Question

Reasoning  The numbers 3.333 . . . and 3.444 . . . are both rational numbers. Which of the following is an irrational number that is between these two rational​ numbers? Use pencil and paper. Show why there has to be an irrational number between any two rational numbers.
4.334333 . . .
3.443444 . . .
3.334
3.334333 . . .
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Part 1
Which of the following is an irrational number that is between the two rational​ numbers? Select all that apply.
A.
3.443444 . . .
B.3.334
3.334
C.
3.334333 . . .
D.
4.334333 . . .

Answer 1

First compare each choice with the interval (3.333..., 3.444...).

- 3.333... = 3.\overline{3} and 3.444... = 3.\overline{4}.
- A: 3.443444... is ≥ 3.443... and < 3.444..., so it lies between 3.333... and 3.444....
- B: 3.334 = 3.334000... lies between the two, but is a terminating decimal (hence rational).
- C: 3.334333... has a repeating block (eventually repeating 3's), so it is rational.
- D: 4.334333... > 4, so it is outside the interval.

Thus the only number from the list that lies between 3.333... and 3.444... and is irrational is A, 3.443444..., because its decimal expansion is nonterminating and nonrepeating (it does not settle into a repeating block), so it is irrational.

Why there must be an irrational between any two rationals a < b:
Take x = a + (b − a)/√2. Since √2 > 1 we have 0 < (b − a)/√2 < b − a, so a < x < b. The number (b − a)/√2 is irrational (a nonzero rational divided by √2 is irrational), and a is rational, so x (rational + irrational) is irrational. Hence there is always at least one irrational number between any two distinct rationals.