Problem 7 - A Primer of Real Analysis

Question

Show that if \(a, b \in \mathbb{Q}\) with \(a

Step-by-Step Solution

Answer

The inequality is proven: \(a < \frac{a+b}{2} < b\).

Step 1: Define Rational Numbers

We start by considering that both \(a\) and \(b\) are rational numbers. This means that \(a\) and \(b\) can be expressed as fractions of integers, i.e., \(a = \frac{m}{n}\) and \(b = \frac{p}{q}\), where \(m, p, n, q\) are integers and \(n, q eq 0\).

Step 2: Calculate the Midpoint

Calculate the midpoint of \(a\) and \(b\) using the formula \(\frac{a+b}{2}\). This will give us \(\frac{a+b}{2} = \frac{\frac{m}{n} + \frac{p}{q}}{2} = \frac{mq + pn}{2nq}\).

Step 3: Compare a to Midpoint

We need to show that \(a < \frac{a+b}{2}\). So our inequality becomes \(a < \frac{mq + pn}{2nq}\). Rewriting \(a\) in terms of the least common denominator gives us \(\frac{2amq}{2nq}\). So the inequality now is \(\frac{2amq}{2nq} < \frac{mq + pn}{2nq}\). This simplifies to \(2amq < mq + pn\).

Step 4: Prove a < Midpoint

We factored \(2amq < mq + pn\) into \(2nm < nq + mp\) (after multiplying everything by the common denominators and simplifying). Given that \(a < b\equiv \frac{m}{n} < \frac{p}{q}\equiv mq < np\), adding \(mq\) to both sides in the expression, we get \(2mq < mq + np\) which holds as \(a < b\). Hence, \(a < \frac{a+b}{2}\) is proven.

Step 5: Compare Midpoint to b

Now we need to show \(\frac{a+b}{2} < b\). Our inequality becomes \(\frac{mq + pn}{2nq} < \frac{p}{q}\), which can be rewritten as \(mq + pn < 2pq\) (by multiplying both sides by \(2nq\)).

Step 6: Prove Midpoint < b

We start from \(mq + pn < 2pq\) and rearrange it as \(mq < 2pq - pn\). Given that \(a < b\equiv mq < np\), this inequality holds since \(np < 2pq\) by subtracting \(pn\) from both sides until \(mq < pq\). This proves that \(\frac{a+b}{2} < b\).

Step 7: Conclusion

Since we have shown both \(a < \frac{a+b}{2}\) and \(\frac{a+b}{2} < b\), the overall inequality \(a < \frac{a+b}{2} < b\) must hold true for all rational numbers \(a\) and \(b\) where \(a < b\).

Key Concepts

Inequality Proof

When tackling an inequality proof involving rational numbers, the main goal is to show that the inequalities reliably hold for any numbers that fit the specified conditions. In our problem, we're given that two rational numbers, \(a\) and \(b\), satisfy \(a < b\). We need to prove that the midpoint of these numbers, \(\frac{a+b}{2}\), lies between \(a\) and \(b\).

• First, remember that rational numbers can be expressed as fractions. Therefore, both \(a\) and \(b\) have the form \(\frac{m}{n}\) and \(\frac{p}{q}\) respectively.
• For the proof, we break the inequality into two parts: \(a < \frac{a+b}{2}\) and \(\frac{a+b}{2} < b\).

To prove the first part, \(a < \frac{a+b}{2}\), we start by expressing it as \(\frac{2amq}{2nq} < \frac{mq + pn}{2nq}\). Simplifying this inequality leads us to \(2amq < mq + pn\), which holds true under the condition \(a < b\).
Similarly, the second part, \(\frac{a+b}{2} < b\), simplifies to \(mq + pn < 2pq\). Rearranging gives \(mq < 2pq - pn\), which also holds as \(a < b\).
Thus, the entire chain of inequalities is established, confirming that \(a < \frac{a+b}{2} < b\).

Midpoint Calculation

Calculating the midpoint of two numbers involves averaging them. This operation is crucial when discussing inequalities between two fractions or numbers. Begin with two rational numbers \(a\) and \(b\), represented as \(\frac{m}{n}\) and \(\frac{p}{q}\). The formula to find their midpoint is \(\frac{a+b}{2}\).

• The calculation for \(\frac{a+b}{2}\) when \(a\) and \(b\) are fractions involves first ensuring a common denominator, thus rewriting \(a = \frac{m}{n}\) and \(b = \frac{p}{q}\) as \(\frac{mq}{nq}\) and \(\frac{pn}{nq}\) respectively.
• Adding these gives \(\frac{mq + pn}{nq}\), and dividing by 2 results in the expression \(\frac{mq + pn}{2nq}\).

This representation of the midpoint not only confirms a balanced point between \(a\) and \(b\) on the number line but also simplifies further inequality proofs by providing a standardized form that can be used in both parts of the inequality \(a < \frac{a+b}{2} < b\).
The midpoint is a significant concept because it offers a way to approach and validate inequalities, showing how they relate to equality at the very center or midpoint.

Fraction Manipulation

Handling fractions involves understanding how to work with the numerators and denominators effectively. In inequality proofs, the ability to accurately manipulate fractions and their terms ensures correct conclusions.

• Firstly, ensure the fractions you are working with have a common denominator. This simplifies additions, subtractions, and comparisons.
• In our example, adding \(a + b\) involves rewriting \(a = \frac{m}{n}\) and \(b = \frac{p}{q}\) as \(\frac{mq}{nq}\) and \(\frac{pn}{nq}\) respectively, giving us a uniform base of \(2nq\) in the expression \(\frac{mq + pn}{2nq}\).
• Fraction manipulation also includes multiplying all terms by the common denominator to clear fractions for easier inequality manipulation. This was shown in expressions like \(2amq < mq + pn\) and \(mq + pn < 2pq\).

Understanding these techniques ensures the consistency of mathematical reasoning, emphasizing that accurate fraction manipulation is vital for establishing proofs involving inequalities in rational numbers.
Fraction manipulation not only aids in simplifying calculations but also supports logical flow in mathematical arguments, highlighting the power of precise algebraic handling.