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Irrational number is non-repeating and non-terminating. Is it possible that an irrational number may contains a sequence of another irrational number? If no, than the rule of non-repeating and non-terminating will not be fulfilled.

If yes, than each irrational number can be transformed into another irrational number with the formula of

y = (10^n)x % 10

For example, in the sequence of PI, there may be 3.14....141421356.... Then It can be transformed into the square root of two by multiplying it with some n to become 314...1.41421356... and modded by 10 to become 1.41421356... which is the square root of two.

Is there any explanation in that if it is wrong? Or if it is correct, how to determine the value of n?

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    $\begingroup$ That can't happen: $\pi$ is transcendental, but $\sqrt2$ isn't. $\endgroup$ – Lord Shark the Unknown Oct 22 '18 at 4:14
  • $\begingroup$ It is just for an example, my question is about any pair of irrational number, not specifically about PI and square root of two. $\endgroup$ – Kuzunoha Oct 22 '18 at 4:15
  • $\begingroup$ This always happens. One property of irrational numbers is that their decimal expansions are non-periodic, so in particular if cut off finitely many digits, the tail of the decimal is still non-periodic, and hence is another irrational number. $\endgroup$ – Dark Malthorp Oct 22 '18 at 4:17
  • $\begingroup$ Given any irrational number $x$ and any positive integer $n$, your formula $(10^n)x % 10$ will also be an irrational number. $\endgroup$ – Dark Malthorp Oct 22 '18 at 4:19
  • $\begingroup$ If you are asking if there exists a specific irrational which can be transformed into another specific irrational, then yes as demonstrated. If you are asking if any arbitrary irrational can be transformed into any other arbitrary irrational, the answer is clearly no. Consider the irrational numbers $0.101101110111101111\dots$ and $0.3233233323333\cdots$ (the previous irrational plus $\frac{2}{9}$) they don't even share any digits in common past the decimal point. Further, a number may only contain at most one of those irrationals I describe as an infinite subsequence at once, never both. $\endgroup$ – JMoravitz Oct 22 '18 at 4:29
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If by "sequence" you mean a finite sequence of digits of another number, then the answer is yes, a number exists that can do this for all real numbers, but nobody know what it is.

As a matter of fact, almost all real numbers will do this, but nobody has found a single example.

Numbers that will do this are numbers that are normal in every base. See https://en.wikipedia.org/wiki/Normal_number

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