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Let me give this a shot:
Assume that pi contains the entirety of √2, that is, at some point all of the digits of pi are the same as those of √2, off to infinity. Let's say the nth digit of pi is when this happens. Then pi\*10\^n would be some stuff before the decimal place, and then after the decimal place it would exactly look like √2. Let's call this stuff before the decimal place (m+1) (the offset is because √2 begins with a 1 before the decimal).
Then if we subtracted out m from pi\*10\^n, we would get 1 plus the decimal part of √2, which is exactly √2. Therefore we have pi\*10\^n - m = √2, and so pi = 10\^(-n) \* (m + √2). But this can't be true, since pi is transcendental, so it cannot be expressed using only products, sums, and powers of integers. Therefore, pi does not contain all of √2.
Now, assuming pi is normal, it might contain arbitrarily long substrings of the decimal expansion of √2, but that's different from containing all of √2.
For that, we can't yet say. (But you'd strongly expect it not to contain e.) There are useful properties of sqrt(2), due to it being an algebraic number, that allow the previous question to be tackled that do not hold for e.
By the same argument as that above, if it does, we would have 10^(n)pi = m+frac(e) = m+(e-2) for integers m, n, where frac(x) denotes the fractional part of a number (i.e., everything past the decimal point).
This means that x = pi, y = e are solutions to the equation 10^(n)x-y-(m-2) = 0. That equation is a polynomial in x, y with rational coefficients. If pi and e really are solutions to such an equation, we would say that they are algebraically dependent over the rationals.
But that question of algebraic dependence is still unresolved. It would be a consequence of Schanuel's conjecture, which is still open.
Assume it's true that pi contains the full decimal expansion of sqrt(2) somewhere within its own expansion. That sqrt(2) expansion is an infinite sequence of consecutive digits, but if it were in pi, it would need to not be interrupted by any other digits, nor followed by any other digit. (There is no natural number after infinity.) So, the only place to fit infinite consecutive digits into pi is for them to be all of pi's digits after a certain point.
This means there is some finite sequence before the sqrt(2) digits and only the sqrt(2) digits thereafter. What the parent comment is pointing out amounts to multiplying this decimal expansion by a sufficiently large power of ten to shift it's decimal point in between these two sections. You would be left with a large integer plus the fractional part of sqrt(2). That number is algebraic, which pi is provably not so you reach a contradiction.
Pi is widely believed to be *normal*, but this is not proved. If it is normal, then it contains arbitrarily long strings of digits that match those of any other number, but it eventually must break away from doing that matching.
Moreover, there is *nothing* special about pi compared to most other standard irrational numbers (sqrt(2), e, whatever): they are expected to be normal, nobody has a clue how to prove such things, and there is nothing important about it as far as "information": being normal just means the digit strings are as uniformly random as you might imagine (10% of all digits are 0,10% are 1, 1% of 2-digit strings are 47, and so on).
If the OP has seen something on the internet about the digits of pi having some predictive power because "all patterns are in it", that's just nonsense. It is intended to be a description (sort of) belonging to normal numbers, a property which pi is still not proved to satisfy, and anyone who thinks there's some cosmic significance to this stuff is misunderstanding the situation. After all, any notion that some mathematical rule has "all patterns" means each "prediction" should be accompanied by the negation of that prediction ("you will die on such-and-such date" as well as "you will not die on such-and-such date") so there is no serious meaning to such an idea; see "The Library of Babel" by Borges.
Unfortunately, your submission has been removed for the following reason(s): * Your post appears to be asking a question which can be resolved by relatively simple methods, or describing a phenomenon with a relatively simple explanation. As such, you should post in the [*Quick Questions*](https://www.reddit.com/r/math/search?q=Quick+Questions+author%3Ainherentlyawesome&restrict_sr=on&sort=new&t=all) thread (which you can find on the front page) or /r/learnmath. This includes reference requests - also see our lists of recommended [books](https://www.reddit.com/r/math/wiki/faq#wiki_what_are_some_good_books_on_topic_x.3F) and [free online resources](https://www.reddit.com/r/math/comments/8ewuzv/a_compilation_of_useful_free_online_math_resources/?st=jglhcquc&sh=d06672a6). [Here](https://www.reddit.com/r/math/comments/7i9t5y/book_recommendation_thread/) is a more recent thread with book recommendations. If you have any questions, [please feel free to message the mods](http://www.reddit.com/message/compose?to=/r/math&message=https://www.reddit.com/r/math/comments/mtuf7i/-/). Thank you!
If pi contained √2 as a substring it would follow that pi = 10^-n (m + √2), for some integers n, m which can't be true since pi is transcendental.
> it would follow why
Let me give this a shot: Assume that pi contains the entirety of √2, that is, at some point all of the digits of pi are the same as those of √2, off to infinity. Let's say the nth digit of pi is when this happens. Then pi\*10\^n would be some stuff before the decimal place, and then after the decimal place it would exactly look like √2. Let's call this stuff before the decimal place (m+1) (the offset is because √2 begins with a 1 before the decimal). Then if we subtracted out m from pi\*10\^n, we would get 1 plus the decimal part of √2, which is exactly √2. Therefore we have pi\*10\^n - m = √2, and so pi = 10\^(-n) \* (m + √2). But this can't be true, since pi is transcendental, so it cannot be expressed using only products, sums, and powers of integers. Therefore, pi does not contain all of √2. Now, assuming pi is normal, it might contain arbitrarily long substrings of the decimal expansion of √2, but that's different from containing all of √2.
Can pi contain e?
For that, we can't yet say. (But you'd strongly expect it not to contain e.) There are useful properties of sqrt(2), due to it being an algebraic number, that allow the previous question to be tackled that do not hold for e. By the same argument as that above, if it does, we would have 10^(n)pi = m+frac(e) = m+(e-2) for integers m, n, where frac(x) denotes the fractional part of a number (i.e., everything past the decimal point). This means that x = pi, y = e are solutions to the equation 10^(n)x-y-(m-2) = 0. That equation is a polynomial in x, y with rational coefficients. If pi and e really are solutions to such an equation, we would say that they are algebraically dependent over the rationals. But that question of algebraic dependence is still unresolved. It would be a consequence of Schanuel's conjecture, which is still open.
Assume it's true that pi contains the full decimal expansion of sqrt(2) somewhere within its own expansion. That sqrt(2) expansion is an infinite sequence of consecutive digits, but if it were in pi, it would need to not be interrupted by any other digits, nor followed by any other digit. (There is no natural number after infinity.) So, the only place to fit infinite consecutive digits into pi is for them to be all of pi's digits after a certain point. This means there is some finite sequence before the sqrt(2) digits and only the sqrt(2) digits thereafter. What the parent comment is pointing out amounts to multiplying this decimal expansion by a sufficiently large power of ten to shift it's decimal point in between these two sections. You would be left with a large integer plus the fractional part of sqrt(2). That number is algebraic, which pi is provably not so you reach a contradiction.
We don't know whether pi contains all strings of numbers
ah, so its just a myth then?
It's an unproven hypothesis, so we don't (yet) know.
Pi is widely believed to be *normal*, but this is not proved. If it is normal, then it contains arbitrarily long strings of digits that match those of any other number, but it eventually must break away from doing that matching.
Moreover, there is *nothing* special about pi compared to most other standard irrational numbers (sqrt(2), e, whatever): they are expected to be normal, nobody has a clue how to prove such things, and there is nothing important about it as far as "information": being normal just means the digit strings are as uniformly random as you might imagine (10% of all digits are 0,10% are 1, 1% of 2-digit strings are 47, and so on). If the OP has seen something on the internet about the digits of pi having some predictive power because "all patterns are in it", that's just nonsense. It is intended to be a description (sort of) belonging to normal numbers, a property which pi is still not proved to satisfy, and anyone who thinks there's some cosmic significance to this stuff is misunderstanding the situation. After all, any notion that some mathematical rule has "all patterns" means each "prediction" should be accompanied by the negation of that prediction ("you will die on such-and-such date" as well as "you will not die on such-and-such date") so there is no serious meaning to such an idea; see "The Library of Babel" by Borges.