# Verify my elementary proof that the nth harmonic number is never an integer

Forgive poor formatting, I don't know how to use mathjax very well. I am a 10th grade student with little experience writing proofs looking for constructive criticism.

I found this problem in the book Solving Mathematical Problems by Terrence Tao. In the book it says that Bertrand's Postulate (for every $$n$$, where $$n$$ is an element of $$Z^+$$, there exists a prime $$p$$, such that $$n) is required to prove it, but--interested--I tried to see if it can be proved in a more elementary manner. I believe I have found one and wish to have it verified and critiqued. I shall proceed:

Theorem. For every $$n, q$$, where $$n$$ and $$q$$ are elements of $$Z^+$$, $$2<=n$$, it is true that $$1/1+1/2+...+1/n$$ does not equal $$q$$. (Could someone please tell me how to use summation notation in mathjax, thanks).

Proof. We will assume that the sum is equal to an integer, and reach a proof by contradiction with a counterexample. Multiplying the individual terms by each other to make them common in denominator then adding them together results in the fraction:

$$2$$ x $$3$$ x ... x $$n+1$$ x $$3$$ x ... x $$n+ ...+1$$ x $$2$$ x ... x $$n-1 /n!$$

Assuming this is equal to an inteqer, $$q$$, then:

$$n!q = 2$$ x $$3$$ x ... x $$n+1$$ x $$3$$ x ... x $$n+ ...+1$$ x $$2$$ x ... x $$n-1$$

$$2(n!/2)q = 2$$ x $$3$$ x ... x $$n+1$$ x $$3$$ x ... x $$n+ ...+1$$ x $$2$$ x ... x $$n-1$$ ($$n$$ is greater than $$2$$, therefore $$n!$$ is divisible by $$2$$).

Therefore, if $$q$$ is an integer, the numerator must be even for all positive integer $$n$$. However, taking the counterexample of $$n=3$$ the numerator equals 11, which is not an even integer. This is a contradiction. Therefore the negation of our assumption must be true, and there do not exist positive integers $$q, n$$ such that the $$nth$$ harmonic number equals $$q$$, as desired.

$$QED.$$

• You can find mathjax tutorial here: math.meta.stackexchange.com/questions/5020/… – Aniruddha Deshmukh Nov 12 '18 at 4:35
• You can also have that $1 \times 2 \times \cdots \times \left( n - 1 \right) = n!q - \left( \sum\limits_{i = 1}^{n - 1} \prod\limits_{j = 1 \\ j \neq i}^n j \right)$ and hence $n | 1 \times 2 \times \cdots \times \left( n - 1 \right)$ which is a contradiction. Usually, for proving something by contradiction, we do not counter examples. Rather, we use a strong fallacy that arises due to our assumption which we then call "contradiction". – Aniruddha Deshmukh Nov 12 '18 at 4:41
• You are just showing, Riley, that for $n=3$ you don't get an integer. What meant to be done is to show that no matter what $n$ is, you don't get an integer. And this can be done without Bertrand, by considering the highest power of two involved. See math.stackexchange.com/questions/2746/… – Gerry Myerson Nov 12 '18 at 4:43
• It is useful to consider the number of factors of $2$ in $\operatorname{lcm}(1,2,3,\ldots,n)$ and note that there is only one number in $1,2,3,\ldots,n$ that has that number of factors of $2$. – robjohn Nov 12 '18 at 7:34
• Thank you for all your comments, and critiques. I understand the errors I have made now and hope to learn from them by considering your critiques. Also, the ord$2$ proof is really cool thanks for showing me it. – Riley Culloty Nov 12 '18 at 11:09

You have your "for all " and "there exists" mixed up. $$q$$ depends on $$n$$ so call it $$q_n$$. And "the numerator" depends on $$n$$ too so call it $$u_n.$$ The 1st sentence of your last paragraph should say: "Therefore if there exists $$n$$ such that $$q_n\in \Bbb Z$$ then there exists $$n$$ such that $$u_n$$ is even."... NOT "for all $$n$$."
For $$n\in \Bbb Z^+$$ let $$S(n)=\sum_{j=1}^n (1/j).$$ Let $$V(n)$$ be the largest non-negative integer $$j$$ such that $$2^j\leq n.$$ That is, $$2^{V(n)}\leq n<2^{1+V(n)}.$$
Prove that if $$S(n)=\frac {A_n}{B_n2^{V(n)}}$$ where $$A_n, B_n$$ are odd positive integers then $$S(n+1)=\frac {A_{n+1}}{B_{n+1}2^{V(n+1)}}$$ where $$A_{n+1}, B_{n+1}$$ are odd positive integers. This implies that $$S(n)\not \in \Bbb N$$ for $$n\geq 2$$ because $$n\geq 2\implies V(n)\geq 1.$$
The proof requires some care. Split it into 2 cases:(i). When $$n+1$$ is a power of $$2.$$ (That is , $$n+1=2^{V(n+1)}),$$ and (ii). When $$n+1$$ is not a power of $$2.$$
• I once looked at the highest power of $3$ that divides the denominator of $S(n)$ when $S(n)$ is in lowest terms and found that for some $n$, it is less than the highest power of $3$ not exceeding $n$. With powers of $2$ it is just an even/odd issue but for powers of $3$ it gets complicated. – DanielWainfleet Nov 13 '18 at 8:17