it be fair to describe ratio AB:BN as a generator of a regular pentagram thus ideally parameterizing it on a ratio?
Could regular n-grams be parametrized with such a measure?
And by extension to regular n-gons after a few more tweaks and twiddles?
It might be a nice feature as a vector generator of shapes in computerized things
EDIT - it looks like the nested pentagram - that is the pentagram contained in enclosed pentagon with a pentagram - appears to rotate by one-tenth or a zero-th of 2π.
Interim observation: enclosed pentagon is rotated by one-fifth. Pentagram enclosed in pentagon is also rotated ambiguously by plus-or-minus one-fifth .
So the composite rotation of a nested pentagram within as enclosed pentagon within a pentagram is exactly zero or one-tenth?
the golden ratio is so golden.
Oh true beauty
Finally, golden ratio..!😊
I don't quite understand the proof by contradiction.
Let's suppose that the scaling factor between two pentagrams was actually rational and not equal to 1/phi as proven.
By merely the infinite reduction alone we do not necessarily end up in a contradiction if we started of with two natural numbered lengths, and showing via infinite reduction that we ended up in non-naturals.
Right? So long as we showed that the quotients of the non-naturals is still rational.
For instance 2 levels down, we would have gotten lengths 1/2 v 1/4.
So despite those two numbers are not naturals, they are both rational and their quotient is rational.
I am missing something the proof step.
The only time i got convinced was when I seen the radical of root(5) when the actual ratio was proven,
Which is then reducing the problem to just proving that the radical is irrational.
But not quite by the infinite reduction step.
**Assume the reduction factor was not 1/phi but say exactly 2**
So 1/2, 1/4, 1/8 ....
Each is non integer, but any quotient between successive numbers in that series is rational
In this specific case it works. The crucial step is that if the initial lengths are natural numbers then it is deduced that a subsequent length is smaller and also a natural number. It's the deduction that each successively smaller length has to be a natural number that results in the contradiction.
@@TinySpongey
This still alludes me. :(
As mentioned lets assume the ratio was 2. Or for that matter lets take an example of a square that I infinitely reduce the length scale 9/10ths, and tilt the smaller square by such an angle such that the edges of the smaller square touch the sides of the immediate outer bigger square.
And I do that infinitely.
By claiming that I start with natural lengths to the sides of the two biggest squares, I can still end up in an infinite reduction where the sides of inner squares at some point get smaller that a natural size length.
But that alone is not sufficient to call this a contradiction hence leading to an irrational quotient.
The ratio is always 9/10ths.
I am not saying the proof in the video is incorrect. I think a crucial element is elusive there in the proof step to make the case that all lengths had to remain natural.
Actually...
I think I might have gotten it now...
Somewhere the video shows that if the length of the largest pentagram is natural so does need the length of the adjacent next smaller pentagram be: natural.
Hence an infinite reduction series is bound to break this...
All good.
This was the key bit.
Can someone help me understand why "infinite sequence of decreasing natural numbers" is impossible?
0.1, 0.01, 0.001, ... seems like a trivial solution to what Michael is saying doesn't exist
Those aren't natural numbers but the rationals 1/10, 1/100 etc.
Naturals are ℕ ={0,1,2,...} (altough someone doesn't count 0)
@@leonardoruzzante2576lol!
how is not constructing a potential continuously ingrained pentagram a contradiction??? im lost. can someone give me a reference to some book/article for this proof. Please?
The contradiction in his proof is not the construction of an infinite number of pentagrams but the construction of an infinite number of smaller and smaller natural numbers.
Natural numbers by essence have a lower bound which is 0. Now let's set for example AB to be 4 cm, since BC is also a natural number which was proven to be strictly smaller than AB, it can only be 3, 2, 1 or 0. If BC = 3 cm then CD can only be 2, 1 or 0 cm and so on and so forth. Inevitably all of this construction will reach 0 cm at one point or another and then you're screwed because there will be another infinity of segments after that and no natural numbers left. Thus contradiction.
To be fair... he glossed over this explanation in the video and it wasn't made very clear.
For reference, it's called a contradiction by infinite descent and while not common it's quite a classic and it's always cool to see it used... even if in this case, it seems quite overkill
Nhsm
Some bozo watching will be like "the scientist are the satan pawns" 😂
Nice...
Also... Hail Satan!
messing with Satan symbol
Devil math
maybe you get yourself a pointer ...