3

I was watching a very-introductory very-basic minicourse (in portuguese, unfortunately) about number theory. In the first minutes of lecture 1, the lecturer (a great one, btw) first defines what is an irrational number: it is a number that cannot be expressed as a quotient of two integers (aka it is not a solution for a polynomial equation of degree 1). Then he starts writing some formulas for $\sqrt 2$ (which would be later proved to be irrational): $$\sqrt 2 = \left(\frac{2\times 2}{1 \times 3}\right)\times\left(\frac{6\times 6}{5 \times 7}\right)\times\left(\frac{10\times 10}{9 \times 11}\right)\times\dots$$

What bugged and I obviously missed something is that this formula says that $\sqrt 2$ is precisely the quotient of two integers. I have no reason to believe that both $2\times2\times6\times6\times10\times10\dots$ is not an integer and the same goes for $1\times3\times4\times5\times9\times11\dots$.

Can anyone please explain what I got wrong about the formula or the theory?

daniel.franzini
  • 159
  • 4
    What integer would that numerator be? There is no infinite integer. – Ian Mar 04 '20 at 14:46
  • 4
    "I have no reason that both $2\times 2\times 6\times 6\times 10\times 10\cdots$ is not an integer..." You should have every reason to believe that it is not an integer. The expression you describe is the limit of the results of all of those infinitely many multiplications of numbers greater than $1$ and is infinity. – JMoravitz Mar 04 '20 at 14:46
  • 4
    Do not confuse "being a rational number" with "being the limit of a sequence of rational numbers." Yes, $\sqrt{2}$ is the limit of a sequence of rational numbers, but that does not in and of itself mean that it is a rational number. This is just like how $0$ is not a positive number but can be seen to be the limit of a sequence of positive numbers $1,.1,.01,.001,.0001,\dots$ – JMoravitz Mar 04 '20 at 14:47
  • 3
    Does this answer your question? Can the product of infinitely many elements from $\mathbb Q$ be irrational? – JMoravitz Mar 04 '20 at 14:51
  • 1
    On a side note: Shouldn't the second parenthesized factor have $5\times 7$ in the denominator, rather than $4\times 5$? (The factor $\frac{36}{20}$ by itself is greater than $\sqrt{2}$, and as far as I can tell the other factors are each greater than $1$.) – paw88789 Mar 04 '20 at 15:03
  • Formula is fixed now. Thanks @paw88789. – daniel.franzini Mar 04 '20 at 15:07
  • Now I know my mistake: I didn't realize that the product of integers is actually a limit in disguise. Studying Wallis formulas and definitions makes it very clear that a ratio of infinite products can be irrational. – daniel.franzini Mar 04 '20 at 22:10

1 Answers1

4

When you see "...", you are looking at an abbreviation for a formal (and sometimes much longer) statement. What the formula means is that if $P(n)=\prod_{j=1}^n\frac {(4j-2)^2}{(4j-3)(4j-1)}$ whenever $n\in \Bbb N,$ then $\lim_{n\to\infty}P(n)=\sqrt 2.$

And $\lim_{n\to \infty}P(n)=\sqrt 2$ is itself an abbreviation for $\forall r>0\,\exists n_r\in \Bbb N\, \forall n\in \Bbb N\, (n\ge n_r\implies |P(n)-\sqrt 2|<r).$

DanielWainfleet
  • 59,529