This theorem is mind blowing, i mean just to think there is a sequence of googolplex consecutive natural numbers which are not prime, is mind blowing to me, really goes to show how rare prime numbers are.
When teaching math, I would define a prime number as “A natural number is prime if it has exactly 2 divisors.”. I know you know this and your audience does as well, but for non-math people, “A number is prime if its only divisors are 1 and itself.” implies that 1 is prime since it meets that criteria.
@@tobypenner9355 You are correct, but my statement stands. Most of the time when primes are defined, the N>1 property is not mentioned. In addition, it hides (some of) the reasoning for why 1 is no longer considered prime. Despite being a positive integer, 1 behaves differently than the prime numbers. One example of this difference is that prime numbers have two divisors whereas 1 is the unique number with 1 divisor. Just an opinion of course.
Numerous terms can be defined in different and equivalent ways. The one you should use depends on your audience and how useful the definition is for your purposes.
@@jimbobago Fair enough. I guess I would advise against using this definition when talking to younger students or non-math people as it hides why we no longer define 1 as prime. The definition I gave makes it clear to students why we separate them (beyond the fundamental theorem of arithmetic).
Wow! I was almost completely lost until the end where everything fell into place. This is a spectacular theorem, and a really simple, well executed proof. Well done!
Great video with a very detailed proof! However, I think there is a conceptually easier way to prove this: Take the first k primes. Let’s call them { 2 , 3 … p_k }. Let N = 2 * 3 * … * p_k be their LCM. Then all integers N + 2, N + 3, … , N + p_k are divisible by at least one of { 2 , 3 … p_k } and thus are not prime. This is an interval of length p_k - 1 and since the number of primes is unbounded, so is the length of these intervals.
By the chinese remainder theorem. the system of equations x = -1 mod p(1) x = -2 mod p(2) … x = -n mod p(n) has a solution mod p(1)•p(2)•..•p(n + 1), where p(i) are primes and thus relatively prime. We can pick a solution x = a, to be arbitrarily large, pick a greater than the largest of the primes (to ensure the numbers aren’t the primes themselves). Rearring the equations we have a + i = 0 mod p(i) for all 1
Pretty nice. If you wanted to build a smaller example, then N=LCM(2,3,...,n+1) would play the same role as (n+1)! for intents and purposes. It might not be the smallest example but it's the smallest refinement that follows the proof without further modifications afaik.
this is crazy to think about in combination with the twin prime conjecture (which may not be true, but weaker forms of it are already proven). could have a gap of something like grahams number between two primes and then get a twin prime not long after
Taking a = 6, b = 3, and c = 2 still satisfies it Although 6 is not a divisor of 3 + 2, we also know that 6 is not a divisor of 3 and 6 is not a divisor of 2 (rather, it's the other way around: 3 is a divisor of 6 and 2 is a divisor of 6). Thus we have false implies false, which is true
![If p ≠ 3 is prime, then 3 divides p^2 - 1 (Proof) [ILIEKMATHPHYSICS]](http://i.ytimg.com/vi/xY9d17vsBJc/mqdefault.jpg)
![If p ≠ 3 is prime, then 3 divides p^2 - 1 (Proof) [ILIEKMATHPHYSICS]](/img/tr.png)
This theorem is mind blowing, i mean just to think there is a sequence of googolplex consecutive natural numbers which are not prime, is mind blowing to me, really goes to show how rare prime numbers are.
When teaching math, I would define a prime number as “A natural number is prime if it has exactly 2 divisors.”. I know you know this and your audience does as well, but for non-math people, “A number is prime if its only divisors are 1 and itself.” implies that 1 is prime since it meets that criteria.
The definition requires N > 1
@@tobypenner9355 You are correct, but my statement stands. Most of the time when primes are defined, the N>1 property is not mentioned. In addition, it hides (some of) the reasoning for why 1 is no longer considered prime. Despite being a positive integer, 1 behaves differently than the prime numbers. One example of this difference is that prime numbers have two divisors whereas 1 is the unique number with 1 divisor. Just an opinion of course.
Numerous terms can be defined in different and equivalent ways. The one you should use depends on your audience and how useful the definition is for your purposes.
@@jimbobago Fair enough. I guess I would advise against using this definition when talking to younger students or non-math people as it hides why we no longer define 1 as prime. The definition I gave makes it clear to students why we separate them (beyond the fundamental theorem of arithmetic).
Wow! I was almost completely lost until the end where everything fell into place. This is a spectacular theorem, and a really simple, well executed proof. Well done!
The exercice itself is quite simple bro
I wish there was a love instead of like on CZcams. You make it look so easy.❤
It is really easy bro
Great video with a very detailed proof! However, I think there is a conceptually easier way to prove this:
Take the first k primes. Let’s call them { 2 , 3 … p_k }.
Let N = 2 * 3 * … * p_k be their LCM.
Then all integers N + 2, N + 3, … , N + p_k are divisible by at least one of { 2 , 3 … p_k } and thus are not prime.
This is an interval of length p_k - 1 and since the number of primes is unbounded, so is the length of these intervals.
Very elegant proof, great video!
BRO. seems hard to believe but i guess it had to be true, otherwise there would be an upper bound on the distance between two primes.
still crazy
Exactly !
This was given as an exercise in a book I was doing. How the hell does someone even think that proof 😭😭
Reminder how you prove there are an infinity of prime numbers ? Well use the same kind of reasoning and this is quite easy
By thinking!
By the chinese remainder theorem. the system of equations x = -1 mod p(1)
x = -2 mod p(2)
…
x = -n mod p(n)
has a solution mod p(1)•p(2)•..•p(n + 1), where p(i) are primes and thus relatively prime. We can pick a solution x = a, to be arbitrarily large, pick a greater than the largest of the primes (to ensure the numbers aren’t the primes themselves). Rearring the equations we have a + i = 0 mod p(i) for all 1
Pretty nice. If you wanted to build a smaller example, then N=LCM(2,3,...,n+1) would play the same role as (n+1)! for intents and purposes. It might not be the smallest example but it's the smallest refinement that follows the proof without further modifications afaik.
this is crazy to think about in combination with the twin prime conjecture (which may not be true, but weaker forms of it are already proven). could have a gap of something like grahams number between two primes and then get a twin prime not long after
Very clean
I am lost here. If a|b and a|c, then a|(b+c). If a=6, b=3, and c=2, then how can 6|(3+2)?
Taking a = 6, b = 3, and c = 2 still satisfies it
Although 6 is not a divisor of 3 + 2, we also know that 6 is not a divisor of 3 and 6 is not a divisor of 2 (rather, it's the other way around: 3 is a divisor of 6 and 2 is a divisor of 6). Thus we have false implies false, which is true
a|b means that a is a divisor of b, so your example should be 2|6 and 2| 10, so 2|6+10 which is 2|16 which is true
@@iliekmathphysics thanks I now understand
@@iliekmathphysics got it
Ugh! Where were you in my Math class? lol Might've actually learned something by cheating off YOUR tests =p
n! + 1, …, n!+ n-1 . Super easy once again 😁
Oh my bad replace n by n+1