Strictly increasing function?

Question

Let f be a strictly increasing function on the interval [0,1]. Can the range of f ever consist of irrational numbers only?

whitesox08,
what about all the numbers between 0 and 1?

how is sqrt(x) irrational for all x in (0,1)?
ex. sqrt(4/9)=2/3

This question is still open. Any help is appreciated.

I'm sorry, but I never assumed f to be continuous. In the continuous case, you are right, but what if f is not continuous?

Okay, in your proof of when f is not continuous, you only defined f(x) for x=0 or 1. What is f(1/2) or f(0.314159...)? Understand that f must be defined for ALL x in [0,1].

whitesox08,
look below you.

There are many counterexamples, let
f(x)=0 for x<1
f(x)=1 x=1
is f continuous?

Maybe you mean that a strictly increasing function f on [0,1] must be continuous at some subinterval of [0,1]?

Okay, your previous statement was
"If f is defined for all x in [0,1] then f is continuous. That is precisely the definition of continuity on an interval"
Can you prove this statement? I don't see how its true with my previous example.

"You can have a completely discontinuous, strictly increasing function on [0,1]"
Isn't this contradicting your previous statement?

btw, I have not taken real analysis. I'm still in high school.

?? I'm sorry but I still don't understand how your proof (when did you prove it anyway?) works. If you believe you are right, then you don't have to keep arguing with me, I will ask somebody else. Thanks for helping anyway.

By strictly increasing, I mean that x>y implies f(x)>f(y).
If it implied f(x)>=f(y), I would say monotone, or nondecreasing.

Answer 1

Such a function does exist. I will describe here how to construct one such function. (The description will be fairly complicated, but this should not be unexpected, as it is certainly a rather bizarre function that is strictly increasing but never assumes a rational value.)

First I'll describe the function, and then I'll show that it has the two properties you want (strictly increasing and never rational).

(I will be more verbose than usual since you have stated that you have not had an analysis course. I apologize in advance to other interested readers who are familiar with more advanced notation or more formal arguments.)

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I'll start by defining f(x) just for numbers *between* 0 and 1 (not 0 and 1 themselves), and then show at the end how I can extend to these numbers.

First, make a list of all the rational numbers between 0 and 1, in the following way:

1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6, 3/6, 4/6, 5/6, ...

(First list the fractions with denominator 2; then those with denominator 3, in order; and so on. Some numbers are listed more than once--for example, 1/2 and 2/4 are listed separately--but this will not matter.)

Call the nth fraction in the list r_n. (So r_1 = 1/2; r_2 = 1/3; and so on.)

Now, I will define f(x) for when x is between 0 and 1. I will define the value of f(x) to always be between 0 and 1, and I'll define each of its digits after the decimal point separately, as follows:

If x > r_n, then the nth decimal place of f(x) should be 1.
Otherwise, the nth decimal place of f(x) should be 0.

As an example, let's find the first few decimal places of f(1/2).

The first decimal place should be 0 (since 1/2 is not greater than 1/2, the first fraction in the list above).

The second decimal place should be 1 (since 1/2 is greater than 1/3, the second fraction).

The third decimal place should be 0 (since 1/2 is not greater than 2/3, the third fraction).

The fourth decimal place should be 1 (since 1/2 is greater than 1/4).

The fifth decimal place should be 0 (since 1/2 is not greater than 2/4).

...and so on.

So f(1/2) starts out as 0.01010...

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I will show that f is a strictly increasing function, by showing that if a > b, then f(a) > f(b).

So suppose a > b.

I'll show that each decimal place in f(a) will be greater than or equal to the corresponding decimal place in f(b).

If a and b are both greater than one of the fractions in the list, both corresponding decimal places will be 1; if a and b are both less than one of the fractions in the list, both corresponding decimal places will be 0; and if a is greater than the fraction but b is less than it, then the corresponding decimal place will be 1 in f(a) but 0 in f(b). (It can't happen that a is larger than the fraction but b isn't, because we know a > b.)

Since each decimal place in f(a) is at least as large as each decimal place in f(b), then f(a) >= f(b); but they can't be equal, because there is *some* rational number in between a and b. So they differ at some decimal place, and f(a) must be larger then f(b) at that decimal place.

So f(a) > f(b).

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Now I'll show that f(x) is irrational for every x between 0 and 1.

Recall that a number is rational just when its decimal expansion is either terminating or repeating. I'll show that the decimal expansion of f(x) can be neither terminating nor repeating.

First, notice that f(x) has infinitely many 0's in it, because x is certainly less than *some* fraction, and if it's greater than, say, 1/10, it's certainly going to be greater than 1/11, 1/12, and so on, which occur later in the list of fractions.

Next, notice that f(x) has strings of 1's as long as you like it in. (It has a string of at least ten zeros somewhere; a string of at least a hundred zeros somewhere; and so on.) This is true because x is bigger than *some* fraction. Say x is bigger than 1/100. Then when we get to the part in the sequence of fractions that goes 1/1000, 2/1000, ..., 9/1000, we have a string of nine 1's in the decimal expansion of f(x). When we get to the part that goes 1/10000, 2/10000, ..., 99/10000, we get a sequence of ninety-nine 1's. And so on.

Since f(x) has infinitely many zeros, but arbitrarily long strings of 1's in it, there's no way it can be repeating. (Can it start repeating after 10,000 decimal places? No, because we haven't found our string of 10,001 1's in a row yet!)

So f(x) is always irrational.

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Thus, f is a function with the properties you seek. So far it's only defined for numbers *between* 0 and 1, but we can easily extend it by defining

f(0) = -square root of 2,
f(1) = square root of 2.

Then the function is still strictly increasing, because f(0) is the smallest value in the range and f(1) is the biggest value, and its range is still entirely irrational numbers.

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(A postscript note to those familiar with analysis: The function I've described is strictly increasing, so it has at most countably many discontinuities. I'm not sure where these are, but my hunch is that it's probably discontinuous at every rational number and continuous at every irrational number. Thoughts?)

Answer 2

Some comments:

whitesox08: no, defined on an interval has nothing to do with continuity. A function can be defined for all x on [0, 1] while being discontinuous at every point in [0, 1]. (Classic example: f(x) = 0 if x is rational, 1 if x is irrational.)

You are obviously correct in that if the function is continuous, it cannot always be irrational. But that is so trivial that it should be obvious that a non-continuous function is intended - particularly since the question contains no mention of continuity.

Scythian: the "strictly" part of "strictly increasing" means that if a ≠ b then f(a) ≠ f(b). And if you don't define f on the rationals it is not defined on [0, 1] as specified. Also, there is no "next irrational number larger than [a given rational]". And personally I don't believe in magic numbers that are less than 1 but arbitrarily close to it. ;-)

Still, the cardinality of the irrational numbers in any interval (say a closed interval with irrational endpoints, so we know the min and max values) matches the cardinality of [0, 1], so there is a bijection between them. It seems that we should be able to rearrange the elements of this to give us an order-preserving function, which would be the required answer.
[In fact this would work for any uncountable set A of irrationals - I chose the "all irrationals in an interval" example because I thought it might make things easier, but it doesn't appear to.]

I originally thought we might be able to get this assuming a well-ordering on the reals, but with further thought I don't see a way to do it. And thinking about the deeper problem I'm beginning to suspect that it may be impossible. I was going to try to explain why, but it's after 5pm and my brain is starting to hurt and I want to go home, so I'm not going to. Sorry, this answer is not up to my usual standards, but there you go.

Answer 3

I've read the posts between you and whitesox08 with interest, and it seems to me that the whole problem has to do with the definition of "strictly increasing function". Do you mean that if f(a) = f(b), then a = b necessarily? And are you seeking an answer expressible in terms of elementary functions? After all, a boneheaded "function" would be a staircase function of all rational values, plus some irrational number like pi. Then it's irrationial everywhere in the interval (0,1), and would be "strictly increasing", unless you say that's not what you mean by strictly increasing

We could construct a function f(x) such that it has indeterminate values whenever x is a rational number, so that abstractly speaking it has a lot of gaps in it, even though they're infinitely small and infinitely few compared to the irrationals, but, again, would that fail to meet your criteria for a "strictly increasing function"? I don't know.

So, finally, we can imagine a function (don't ask me how it may be expressed in mathematical terms!) whereby for every rational value a function f(x) has, it is replaced by the next irrational number larger than it, and the entire rest of the function is shifted to the left. I think this could be of great interest in set theory, but, again, I'm not sure if that's what you're interested in. You just want a expressible function of some kind? Let me think about it, I'm not so sure if it's true it's really impossible for such a function to exist.

Addendum: Okay, I'm going to throw this out there. Let D(x) be the Dirichlet function. Consider the straight line y = x, so that

x D(x) = x whenever x is rational
x (1 - D(x)) = x whenever x is irrational

Let's propose the existence of a magic irrational factor k infinitely close to 1 but that k < 1, so that k x is an irrational number for any rational x, but infinitely close to it. Then we can construct the function you want:

y = x (1 - (1-k) D(x))

which I believe meets all of your criteria. I'll leave it as an exercise to the reader whether or not k can exist.

Addendum: In response to newer answers, this problem seems to have 2 levels to it, the first being whether or not such a function can even exist abstractly, the second being how can such a function be expressed in terms of mathematical functions. As of right now, my GUESS is that 1) is probably possible, I can't see why it can't be done, if it doesn't have to be continuous, and 2) I have no idea. Anyone?

Answer 4

Well since you didn't define f to be continuous, of course it can.

Let
f = sqrt(2) for x = 0
and sqrt(3) for x = 1

Now if f has to be continuous, the answer is no because between any two irrational numbers lies a rational number.


EDIT: Nope Chris. sqrt(1/4) = 1/2 which is rational.

EDIT: Ok here's a rigorous proof.
Let f(x) be a continuous, strictly increasing function on the interval [0,1].
Let f(0) = a and f(1) = b where a and b are both irrational. Since the function is strictly increasing, a < b. Also, since there are an infinite number of rationals between any two irrationals, there exists some rational c such that a < c < b (ie f(0) < c < f(1)). Since f is continuous and c is between f(0) and f(1), there exists some x* such that f(x*) = c. Therefore the range of the function does not consist of irrational numbers only.

EDIT: Ok I erased all my edits and starred the question so that someone else can show you that my proof is accurate. I'm not trying to argue with you. I'm trying to teach you. You are arguing that I am wrong. Good luck.

EDIT: Ok Moshi I get it now. I think I know what you MEANT to ask now: Is it possible for there to be a strictly increasing function on the interval [0,1] that maps every real number on that interval to an irrational. Dr. D gave the correct answer to this.
PLEASE know what you're asking next time you ask it. I wasted a lot of time pondering the wrong question. I should add, however, that the question you intended on asking is a good one.

Answer 5

In this example you gave in the comments:
f(x)=0 for x<1
f(x)=1 x=1
is f continuous?

f(x) is not strictly increasing.

I understand strictly increasing to mean that for a,b in [0,1]
if f(a) = f(b), then a = b.

So clearly f(x) cannot be continuous otherwise it will include a rational number, because as whitesox said, there is a rational number between any two irrationals. f(x) cannot even have continuous segments for the same reason.

The only way this would work is if f(x) consists of an infinite series of disconnected points. Once you connect any two of those points, f(x) will be rational at some point. Certainly the function will have no simple definition.

Answer 6

A similar construction to The Mathemagician. Maybe, somewhat simpler.

Rational numbers are countable, and we can enumerate all rationals q in the interval [0,1], as q_1, q_2, q_3, ..etc. Consider another sequence of rational numbers:

r_1 = 0.1, (1 at 1st position)
r_2 = 0.0001, (1 at 4th position)
r_3 = 0.000000001, (1 at 9th position)
..................................
Digit 1 in r_n appears at position n^2. This 10^(-n^2). You can take another sequence of this type. What you need is that the length of r_n increases faster than linearly.
Introduce the function R:

R (q_i) = r_i.

For x=0, take f(0) to be any negative, irrational number, i.e. f(0) = -π. For x>0, define the function f(x) as:

f(x) is the sum of R(q_i)=r_i for all those q_i, which are less than x.

Clearly, f(x) > 0 for x>0. If x2 > x1> 0, then f(x2) = f(x1) + { sum of R(q_j) for q_j in the interval [x1, x2) } . The term {...} is strictly positive, because there are rational numbers between any two distinct numbers, and all R(q_j) are positive. Hence, f(x) is strictly increasing.

The function f(x) for x>0 always contains infinitely many terms R(q_i), and hence it is represented by a decimal fraction of the type 0. 0...1...0001...000000001..., which never finishes. Suppose, that this fraction is periodic, the period is P and it starts after digit N. Take a term R(q_M) such as M is larger than N. Existence of the period P means that we have digit 1 at positions M^2 and M^2 + N. This presumes that M^2 + N is a square number. However, this is impossible, because for any K>M: K^2-M^2 >= (M+1)^2 - M^2 = 2M+1 > N.

Consequently, the function f(x) for x>0 is represented by an infinite, non-periodic decimal fraction, so that f(x) is irrational.

Answer 7

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