Question 22 of 58intermediate💡 UnderstandShort Answer2 marks

Try to explain why the average of two rational numbers aa and bb, which equals (a+b)2\frac{(a + b)}{2}, is always a rational number between aa and bb.

Correct Answer

The average of two rational numbers aa and bb is a+b2\frac{a+b}{2}. Since aa and bb are rational, their sum a+ba+b is also a rational number by the rules of addition of rational numbers. Dividing this sum by 2 (a non-zero rational number) using the rules of division yields another rational number, a+b2\frac{a+b}{2}.

Furthermore, this average always lies between aa and bb. If a<ba < b, adding aa to both sides gives 2a<a+b2a < a+b, which simplifies to a<a+b2a < \frac{a+b}{2}. Similarly, adding bb to both sides of a<ba < b gives a+b<2ba+b < 2b, which simplifies to a+b2<b\frac{a+b}{2} < b.

Therefore, a+b2\frac{a+b}{2} is always a rational number strictly between aa and bb, illustrating the density of rational numbers.

Exercise: 3.4.2 The Density of Rational Numbers | Q: 1 | (Page 12)
For More Understanding

Explanation

The textbook explains that rational numbers are dense, meaning between any two rational numbers, there is always another. By taking the average, we perform addition (which results in a rational number) and division by a non-zero rational number (which also results in a rational number). Algebraically, the average must fall between the two numbers because it is equidistant from both on the number line.

Solution Steps

  1. Step 1: State that the sum a+b is rational because rational numbers are closed under addition.

  2. Step 2: State that dividing (a+b) by 2 gives a rational number because rational numbers are closed under division by non-zero rational numbers.

  3. Step 3: Show algebraically that if a < b, then a < (a+b)/2 < b, proving it lies between a and b.