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this is hypnotizing

genius

great topic!

I have a proof for the cube root. Assume for sake of contradiction that ³√2 = m/n. This would imply m³/n³ = 2, or m³=2n³, or m³=n³+n³, but Fermat’s last theorem prohibits this, QED.

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Personally my favorite proof in this line is that the Rational Root Theorem (RRT) directly tells you that all roots of integers are either integers or irrational, i.e. there are no roots of integers that have rational fractional parts. As a quick reminder, RRT says that if you have a polynomial equation with all integer coefficients of the form aₙxⁿ + ... + a₀ = 0, then if x is a rational root of reduced form p/q it must be that p is a factor of a₀ and q is a factor of aₙ . Now consider the special case xⁿ - a = 0 . If x=p/q is a reduced form rational root of that, then the denominator q is a factor of 1, but that means p/q = ±p , and therefore x is an integer. x is also the n-th root of a, so therefore the n-th root of any integer a must either be an integer or an irrational number.

To introduce the Dedikindes's cuts would now be only a "small step", you already introduced a quite illustrativ exampel. (I enjoy your lecture, it's giving insight... thanks!)

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I usually test more than just a base case so I “know” that induction might be the way to go in proving a conjecture. Looks can be deceiving in mathematics.

I have found some more hidden gems!

I don't like the proof, because it has a key step where some out of the blue, contrived trinomial is added to each side to make it work thereafter.

Fun thing about the “classical proof”: It doesn’t use anything about the number 2 other than it is prime. Very easy to rewrite the proof to show that the square root of every prime number is irrational.

First observe that 1<2<4, so 1<√2<2. Suppose that √2 was rational. Choose the smallest positive q so that p/q = √2. Then p² = 2q², q<p<2q, and thus {(2q - p)/(p - q)}² = {4q² - 4pq + p²}/{p² - 2pq + q²} = {6q² - 4pq}/{3q² - 2pq} = 2. So (2q-p)/(p-q)=√2 & 0<p-q<q. But this contradicts q was least possible denominator.

interesting

The easiest way to prove this kind of result is to use fundamental theorem of arithmetic and notice that you can extend the theorem to rational numbers.(it is not a difficult corollary of the original result). The fundamental theorem of arithmetic is not a difficult theorem to prove.

ive been watching your videos for so much time now and its so useful ugh I cant be more thankful to you and math

The last step of the proof skips a lot of steps, but writing it out formally would take too long, so I think you did a great job.

1:38 for a more complete version of (p|ab -> p|a or p|b): ((p|ab) implies ((p|a) or (p|b))) *if and only if* ((p is prime) or (gcd(p,a) = 1) or (gcd(p,b) = 1) or (gcd(a,b) > p)) The statement above is true, because it can be broken down into the following cases: ((p|ab) and (p is prime)) implies ((p|a) or (p|b)) ((p|ab) and (gcd(p,a) = 1)) if and only if ((not (p|a)) and (p|b)) ((p|ab) and (gcd(p,b) = 1)) if and only if ((p|a) and (not (p|b))) ((p|ab) and (gcd(a,b) > p)) if and only if ((p|a) and (p|b)) The above statements are written using formal logic. By the way, "implies" refers to the right arrow operator "->", and "if and only if" is the 2-way version of it, "<->".

I lost you when said, let's add thus to both sides. It is not crystal clear which equation you are referring to.

This is just the same classical proof done in an aroundabout way.

Consider the polynomial x^2-2. The roots of the polynomial are both the negative and positive square root of 2. Notice that the polynomial has integer coefficients. From the rational roots theorem, any such polynomial's rational roots must be the possible to express as a factor of the constant over a factor of the leading term. We can list all such fractions: -1, 1, -2, 2. The square root of 2, does not belong to this set of possible rational roots, yet, it is a root of the polynomial in question. From this, it's possible to conclude that square root of 2 must be a non-rational root.

One problem I've always had with the "classic" proof is that we aren't using any properties of m,n being coprime while constructing this proof. We could just as well have assumed that gcd(m,n) = anything and we woulld still get the same result with no contradiction.

you proved everything very meticulously, but in the last step you assumed cancellation must be going on. why couldnt the denominators be different primes for example?

This proof is exactly the same as the classical proof. The best proof is to note that if you square (m/n) and you get an integer, you must get a square, by unique factorization of m and n into primes. The only reason people give more complicated proofs is because unique factorization into primes is a complicated theorem, so they want to present a proof that sidesteps it. An essentially different proof would be, for instance, using the fact that sqrt(2) has a periodic continued fraction, while a rational number must have a terminating continued fraction.

Overall good video, the main points are done well, however i have a few problems. First, the leading question in the beginning, "...they're equal to what? I mean, they have to be equal to something right?" is not something you would normally ask when learning a theorem that establishes equalities. It's not the case that teachers around the world somehow fooled people by delibiretaly not telling this detail, it's just in pretty much almost all cases, there's no "nice" formula involving some other quantiy (in this case R) relating to the ones that are equal, meaning they don't "have to" equal to anything. You'd never ask the same question about for example the Pythagorean theorem and many others. The 2R result naturally arises because you have to use it in the proof, not as an "extra". Second, and this is the bigger issue, the whole section about the "Thales's Circle Theorem". There's no theorem with this name. What you described in that section is the inscribed angle theorem, for which the actual Thales's theorem (which DOES involves circles, that's the whole point) is a special case of, when the angle is 90° (in other words when the chord is the diameter). You used this special case in the proof. As a side note, this theorem can be proven without using the inscribed angle theorem, instead making use of the central angle.

Ohhh I really love that proof!!! :3 I think the one by Euclid is still my favorite one though. ^^

Let's say it's trivial that (1) each positive number has one unique positive square root and (2) each positive rational number has exactly one unique simplest form where the nominator and the denominator are relatively prime and (3) if m and n are relatively prime then (m - n) and n are also relatively prime. Then you're basically saying that sqrt(2) = (2 - sqrt(2)) / (sqrt(2) - 1), so if sqrt(2) = m/n then sqrt(2) = (2 - m/n)/(m/n - 1) = (2n - m) / (m - n) and this is a contradiction because (m - n) and n are relatively prime, so once this new fraction is simplified into its simplest form it is clearly won't be m/n. And in general, this is applicable for all n because sqrt(n) = (n - sqrt(n)) / (sqrt(n) - 1) so if sqrt(n) = p / q then sqrt(n) = (n - p/q) /(p/q - 1) = (nq - p) / (p - q) and the only way to avoid contradiction is having a square number and q = 1, because then the relatively prime argument falls apart. Probably my high school textbook had a variant of this, so probably I'm also overcomplicating it.

Hello! Such a good video. I'm in the 11:th grade but interested in more difficult math than the ones we do now:)

Very interesting alternative proof. For the lemma about divisibility of a square by 2, I like the conciseness of this proof: _n(n - 1) ≡ 0 (mod 2)_ for any pair of consecutive integers _(n - 1)_ and _n_ ⇒ _n² - n ≡ 0 (mod 2)_ ⇒ _n ≡ n² (mod 2)_ ∴ _2|n² ⇒ 2|n ⇒ (2×2)|(n×n) ⇒ 4|n²_

Very nice proof - another super neat proof is as follows: Consider any polynomial with integer coefficients of the form a1 + a2\sqrt{2} + a2\sqrt{2}^2 + ... These expressions can all be written in the form a + b\sqrt{2} for integer a, b. Now consider the geometric series (\sqrt{2} - 1), (\sqrt{2}-1)^2, ... We have a common ration 0 < r < 1 thus we can choose arbitrarily small numbers of the form a + b\sqrt{2}. If root 2 is rational -> \sqrt{2} = m/n for positive integers m, n. a + b\sqrt{2} = (ay + bx)/n which must be greater than or equal to 1/n since but clearly we can choose a + b\sqrt{2} < 1/n so we are done. Just a really nice proof - thought I'd share!

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Thank you so much for taking the time to make these videos and help math as a whole

Really enjoy the showcasing how to prove inequalities true by replacing them with a tighter, more narrow case and showing that if they are true for the narrower case, they. Must be true for the broader case. I know that this technique is possible but it was a nice review of that These videos are excellent. Thanks for currating such nice problems. Are there any cases of induction where something is proved true for all n in the rationals?

Great video! Excellent choice of examples and a really good pace. Looking forward to more such work! And yes, I would very much like to see a video on all the popular inequalities in mathematics.

Helpful

Absolutlly loved the video. Great "oddball" examples! Jajajaja

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Would love to see more esoteric applications of induction, this is a great series.

You could have started with a circumscribed pentagon with a side length of 1 and then derived the ratios for the 18°-72°-90° triangle. After that you then scale it down by a factor of the circumradius so the circumradius becomes 1 thus making your pentagon be fit literally inside your unit circle. That method may or may not be as algebraically intensive but it allows for use of Ptolemy’s theorem to derive the equation for the golden ratio. I’ve never been able to hack the intense algebra that goes with roots of unity and complex numbers.

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Brilliant video as always

Brilliant video as always

I learned induction about 10 years ago and this finally made the second step "click."

We just got to this in my trig class... my feed is listening to me

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I'm a retired electronic engineer, so I'm very familiar with all the maths used, but nevertheless I enjoyed watching the way everything fits together. Had I been tasked with finding the 5 fifth roots of unity, I would have tackled it from the other direction - draw a unit circle then divide the 360º into 5 to get 72º and then construct the 5 vectors (including the unit vector). A few applications of Pythagorus later, we've got the roots.

Try to find a more clever argument for why the roots x1, x2, x3, x4 are on the unit circle! See if you can avoid the algebra of explicitly computing norms. Like, comment, SUBSCRIBE. Follow me on FB: facebook.com/profile.php?id=61559517069850