I have been wondering what I am going to do with this blog for quite some time now. Should I ramble about the little things that occur in my life. Should I focus on academic things? These are some of the questions I have been asking myself. However, they lead to more questions. Are other people interested in reading about my life? If I focus on academic things, is there an audience here with the background knowledge to understand everything I am writing about? To answer the last question, probably not, as is evident from my last post. So for now, I am going to experiment a little bit and see what happens.
I thought I would share an elegant math proof with everyone. This is a proof that necessarily all undergraduates in mathematics have seen at least once in their lives. It proves something that is rather intuitive: there are infinitely many prime numbers. This should be intuitive to everyone, since there are infinitely many integers. However, you cannot simply say that there are infinitely many prime numbers; you have to prove it.
Since I am writing this for non-math people, there is some background we need to cover first. Here is the formal definition for divisibility. Let a and b be integers with b not equal to 0. We say that b divides a if a=bc for some integer c. In symbols, "b divides a" is written as b|a. This is essentially division with a remainder of 0. To give some examples, 5|10 because 10=2*5. However, 2 does not divide 5 because there is not an integer c such that 5=c*2.
The second preliminary is a theorem that I will not prove. The theorem states that every integer except 0, +1, and -1 is a product of primes. You might be familiar with prime factorization. This theorem essentially proves prime factorization.
Now for the proof! The method of the proof is as follows. We will assume that there are finitely many prime numbers and arrive at a contradiction. Since I cannot do subscripts here, I will write P1 as one prime number, P2 as another, and Pn as another, where n is a finite integer. Assume there are finitely many prime numbers P1, P2, P3, ..., Pn. The product of all of the prime numbers is then P1P2P3...Pn. Now consider the integer N=P1P2P3...Pn + 1. Then none of the prime numbers in our finite list divide N. But from the previous theorem, the integer N is a product of primes. Therefore, there must be a prime number not in our list that divides N. That is it. There is a more formal way to do it, but I want to leave this as informal as possible.
I wrote this in hopes that someone would find it interesting. Most people dislike mathematics, and I can understand why. One of my professors always asks, "what were you in Calculus?" She then replies, "little machines." In lower math, all we do is memorize formulas and apply them hundreds of times. We are in fact machines in lower-level math. Real mathematics is not like this. Real mathematics requires creativity, something that many people might not ever realize.
I thought I would share an elegant math proof with everyone. This is a proof that necessarily all undergraduates in mathematics have seen at least once in their lives. It proves something that is rather intuitive: there are infinitely many prime numbers. This should be intuitive to everyone, since there are infinitely many integers. However, you cannot simply say that there are infinitely many prime numbers; you have to prove it.
Since I am writing this for non-math people, there is some background we need to cover first. Here is the formal definition for divisibility. Let a and b be integers with b not equal to 0. We say that b divides a if a=bc for some integer c. In symbols, "b divides a" is written as b|a. This is essentially division with a remainder of 0. To give some examples, 5|10 because 10=2*5. However, 2 does not divide 5 because there is not an integer c such that 5=c*2.
The second preliminary is a theorem that I will not prove. The theorem states that every integer except 0, +1, and -1 is a product of primes. You might be familiar with prime factorization. This theorem essentially proves prime factorization.
Now for the proof! The method of the proof is as follows. We will assume that there are finitely many prime numbers and arrive at a contradiction. Since I cannot do subscripts here, I will write P1 as one prime number, P2 as another, and Pn as another, where n is a finite integer. Assume there are finitely many prime numbers P1, P2, P3, ..., Pn. The product of all of the prime numbers is then P1P2P3...Pn. Now consider the integer N=P1P2P3...Pn + 1. Then none of the prime numbers in our finite list divide N. But from the previous theorem, the integer N is a product of primes. Therefore, there must be a prime number not in our list that divides N. That is it. There is a more formal way to do it, but I want to leave this as informal as possible.
I wrote this in hopes that someone would find it interesting. Most people dislike mathematics, and I can understand why. One of my professors always asks, "what were you in Calculus?" She then replies, "little machines." In lower math, all we do is memorize formulas and apply them hundreds of times. We are in fact machines in lower-level math. Real mathematics is not like this. Real mathematics requires creativity, something that many people might not ever realize.
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thanks mister, the support is very much appreciated