Our Daily Story #12 (Final): The Vagabond Mathematician Paul Erdős



Paul Erdős (\er-dish) was one of the greatest mathematicians in 20th Century, a Jewish Hungarian, single, no job and and no home. He traveled around the world in one small suitcase containing his mathematical papers. He would knock impromptu at the door of his former students or math collaborators, started working and published the papers, then moved on to next destination the moment his overstay became unwelcome by the host’s wife. In this way he published 1,500 articles in his lifetime. He was awarded the International Wolf Prize in 1983/84 together with the Chinese S.S. Chern 陈省身.

What is Erdős Number ?



Einstein had the Erdős Number 2 , through his assistant who co-wrote a paper with Paul Erdős.

As a side remark, the Chinese would like to know that Paul Erdős was a good friend of Hua Luogeng 华罗庚 (Story #10). There are 2 anecdotes of their friendship below:

1. During the WWII when China was invaded by the Japanese, Paul’s students wanted to raise fund in the Berkeley university campus for Chinese war victims. They knew that leveraging on Paul Erdős reputation would be a great appeal to the fund-raising campaign. Someone put up a postal announcing a nude girl performance to be attended by Paul himself. Everybody knew Paul was never interested in girls (in the video he admitted he had a rare sensitivity against sexual pleasure). Paul obliged to honor the event by attending personally – wearing a pair of completely no-see-through dark glasses.

2. Paul got into trouble for the habit of writing postcards where the texts were openly visible to the FBI, especially he was from the communist Hungary. Once in his postcard to the communist China’s Hua Luogeng, his texts written as :”Dear Hua, …” followed immediately by all the ‘encrypted codes’ to the FBI agents (who were ‘mathematical idiots’ ): \forall \varepsilon > 0,  \exists \delta > 0,  \nabla \partial x
Paul was interrogated by the FBI on several occasions for spying. Later he chose the freedom by denouncing the USA citizenship, beginning his vagabond life till death.

In the Chinese greatest pugilistic Classic, read by fans globally including Deng Xiaoping, “射鵰英雄传” (The Legend of the Condor Heroes) by 金庸 (Jin Yong, aka Dr. Louis Cha), in which the Kungfu world (武林) was dominated by 4 grand masters:
(A) The northen vagabond beggar clan chief 北丏洪七公
(B) The western egoistic toxic master 西毒欧阳丰
(C) The southern seclusive monk 南僧段誉
(D) The eastern eccentric pharmacist 东邪黄药师

Interestingly the mathematical world is also similarly characterized by the 4 grand masters: (1-to-1 mapping)

\text {Mathematics} : \to \text { Kungfu }
Paul Erdős \mapsto (A)
Cauchy \mapsto (B)
Fermat \mapsto (C)
Gauss \mapsto (D)

\boxed {\text {EPILOGUE}}

This ends the 12 “Our Daily Story in Math” (#1 – #12), all originated from “The Last Theorem of Fermat” – the Math since ancient Greek 3,000 years ago till proven in 1994 – a long human mathematical journey to understand the language of God.

I hope you enjoy this revolutionary form of eReading (blog, texts, hyperlinks, videos, images), piecing all the ‘gems’ available freely on the Internet to form a precious ‘necklace’ (eBook).

Please share this freely via the buttons below to “Twitter” or “Facebook”. Press “Like” if you like the stories.

Thank you.



Books: (available for loan at the Singapore National Library)

1. The man who loved only numbers

2. “My Brain Is Open

Erdös and Posa: coprime

Paul Erdös asked the 12-yr old Hungarian Math prodigy Posa:

Take 51 numbers from 1 to 100, why there will be at least 2 numbers relative prime?

Posa took few coffee cups, imagine 50 of them.
1. Put consecutive numbers (1, 2) in 1st cup, (3, 4) in 2nd cup,… (99, 100) in 50th cup.
2. Take out all odd numbers (1, 3, 5, k…, 99) from each cup => total 50 odd numbers
3. Take 51st number from any remaining (say kth) cup => the number = k+1
4. (k+1) & k from (step 2) co-prime    [QED]

Two consecutive numbers co-primes

Prove by reductio ad absurdum:
Let a, b consecutive numbers: a – b = 1
Assume a, b not co-prime,
There is g.c.d. (a, b) = c divides both a, b
=> a = c.a’, b = c.b’
=> a – b = c(a’-b’)
Since a-b = 1 (consecutive),
c(a’-b’) = 1
=> c |1 impossible
=> a , b must be co-prime [QED]