Sure! Let's tackle the problem step-by-step:
Given: Two rational numbers [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that [tex]\( a < b \)[/tex].
We want to prove that:
[tex]\[ a < \frac{a+b}{2} < b \][/tex]
### Step 1: Understand the expression [tex]\(\frac{a+b}{2}\)[/tex]
The expression [tex]\(\frac{a+b}{2}\)[/tex] represents the arithmetic mean (average) of the two numbers [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
### Step 2: Prove the first part [tex]\( a < \frac{a+b}{2} \)[/tex]
1. Starting with [tex]\( a < b \)[/tex]:
[tex]\[ a < b \][/tex]
2. Add [tex]\( a \)[/tex] to both sides of the inequality:
[tex]\[ a + a < a + b \][/tex]
[tex]\[ 2a < a + b \][/tex]
3. Divide both sides by 2 to isolate [tex]\(\frac{a+b}{2}\)[/tex]:
[tex]\[ \frac{2a}{2} < \frac{a+b}{2} \][/tex]
[tex]\[ a < \frac{a+b}{2} \][/tex]
### Step 3: Prove the second part [tex]\(\frac{a+b}{2} < b\)[/tex]
1. Starting with [tex]\( a < b \)[/tex]:
[tex]\[ a < b \][/tex]
2. Add [tex]\( b \)[/tex] to both sides of the inequality:
[tex]\[ a + b < b + b \][/tex]
[tex]\[ a + b < 2b \][/tex]
3. Divide both sides by 2 to isolate [tex]\(\frac{a+b}{2}\)[/tex]:
[tex]\[ \frac{a+b}{2} < \frac{2b}{2} \][/tex]
[tex]\[ \frac{a+b}{2} < b \][/tex]
### Conclusion
We have shown that:
[tex]\[ a < \frac{a+b}{2} < b \][/tex]
Thus, the inequality [tex]\( a < \frac{a+b}{2} < b \)[/tex] holds true for any two rational numbers [tex]\( a \)[/tex] and [tex]\( b \)[/tex], where [tex]\( a < b \)[/tex].
