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2. Let A be as in the previous question. A number M is said to be an upper bound of A if a ≤ M for all a € A. Consider the set of rational upper bounds of A. That is, define B = {q: q rational number and q is an upper bound for A} (a) Show B is non-empty. (b) Show B doesn't have a smallest element. Hint: q² = 2, < 2 or > 2. Argue first two can't happen. Then follow similar idea as in previous question.

2. Let A be as in the previous question. A number M is said to be an upper bound of A if a ≤ M for all a € A. Consider the set of rational upper bounds of A. That is, define B = {q: q rational number and q is an upper bound for A} (a) Show B is non-empty. (b) Show B doesn't have a smallest element. Hint: q² = 2, < 2 or > 2. Argue first two can't happen. Then follow similar idea as in previous question.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you do number 2 please? Thanks 

1. Let A = {r: r is a rational number and r² <2}. Prove that A has no largest member. Hint. To show r E A is not the largest, we must exhibit a number in A that is greater than r. We have to find a positive, rational d such (r + 5)² < 2. The 8 needs to be described in terms of r and will use the knowledge that r² < 2. This may be useful: for 0 < 5 < 1, we have 0 < 6² <§. This may be useful: Given any two distinct rational numbers, there is another rational number between them. 2. Let A be as in the previous question. A number M is said to be an upper bound of A if a ≤ M for all a E A. Consider the set of rational upper bounds of A. That is, define B = (a) Show B is non-empty. (b) Show B doesn't have a smallest element. Hint: q2 = 2, < 2 or > 2. Argue first two can't happen. Then follow similar idea as in previous question. = {q: q rational number and q is an upper bound for A}
Transcribed Image Text:1. Let A = {r: r is a rational number and r² <2}. Prove that A has no largest member. Hint. To show r E A is not the largest, we must exhibit a number in A that is greater than r. We have to find a positive, rational d such (r + 5)² < 2. The 8 needs to be described in terms of r and will use the knowledge that r² < 2. This may be useful: for 0 < 5 < 1, we have 0 < 6² <§. This may be useful: Given any two distinct rational numbers, there is another rational number between them. 2. Let A be as in the previous question. A number M is said to be an upper bound of A if a ≤ M for all a E A. Consider the set of rational upper bounds of A. That is, define B = (a) Show B is non-empty. (b) Show B doesn't have a smallest element. Hint: q2 = 2, < 2 or > 2. Argue first two can't happen. Then follow similar idea as in previous question. = {q: q rational number and q is an upper bound for A}
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