(open) The most basic math proof I've ever seen

[fresh_42]

Well, it is what I consider the most basic proof I've ever seen. I like it for several reasons: it is so simple. that you can convince a preschool kid, it uses one of the most fundamental principles at all: you cannot derive false from true, and it is old enough to count as such:

No, sorry kids, not Pythagoras. Euclid's ($\approx 300\, BC$) proof about the number of primes.

For any finite set $\{p_1,\ldots,p_r\}$ of primes, consider the number $n = p_1p_2 \cdots p_r + 1.$ This $n$ has a prime divisor $p$. But $p$ is not one of the $p_k$ : otherwise $p$ would be a divisor of $n$ and of the product $p_1p_2 \cdots p_r$, and thus also of the difference $n − p_1p_2 \cdots p_r = 1,$ which is impossible. So a finite set $\{p_1,\ldots,p_r\}$ cannot be the collection of all prime numbers.

Can you phrase the statement which is proven, and rephrase this proof, such that

• it is clearly an indirect proof
• it is clearly a proof by contradiction

Can you tell the tiny difference? It is not a difference in logic, but in wording.

Can you find all the hidden assumptions in the above proof?
Hint: There are many.

 

[jedishrfu]

One thing I noticed is a preschool kid is going to ask you what a prime number is but maybe a fourth-grader will understand.

The set of primes is assumed to have all primes in the interval of 2 thru n right? so in this case, if 2 were missing then the prime product+1 would be even implying 2 is missing from the set.

 

[fresh_42]

One thing I noticed is a preschool kid is going to ask you what a prime number is but maybe a fourth-grader will understand.

The set of primes is assumed to have all primes in the interval of 2 thru n right? so in this case, if 2 were missing then the prime product+1 would be even implying 2 is missing from the set.

Yes, preschool was a bit ambitious. I didn't know the English term for the first 4 school years. As soon as they know how to multiply small numbers, this proof can be taught. Maybe with a bit more effort than the one above, but it is basically possible,

No, the set of primes could be any. Only finiteness is required.

 

[jedishrfu]

In the US, multiplication tables up to 10x10 are taught in fourth grade (my own experience circa 1960s)

However, according to this article its now 3rd grade that gets kids into multiplication.

https://www.verywellfamily.com/what-your-child-will-learn-grade-guide-620869

 

[fresh_42]

In the US, multiplication tables up to 10x10 are taught in fourth grade (my own experience circa 1960s)

However, according to this article its now 3rd grade that gets kids into multiplication.

https://www.verywellfamily.com/what-your-child-will-learn-grade-guide-620869

The age doesn't play a role as long as curiosity on the child's side and patience on the adult's sde is there. I once taught a six year old negative numbers and by seven she could solve $\sqrt{-4}$.

 

[Rfael]

Bayes theorem :D :D: https://en.wikipedia.org/wiki/Bayes'_theorem just use 3 equations to get it

Taylor series: uses only derivatives and polynomials

 

[Hill]

Cantor's diagonal argument.

 

[jedishrfu]

John Conway and Alexander Soifer proof is pretty short and hard to beat

https://fermatslibrary.com/s/shorte...ous-math-journal-john-conway-alexander-soifer

 

[martinbn]

Every number has a prime divisor is one of the assumptions.