Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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].
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].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.