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.
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_
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.
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!
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, "".
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.
Assume that GCD(m,n)>1. Because m and n are both positive integers, we know that their common divisor must be a natural number. Because the natural numbers have a lower limit (1 or 0 depending on your definition), there must be a finite amount of common factors between m and n. Because of how the classic argument is constructed, the contradiction arises as a proof by infinite descent. Even assuming a large common divisor, we should be able to only extract finitely many common factors, but we can find a factor of 2 infinitely many times, which implies we can get arbitrarily small pairs of natural numbers m,n, descending "infinitely," contradicting the well ordered nature of the natural numbers.
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.
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?
Let a/b and c/d be rational numbers, where a/b = c/d, and neither are 0. Now, swap the variables if c>a. Example: with 2/4 = 6/3, we swap both fractions, getting 6/3 = 2/4. We now know what a>=c. If b c/d, which causes a contradiciton. Therefore, b>=d. The only way for a/b = c/d, regardless of whether the numbers are prime or not, is for there to exist some integer e>=1 such that a=ce and b=de, so a/b = (c/d)(e/e). "Cancelling" term is mentioned because a/b = (ce)/(de), and the common term e can obviously be cancelled on both sides. But that means that a and b must have a common factor of e. In the video, a = m, b = n, c = 2n-m, and d = m-n. If m and n share a common factor of e, then gcd(m,n) must be a multiple of e. gcd(m,n) = 1 though, so the only valid value of e is 1. However, that means m = 2n-m, which is not possible since m > 2n-m. This causes a contradiction, so m/n can not = (2n-m)(m-n).
@@simonwillover4175 "The only way for a/b = c/d, regardless of whether the numbers are prime or not, is for there to exist some integer e>=1 such that a=ce and b=de" I would really apreciate a proof of that
The set of rational numbers is defined to be the set of ordered pairs (m,n) in Z where n != 0 and equivalence is defined such that (a,b)=(c,d) if and only if ad=bc. Assume a and b share no factors, assume c and d share no factors. In order for ad and bc to be equal the Union of the set of prime factors of a and d must equal the Union of the set of prime factors of b and c. Since a and b share no factors, the prime factors of ad contributed by a must all be contributed by c in bc, and therefore a=c. Similarly b=d. Therefore the only way for 2 irreducible rationals to be equal is for them to be the same rational. Now assume c and d share no factors, assume a !=c and b !=d. Again we have the Union of the factors of ad equal to the Union of the factors of bc. Since c and d share no factors, the factors contributed by c in bc must be contributed by a in ad, but since a != c there must be some factor in a that’s not in c or some factor in c that’s not in a. But since all factors of c must be in a, the only possibility is that there’s some factor in a that’s not in c. Similarly all factors of d belong to b and there’s some factor in b that’s not in d. Since the set of factors of ad is equal to the set of factors bc, the factor in a but not c must be the same as the factor of b not in d. Therefore we have some factor e such that ea=c eb=d. Q.E.D
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.
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.
Thank you. I encourage you to continue with your interest, because there's so much fascinating math stuff accessible to 11th graders which is never touched on in class.
Same, I'm in ninth grade, we are doing basic linear equations right now in school, it is pretty boring to me. X3 Most of my teachers (!!!) don't even know what irrational means for numbers and call me stupid when I show them. qwq
@@DoxxTheMathGeek If you want harder equations try learning some quadratic formulas, like the abc formula. Pythagorean theorem is a good one to learn. If you don’t know them that is. Start with pythagorean theorem:) If you know that try trigonometry! Cosine, sine, tangent. Very important in highschool here in Sweden at least.
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.
Yes, rational roots theorem is a nice way!
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_
Cute proof of the lemma!
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.
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!
Thanks. I'll have to sit down with a nice hot cup of green tea (no coffee) and think about this.
This is equivalent to the continued fraction proof, except folding in the theorem that best-approximations are in the continued fraction.
First observe that 1
Ohhh I really love that proof!!! :3
I think the one by Euclid is still my favorite one though. ^^
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, "".
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.
Assume that GCD(m,n)>1. Because m and n are both positive integers, we know that their common divisor must be a natural number. Because the natural numbers have a lower limit (1 or 0 depending on your definition), there must be a finite amount of common factors between m and n. Because of how the classic argument is constructed, the contradiction arises as a proof by infinite descent. Even assuming a large common divisor, we should be able to only extract finitely many common factors, but we can find a factor of 2 infinitely many times, which implies we can get arbitrarily small pairs of natural numbers m,n, descending "infinitely," contradicting the well ordered nature of the natural numbers.
@@ProactiveYellow That makes sense actually, yeah. I never thought of it that way.
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.
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?
Let a/b and c/d be rational numbers, where a/b = c/d, and neither are 0. Now, swap the variables if c>a. Example: with 2/4 = 6/3, we swap both fractions, getting 6/3 = 2/4. We now know what a>=c. If b c/d, which causes a contradiciton. Therefore, b>=d.
The only way for a/b = c/d, regardless of whether the numbers are prime or not, is for there to exist some integer e>=1
such that a=ce and b=de, so a/b = (c/d)(e/e). "Cancelling" term is mentioned because a/b = (ce)/(de), and the common term e can obviously be cancelled on both sides. But that means that a and b must have a common factor of e.
In the video, a = m, b = n, c = 2n-m, and d = m-n. If m and n share a common factor of e, then gcd(m,n) must be a multiple of e. gcd(m,n) = 1 though, so the only valid value of e is 1. However, that means m = 2n-m, which is not possible since m > 2n-m. This causes a contradiction, so m/n can not = (2n-m)(m-n).
@@simonwillover4175 The only way ... why? Thats exactly my point. It is of course true, but the proof is missing.
@@simonwillover4175 "The only way for a/b = c/d, regardless of whether the numbers are prime or not, is for there to exist some integer e>=1
such that a=ce and b=de" I would really apreciate a proof of that
The set of rational numbers is defined to be the set of ordered pairs (m,n) in Z where n != 0 and equivalence is defined such that (a,b)=(c,d) if and only if ad=bc. Assume a and b share no factors, assume c and d share no factors. In order for ad and bc to be equal the Union of the set of prime factors of a and d must equal the Union of the set of prime factors of b and c. Since a and b share no factors, the prime factors of ad contributed by a must all be contributed by c in bc, and therefore a=c. Similarly b=d. Therefore the only way for 2 irreducible rationals to be equal is for them to be the same rational. Now assume c and d share no factors, assume a !=c and b !=d. Again we have the Union of the factors of ad equal to the Union of the factors of bc. Since c and d share no factors, the factors contributed by c in bc must be contributed by a in ad, but since a != c there must be some factor in a that’s not in c or some factor in c that’s not in a. But since all factors of c must be in a, the only possibility is that there’s some factor in a that’s not in c. Similarly all factors of d belong to b and there’s some factor in b that’s not in d. Since the set of factors of ad is equal to the set of factors bc, the factor in a but not c must be the same as the factor of b not in d. Therefore we have some factor e such that ea=c eb=d. Q.E.D
@@harsinsinquin4032 yes, the uniqueness of the prime factor decomposition is the answer.
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.
Very nice, I like very much this idea of sqrt(2) = (2 - sqrt(2)) / (sqrt(2) - 1) and then putting in m/n for sqrt(2) on the right-hand side!
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.
interesting
Hello! Such a good video. I'm in the 11:th grade but interested in more difficult math than the ones we do now:)
Thank you. I encourage you to continue with your interest, because there's so much fascinating math stuff accessible to 11th graders which is never touched on in class.
Same, I'm in ninth grade, we are doing basic linear equations right now in school, it is pretty boring to me. X3
Most of my teachers (!!!) don't even know what irrational means for numbers and call me stupid when I show them. qwq
@@DoxxTheMathGeek If you want harder equations try learning some quadratic formulas, like the abc formula. Pythagorean theorem is a good one to learn. If you don’t know them that is. Start with pythagorean theorem:) If you know that try trigonometry! Cosine, sine, tangent. Very important in highschool here in Sweden at least.
@@MrViktorWahrb I'm doing fractional calculus, I already know them. X3
Thanks though! :3
oh brother college math is a roller coaster. you’ll have fun
I lost you when said, let's add thus to both sides. It is not crystal clear which equation you are referring to.
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This is just the same classical proof done in an aroundabout way.