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 want an irrational between 7/3 and 22/9.
- A. 2.334 is a terminating decimal → rational, so not acceptable.
- B. 2.334333... has a repeating tail (eventually all 3's) → rational.
- C. 3.334333... > 3, so it is not between 2.333... and 2.444....
- D. 2.443444... lies between 2.333... and 2.444... (compare digit by digit: at the third decimal place 3 < 4, so it is < 2.444...; and it is > 2.333...). Its decimal expansion is nonterminating and nonrepeating (the pattern of 4's does not settle into a fixed repeating block), so it is irrational. Therefore D is the only choice that is an irrational between the two rationals.

Why there must always be an irrational between any two rationals a < b:
- Let r = (a + b)/2 (a rational midpoint). Take an irrational number t (e.g. √2) and set s = ((b − a)/4)·t. Since (b − a)/4 is a nonzero rational and t is irrational, s is irrational. Also 0 < s < (b − a)/2, so r − s > a and r + s < b. Thus r + s is between a and b. Finally, the sum of a rational and an irrational is always irrational, so r + s is an irrational number in (a, b). This proves there is at least one irrational between any two distinct rationals.