# If x<1 what is the maximum value of x?

1. Jul 15, 2013

### I_am_learning

It appears the maximum value of x = 0.999... . But 0.999... = 1. But max value of x cannot be 1 based on the criteria. What is the maximum value of x then? It's puzzling me.

2. Jul 15, 2013

### hilbert2

An open interval does not have a maximum number. We can say, though, that the supremum of the set $\left\{x\in\Re|x<1\right\}$ is 1.

3. Jul 15, 2013

### Number Nine

It doesn't have one. For any a < 1, there is some b such that a < b < 1.

Last edited: Jul 15, 2013
4. Jul 15, 2013

### I_am_learning

Is it that it does not have maximum or is it that we cannot numerically represent it?
If we are asked to compare "the maximum value" and 1, we still say 1 is the larger number, won't we?
Thanks for the feedback.

5. Jul 15, 2013

### micromass

Staff Emeritus
There is no maximum value. That is, there is no real number $a$ such that $a<1$ and such that $b\leq a$ for all $b<1$.

6. Jul 15, 2013

### I_am_learning

Thank you micromass.
I now get that there isn't any number which you can claim to be the maximum, because whatever number (<1) you claim, I always got another number (<1) but greater than yours. Since there isn't any number, the comparison cannot be done. But I have this gut feeling that even if we can't have any number as 'the maximum', we can still compare and say, 'the maximum' is less than 1.

If there is no maximum for such open intervals, is someone 'correct' to ask this question? If yes what is your answer?

Provided |x - 3| < 5
Quantity A: Least value of x
Quantity B: -2
Compare Quanity A and B and mention which of them is greater or if they are equal or if the relation cannot be determined.

P.S.: You mentioned 'real'. Is there a complex solution?

7. Jul 16, 2013

### micromass

Staff Emeritus
No, the question doesn't make any sense, since there is no maximum value.
In your second example, the "Quantity A" does not exist, so we can't compare it with -2.

No, there is no complex solution.
However, I would think that it is possible to extend the real numbers. So we add new elements to $\mathbb{R}$. Then we get some kind of new number system in which a solution does exist. This is in principle possible. But I know of nobody that has studied this.