Problem 48 Explain why every open interval ... [FREE SOLUTION]

91影视

Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 0: Problem 48

Explain why every open interval containing 0 contains an open interval centered at $0 .$

Short Answer

Expert verified

For any open interval (a, b) containing 0, with a < 0 and b > 0, we can find two positive numbers m and n such that m = |a| and n = b. Let p be the minimum of m and n, i.e., $p = \min(m, n)$. Since -p < 0 < p, the open interval (-p, p) is contained within (a, b) and is centered at 0, as $\frac{-p + p}{2} = 0$. This demonstrates that every open interval containing 0 contains an open interval centered at 0.

Step by step solution

Define an open interval

An open interval is a set of all points between two real numbers, not including the two endpoints. We can represent an open interval as (a, b), where a and b are the endpoints of the interval, and any value x that lies in the interval follows the inequality a < x < b.

Understanding the problem

We have an open interval containing 0, let's call it (a, b), with a < 0 and b > 0. We need to prove that there exists an open interval centered at 0 that is entirely contained within this given interval (a, b).

Finding the open interval centered at 0

To find an open interval centered at 0, we need to identify two points, let's call them c and d, such that 0 is the center (midpoint) of this new interval, and both c and d are contained within the given interval (a, b). In formal terms, we need to find (c, d) 鈯� (a, b) and (c+d)/2 = 0.

Proving the existence of the open interval centered at 0

Since a < 0 and b > 0, there must be two positive numbers, let's call them m and n, such that m = |a| and n = b. Now, let's take the minimum of m and n. Call it p: \[ p = \min(m, n) \] By taking p as the minimum of m and n, we guarantee that -p < 0 < p, which means that (-p, p) 鈯� (a, b). Moreover, as a simple calculation shows, the midpoint of (-p, p) is (0): \[ \frac{-p + p}{2} = \frac{0}{2} = 0 \] Therefore, the open interval (-p, p) is centered at 0 and is entirely contained within the given interval (a, b).

Conclusion

We have shown that for any open interval (a, b) containing 0, there exists an open interval centered at 0, namely (-p, p), that is entirely contained within (a, b). This proves that every open interval containing 0 contains an open interval centered at 0.

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91影视

Explain why every open interval containing 0 contains an open interval centered at \(0 .\)

Short Answer

Step by step solution

Define an open interval

Understanding the problem

Finding the open interval centered at 0

Proving the existence of the open interval centered at 0

Conclusion

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