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Assume that there is a positive rational number representable by a finite series of digits (I'll use 3 as an example). What is the largest rational number less than 3? We generate the sequence P = {2, 2.9, 2.99, 2.999, ...}. The limit of the sequence is 3, but we're looking for a smaller number, so it is 2.9~, where ~ means 'repeater but don't take the limit'. :-) There must exist some irrational number x, such that $p_i$ < x < 3 for all i. What is the decimal expansion of x? Applying the normal method of generating a decimal expansion will produce 2.9~, but that is rational. It appears that x does not have a decimal expansion.

Can someone please show me the error in this argument? Thanks in advance.

Edit: Just to clarify. I'm happy to accept that 2.9~ is not a real thing. Every $p_i$ is a rational number, so there must be an irrational number between it and 3 (you can find the proof of that elsewhere). x is just the number that meets that requirement for every element of the sequence (you can demonstrate that with induction). How do you show that x has an infinite non-repeating expansion? It seems to me that inducing the digit sequence for x must produce an infinite sequence of nines.

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    " What is the largest rational number less than 3?" There is no such smallest rational number less than 3. That is a fundamental concept you'll simply have to stumble onto. It's a growing pain all mathematicians face. "There must exist some irrational number x, such that $p_i < x < 3$ for all i". Why? No, there doesn't have to be any such number. There is no such number.... welcome to the world of grown-up mathematics. There's lot's of brick walls to slam our faces into but... they eventually break down. – 2017-01-04
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    Your infinite sequence of rational numbers $2.9, 2.99, 2.999, \dots$ proves already that there is no largest rational number less than $3$. – 2017-01-04
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    The error is that, even if $2.99\ldots 99$ is in your sequence for any finite sequence of $9$'s, that means absolutely nothing about infinite sequences. That is, you are presuming that there is some element $2.99\ldots$ in your sequence and that it is rational and not $3$ - which doesn't work out, because none of the steps in that leap are valid. For instance, it doesn't answer questions like, "So, how is $2.9$~ written as a ratio of integers?" – 2017-01-04
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    "There must exist some irrational number x, such that p_i < x < 3 for all i". Actually, I *would* like to know why you think this. In fact, I'd say as the $p_i$ can get "as close as we like" to 3. There must *not* be such an x (rational or irrational).... because we can always find a $p_i$ that is closer to 3. – 2017-01-04
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    "There must exist some irrational number x, such that $p_i < x < 3$ for all i". This isn't true. What you meant to say was "For each $i$ there must exist some irrational number $x$ that $p_i < x < 3$". That statement is true but it is a different statement. That $x$ only exists for *one* $p_i$. For *another* $p_k$ there is another $y$ so that $p_k < y < 3$. But there is no such irrational that is larger than *all* $p_n$. – 2017-01-05

5 Answers 5

1

Suppose that we have

$$p_i=2.\underbrace{999\dots999}_i$$

And we want

$$p_i

By the squeeze theorem, as $i\to\infty$,

$$x=3$$

So there does not exist a largest rational $x$ less than $3$, for it would have to equal $3$ and still behave like a number.

5

There is no such $x$. For any real number $x<3$ there is an $n\in\Bbb Z^+$ such that $\frac1{10^n}<3-x$, and it follows immediately that

$$x<3-\frac1{10^n}=2.\underbrace{99\ldots99}_n<3\;.$$

That such an $n$ exists is a consequence of the Archimedean property of the real numbers.

4

Given a rational $q$ there is no largest rational number less than $q$

Suppose there was, called $x$ with $x \lt q$. Then $x \lt \frac{x+q}{2} \lt q$, where $\frac{x+q}{2}$ would be rational too.

The same argument applies if $q$ and $x$ have a finite decimal representation as $\frac{x+q}{2}$ would too.

3

" What is the largest rational number less than 3?"

Answer: There is none.

Just as there is no answer to "what is the largest number" has no answer because "for any number $n$ you can always have $n + 1 > n$", we have "for any rational number $r < 3$ we have $r < \frac {r+3}2 < 3$".

"There must exist some irrational number x, such that $p_i < x < 3$ for all $i$"

----- edit ----

Ah! Now I understand why you said that.

You mean for each $p_i$ there is an irrational $x_i$ so that $p_i < x_i < 3$. That is true. But it isn't true that there is one irrational $x$ so that $p_i < x < 3$ for all $i$.

Indeed, by definition $\lim p_i = 3$ means that $3$ is the smallest number that is larger than all $p_i$. For any number $y < 3$ (whether $y$ is irrational or rational) there will be some $p_i$ (for $i$ large enough) so that $y < p_i < 3$.

For any $0 < w$ there is a $0 < \frac {1}{10}^n < w$ sp for any $x < 3$ there will be a $0 < \frac {1}{10}^n < 3-x$ so $x < 3-\frac 1{10}^n = p_n$.

So no single $x < 3$ is greater than all $p_i$.

BUT you are correct that for every $\frac 1{10}^i > 0$ there is a $y_i$ so that $\frac 1{10}^i > y_i > 0$ so there is an $x_i = 3 - y_i$ so that $p_i < x_i < 3$. ($y_i$ may be rational or irrational; it doesn't matter.)

BUT that $x_i$ can be different for every different $p_i$. And there will never be a single number $x$ so that $p_i < x < 3$ for all $p_i$.

------ end of edit and back to my original post -------

Response: Na-hnnh!!!, does not!

No. There doesn't have to be such an irrational number. In fact there can not be any such number.

For any $x < 3$, $3 -x > 0$. If $3-x > .1$ let $i = 1$. If $3-x > .01$ let $i = 2$. There must be some $i$ where $3-x > 10^{-i}$ because if it didn't $3-x \le .00000 .... = 0$. But $3-x > 0$. So$3-x > 10^{-i} = 0.0000.......1$ so $x < 3- 0.000000....1 = 2.999999.......9 = p_i$.

Remember: $2.99999......$ with infinite number of nines IS equal to exactly $3$. Not, "it gets as close as you like to $3$" or "it gets infinitely close to $3$". No. It IS $3$. Exactaly $3$. So obviously for any $x < 3$ then $x < 3 = 2.999999......$ so "obviously" there is some $i$ so that $x < 2.999.....999 < 3$.

Welcome to grown-up mathematics (and childish mathematicians).

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    So many downvotes today :( And the flavor of this post gives me a good laugh. – 2017-01-04
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    "There must be some i where 3−x>$10^{−i}$ because if it didn't 3−x≤.00000....=0. " 3-x is only zero in the limit. Any element actually in the sequence must be a finite rational less than 3, and so the required irrational x must exist. You can find the proof elsewhere on this site. – 2017-01-05
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    @Jim "and so the required irrational number $x$ must exist" No, that's not correct. While every element in the sequence is $<3$, every irrational $<3$ is $<$ some element of the sequence - there is no number, irrational or otherwise, which is both $<3$ *and* bigger than every element of the sequence. For each *specific element* $p_i$ of the sequence, there is an irrational between $p_i$ and $3$ - but as $p_i$ moves closer to $3$, that irrational has to move too. There is no *single* irrational that works for *every* $p_i$. – 2017-01-05
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    But that is a require irrational for *each* p_i. It is not a required rational for *all* p_i. For $p_i < 3$ there is an irrational $x_i$ so that $p_i < x_i < 3$ but $x_i$ need not be the same as $x_{i+1}$. There might be an $x > p_i$ for a *lot* of $p_i$ but there will always be a $p_n > x$ and then *another* $y > p_n$ but there is *NO* $x > $ *all* $p_i$. That would be impossible. – 2017-01-05
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    Now that does answer why you made that statement which I simply could not understand. For *each* $p_i$ there is an $p_i < x_i < 3$ but there is no *one* irrational $x$ that is larger than *all* $p_i$. In fact $3$ is the smallest number that is larger than all $p_i$. – 2017-01-05
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    "For each specific element pi of the sequence, there is an irrational between pi and 3 - but as pi moves closer to 3, that irrational has to move too." If there is an irrational x that is in ($p_i$,3), it must also be in all ($p_j$,3), j < i. It doesn't move, it's always there. ($p_i$,3) can be bijected to R, so there's not just one possible x, there's an infinite number of them. The sequence is strictly monotonic, so $p_i$ is never equal to 3 unless you take the limit. To generate the digit sequence you don't take the limit, you induce. – 2017-01-05
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    If this helps, I'm not picking a number x and trying to show that it is greater than every $p_i$, I'm defining x by the fact that it is greater than every $p_i$. Because every $p_i$ is rational, and the sequence cannot contain 3, an irrational x defined that way must exist. "For any number y<3 ...". x is not y, because there is some $p_i$ that is bigger than y. You're positing a number that doesn't meet the definition. – 2017-01-05
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    "I'm defining x by the fact that it is greater than every p_i" Then you are definining $x \ge 3$. $3$ is the *smallest* number (rational or otherwise) that is larger than all $p_i$. That is what the definition of limit means. I can define $k$ to be the smallest odd number that is divisible by two. That doesn't mean that $k$ can actually exist. – 2017-01-05
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    "Because every p_i is rational, and the sequence cannot contain 3, an irrational x defined that way must exist." NO. IT. DOES. NOT. There is no largest $p_i$. Every p_i has another p_i+1 larger than it and any number between $p_i$ and $x$ and $3$ there is a bigger $p_k$ that is between $x$ and $3$. The SMALLEST number in the ENTIRE MUFFIN FREEZING UNIVERSE the SMALLEST number larger than all $p_i$ is $3$. For any number, x, less than $3$ there will be *some* $p_n$ so that $x < p_n < 3$ – 2017-01-05
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    $p_1 < p_2 < p_3 < p _ 4 < .......... < p_{4378932789} < p_{4378932790} < .....<3$ For *any* irrational $x < 3$ there *will* be some $p_k < x < p_{k+1}$. There is *no* "space" between the last $p_i$ or between "all" $p_i$ and $3$ because there is no last $p_i$. The $p_i$s get infinitely close to $3$ there is no "space" for the x to exist between the $p_i$s and the $3$. – 2017-01-05
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    "If there is an irrational x that is in (pi,3), it must also be in all (pj,3), j < i" But there *is* some n > i so that x is *not* in (pn,3). For any $x$ I chose to be the magic x there will be an interval (x, 3) and there *will* be some $p_i$ in (x,3) and .... x < p_n. – 2017-01-05
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    "not just one possible x, there's an infinite number of them" Yes, that's the whole point. There are an infinite number of $x$s and ***NONE*** of them are larger than all $p_i$. All of them are larger the *some* $p_i$ and for every $p_i$ I will find an $x$ larger than it, but I will NEVER find an x that is larger than *all* of them. Nor will I find a $p_i$ larger than all xes. There is *no* last x and there is no last p_i. There is no last number, rational or irrational that is smaller than 3. There is only a first number that is larger than all numbers smaller than 3. That number is ... 3. – 2017-01-05
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    Notice: $2.9 < x_1 < 2.99 < x_2 < 2.99 < x_3 < p_3 < x_4 < ......< x_k < p_k < x_{k+1} < p_{k+1} < ....$ The series never ends and no $x_n$ is larger than *all* $p_i$ and no $p_n$ is greater than *all* $x_i$. – 2017-01-05
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    Okay... you seem to think that because the sequence of all the {p_i} are all each less than 3 that somehow the sequence itself is less than 3; that there is some master super element that all the p_i are less than but itself is less than 3. There is no such element. There is no maximum p_i. The p_i are all smaller than 3 but there is no number smaller than 3 that is bigger than all of them. Infinite bounded sequences don't have to have a maximum. The have a limit. The limit can be bigger than all of them. If so, the limit is the smallest number that is. – 2017-01-05
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    I already +1'ed, but I'd do it a second time if I could for "muffin freezing universe". – 2017-01-05
2

You're correct that since each $p_i$ is less than $3$, there must be some irrational $x$ so that $p_i < x < 3$. The problem is, this is true for each $p_i$ individually. So there's an irrational between $2.9$ and $3$, and there's an irrational between $2.99$ and $3$, and so on - but they need not be the same irrational. In fact, as your argument demonstrates, they cannot be the same irrational - as you've observed, any number that lies between every $p_i$ and $3$ must be $2.9999\ldots$, which is simply $3$ ("do not take the limit", note, is not a valid instruction - it's like talking about a number "$1+1$, but don't do the addition").

EDIT: (To address Jim's comment below.) The argument that there is no number greater than every $p_i$ and less than $3$ has nothing to do with decimal representations, but instead depends on the Archimedean Property of numbers - given $0 < r < s$, there is a positive integer $m$ so that $mr > s$. In particular, for any positive $r$, there is a whole number $m$ so that $mr > 1$. Since there is an $n$ such that $10^n > m$, we have that $10^nr > 1$; in particular, $r > \frac{1}{10^n}$ for some $n$. Now, suppose we have a number $x$ that is larger than every $p_i$ and smaller than $3$. $3 - x$ is a rational number $q$; $q > \frac{1}{10^n}$ for some $n$. So $x < 3 - 10^{-n}$ for some $n$. But the $p_i$ are $3 - 10^{-1}, 3 - 10^{-2}, 3 - 10^{-3}$ and so on; so $x < p_n$ for some $n$, contradicting our supposition.

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    "any number that lies between every pi and 3 must be 2.9999…, which is simply 3" OR. The number x doesn't have a decimal representation. The proof that every irrational number has a decimal representation is (I believe) inductive, and this looks like a case where that induction cannot produce a proper irrational representation. – 2017-01-05
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    @Jim My response was too long, so I added it to my answer as an edit. – 2017-01-05
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    Thanks. I think you've answered my question, but I think I can frame it differently to bypass the explanation. What's the protocol? Should I start a new thread or edit my post? – 2017-01-05