Question

Reasoning  The numbers 2.333 . . . and 2.444 . . . 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.
3.334333 . . .
2.443444 . . .
2.334
2.334333 . . .
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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.2.334
2.334
B.
2.334333 . . .
C.
3.334333 . . .
D.
2.443444 . .

Answer 1

Answer: D only.

Reasoning:
- 2.333... = 7/3 and 2.444... = 22/9, so we are looking for an irrational number strictly between 7/3 and 22/9 (≈ 2.3333… and 2.4444…).

Check the choices:
- A. 2.334 is a terminating decimal → rational. It lies between the two rationals but is not irrational.
- B. 2.334333... has an eventually repeating decimal expansion (after some place the digit 3 repeats) → rational.
- C. 3.334333... > 3, so it is greater than 2.444... and not between the two numbers.
- D. 2.443444... is less than 2.444... (its third decimal digit is 3 while 2.444... has 4 there) and greater than 2.333...; its decimal expansion is not eventually periodic (it is not a repeating decimal), so it is irrational. Thus D is an irrational number between the two given rationals.

Why there must always be an irrational between any two rationals:
Let p and q be rationals with p < q. Take r = p + (q − p)/√2. Since (q − p) > 0 and √2 is irrational, (q − p)/√2 is irrational; adding the rational p gives r irrational. Also 0 < (q − p)/√2 < q − p (because √2 > 1), so p < r < q. Hence between any two rationals there exists an irrational.