We know all the irrational numbers cannot be written as the ratio of two rational numbers like all the square roots of not perfect squares 2,3,5,... Not all real numbers are equal to a square number. In general, every number not equal number the power $n$ for all $n$ in rational numbers. In addition Euler number $e$ and $\pi$ and the golden ratio is $φ$ always if want to to prove any result we used these numbers as examples. Are there other irrational numbers that are not mentioned?
Asked
Active
Viewed 162 times
4
-
10Yes, uncountably many. The first sentence says it all. A real number other than the quotient of two integers is irrational. It's hopeless to try to list them. – saulspatz Apr 10 '21 at 12:23
-
Most numbers are irrational, this can be put in rigorous terms using measure theory. Think of this as saying, if you randomly select a real number you will pick an irrational number with probability equal to 1 – Nick Castillo Apr 10 '21 at 12:25
-
3It can be proven that the number is irrational if and only if its decimal expansion is not finite and does not "repeat" (as a finite cycle) after some point. Thus, if you can imagine a number such as $0.12345\ldots$ and make sure the digits never repeat, you've found a way to generate irrational numbers. (Also, this is the way you can generate all the rational numbers: just make the decimal expansion finite, or infinite where digits repeat after some point, e.g. $3.14159292929292\ldots$.) – Apr 10 '21 at 12:27
-
Related: https://math.stackexchange.com/questions/1443877/how-to-prove-that-x-0-1234567891011-dots-is-irrational – Apr 10 '21 at 12:29
-
There are uncountably many numbers with even a stronger property , so-called transcendental numbers. They are not even the root of a polynomial with integer coefficients. Examples include $\pi$ , $e$ and the solutions (also the complex solutions) of $x=\cos(x)$ The algebraic numbers (numbers that are a root of a polynomial with integer coefficients) are not only countable, they form a field (which is even algebraically closed) – Peter Apr 10 '21 at 12:34
-
2You are also advised to make a difference between: (a) The definition of irrational numbers: "A number $x$ is irrational if it is not equal to any fraction $p/q$ with $p, q$-integers ", and (b) The examples of irrational numbers (such as $\sqrt{2}, \sqrt[3]{5}, \varphi, e, \pi$ etc.). The statement (a) fully characterizes the irrational numbers. The examples (b) can never exhaust that definition - there are just too many irrational numbers! – Apr 10 '21 at 12:37
-
@StinkingBishop Very good point ! In fact, "almost all" irrational numbers are even uncomputable ! – Peter Apr 10 '21 at 12:40
-
2"Are there other irrational numbers that are not mentioned?" As alluded to already, there are uncountably infinitely many more. That said, proving that a particular number is irrational can be particularly challenging. It is obvious that $q+x$ where $q$ is rational and $x$ is irrational will always be irrational (so things like $3+\pi$ will still be irrational) but it is not so obvious when talking about $x+y$ where $x$ and $y$ are both irrational. There are examples in both directions and many unknowns. We don't even have a proof yet for if $\pi+e$ is irrational – JMoravitz Apr 10 '21 at 12:41
-
Since i can write any number randomly by infiniting and repeating decimal expansion to get an irrational number so i can not say all irrational number must be limit of sequences of rational numbers, like when i can that all rational number must be limitd of sequence of irrational numbers. What u think? – EBTISAM ALGASEM Apr 10 '21 at 12:51
-
1@Bsmah this is exactly right. Every irrational number is a limit of a sequence of rational numbers. What you say is also my preferred way of showing that. Indeed, for example, $\pi$ is the limit of the sequence of rational numbers: $3, 3.1, 3.14, 3.141, 3.1415, 3.14159,\ldots$. However, the digits of $\pi$ don't start repeating at any point, as $\pi$ itself is irrational. – Apr 10 '21 at 12:54
-
@Bsmah this is exactly right. Every irrational number is a limit of a sequence of rational numbers. What you say is also my preferred way of showing that. Indeed, for example, π is the limit of the sequence of rational numbers: 3,3.1,3.14,3.141,3.1415,3.14159,…. However, the digits of π don't start repeating at any point, as π itself is irrational. – Stinking Bishop 4 mins ago – EBTISAM ALGASEM Apr 10 '21 at 13:02
-
When it comes to (non-)rationality, rather than thinking in terms of decimal expansion, I like to consider continued fractions instead. If the continued fraction representation is finite, it represents a rational, if it's infinite, it represents an irrational. Take some infinite sequence of naturals, anything, repeating or not, combine it with some $a_0 \in \mathbb{Z}$, you get the continued fraction representation of an irrational. – user3733558 Apr 10 '21 at 13:04
-
Yah you pick especially pi as example to prove there is sequence of rational convergent to pi. That is right but there are infinitn incontable irrational number. My question is all irrational number in real numbers is limited of sequence of rational numbers – EBTISAM ALGASEM Apr 10 '21 at 13:10
-
1Yes, every real number, irrational or otherwise, can be seen as the limit of a sequence of rational numbers. Following from the earlier example, given a number $x$ (for example $\pi$ but doesn't need to be $\pi$) you have $x=\lim\limits_{n\to\infty}\lfloor 10^n\cdot x\rfloor \cdot 10^{-n}$. There are of course other ways you can express numbers too, you can have an irrational be the limit of a sequence of irrationals, or a rational the limit of a sequence of irrationals, etc... these aren't mutually exclusive. – JMoravitz Apr 10 '21 at 13:56
-
@Bsmah You understand that, whatever I did for $\pi$, I can do for any real number! – Apr 11 '21 at 09:25
1 Answers
3
$$\sum_{k=1}^\infty \frac{a_k}{2^{k!}}$$ where $\{a_k\}$ is any non-eventually vanishing binary sequence, already provides you with an uncountable number of irrational numbers.
amWhy
- 210,739
mathcounterexamples.net
- 71,758