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(a) ( Consider the following proof with missing parts (indicated by bold, underlined digits 1. 2. 3. 4 and 5). These missing parts should be selected from the list of statement options below. Note that parts may be re-used more than once and with correct capitalisation. So, if you use option A after a full stop, it would be correctly capitalised to "Proof by contradiction" (i.e., do not worry about capitalisation aspects). Proposition: 2 is irrational 10 16

(a) ( Consider the following proof with missing parts (indicated by bold, underlined digits 1. 2. 3. 4 and 5). These missing parts should be selected from the list of statement options below. Note that parts may be re-used more than once and with correct capitalisation. So, if you use option A after a full stop, it would be correctly capitalised to "Proof by contradiction" (i.e., do not worry about capitalisation aspects). Proposition: 2 is irrational 10 16

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Q4 ( a ) Plz do exactly according to question requirements plz and take a thumb up
Question 4: (a) ( Consider the following proof with missing parts (indicated by bold, underlined digits 1. 2. 3. 4 and 5). These missing parts should be selected from the list of statement options below. Note that parts may be re-used more than once and with correct capitalisation. So, if you use option A after a full stop, it would be correctly capitalised to "Proof by contradiction" (i.e., do not worry about capitalisation aspects). Proposition: 2 is irrational Proof: 1. Suppose that 2 is 2. Therefore, 2 can be expressed as the ratio of two integers, where d# 0. Any ratio can be simplified into lowest terms. Therefore, we can assume that 3. Cubing both sides of the equation 2 gives 2. Then multiplying both sides of the equation by d³ gives 2d³ = n³. The fact that n³ is 4. Since 5, n = 2k for some integer k. Plugging in the expression 2k for n into n³: n³ = (2k)³ = 8k³. Since 2d³= n³ = 8k³, dividing by 2 gives d³= 4k³= 2(2k³). Since ď³ 6. We have established that 7. which 8. Thus, we must conclude that the assumption that 2 is a rational number is false. A. we proceed by direct proof B. proof by contradiction C. proof by contrapositive D. irrational E. rational F. there is no integer greater than 1 that divides both n and d G. there is integer greater than 1 that divides both n and d H. a multiple of 3 means that n³ is odd, and therefore n is even I. a multiple of 3 means that n³ is odd, and therefore n is odd J. a multiple of 2 means that n³ is even, and therefore n is odd K. a multiple of 2 means that n³ is even, and therefore n is even L. n is even M. nis odd N. is a multiple of 3, d³ is odd and therefore d is also odd O. is a multiple of 3, d³ is odd and therefore d is also even P. is a multiple of 2, d³ is even and therefore d is also odd Q. is a multiple of 2, d³ is even and therefore d is also even R. n and d are even S. n and d are odd T. n and d are rational U. n and d are irrational V. confirms the assumption that no integer greater than I can divide both n and d W. confirms the assumption that there is an integer greater than I can divide both n and d X. contradicts the assumption that there is an integer greater than 1 can divide both 7 and d Y. contradicts the assumption that no integer greater than I can divide both n and d Z. (nothing, we leave this part blank)
Transcribed Image Text:Question 4: (a) ( Consider the following proof with missing parts (indicated by bold, underlined digits 1. 2. 3. 4 and 5). These missing parts should be selected from the list of statement options below. Note that parts may be re-used more than once and with correct capitalisation. So, if you use option A after a full stop, it would be correctly capitalised to "Proof by contradiction" (i.e., do not worry about capitalisation aspects). Proposition: 2 is irrational Proof: 1. Suppose that 2 is 2. Therefore, 2 can be expressed as the ratio of two integers, where d# 0. Any ratio can be simplified into lowest terms. Therefore, we can assume that 3. Cubing both sides of the equation 2 gives 2. Then multiplying both sides of the equation by d³ gives 2d³ = n³. The fact that n³ is 4. Since 5, n = 2k for some integer k. Plugging in the expression 2k for n into n³: n³ = (2k)³ = 8k³. Since 2d³= n³ = 8k³, dividing by 2 gives d³= 4k³= 2(2k³). Since ď³ 6. We have established that 7. which 8. Thus, we must conclude that the assumption that 2 is a rational number is false. A. we proceed by direct proof B. proof by contradiction C. proof by contrapositive D. irrational E. rational F. there is no integer greater than 1 that divides both n and d G. there is integer greater than 1 that divides both n and d H. a multiple of 3 means that n³ is odd, and therefore n is even I. a multiple of 3 means that n³ is odd, and therefore n is odd J. a multiple of 2 means that n³ is even, and therefore n is odd K. a multiple of 2 means that n³ is even, and therefore n is even L. n is even M. nis odd N. is a multiple of 3, d³ is odd and therefore d is also odd O. is a multiple of 3, d³ is odd and therefore d is also even P. is a multiple of 2, d³ is even and therefore d is also odd Q. is a multiple of 2, d³ is even and therefore d is also even R. n and d are even S. n and d are odd T. n and d are rational U. n and d are irrational V. confirms the assumption that no integer greater than I can divide both n and d W. confirms the assumption that there is an integer greater than I can divide both n and d X. contradicts the assumption that there is an integer greater than 1 can divide both 7 and d Y. contradicts the assumption that no integer greater than I can divide both n and d Z. (nothing, we leave this part blank)
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