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. 4and 5). These missing parts should be selected from the list of statement options below. Note that partsmay be re-used more than once and with correct capitalisation. So, if you use option A after a full stop, itwould be correctly capitalised to "Proof by contradiction" (i.e., do not worry about capitalisation aspects).Proposition: 2 is irrationalProof: 1. Suppose that 2 is 2. Therefore, 2 can be expressed as the ratio of two integers, whered# 0. Any ratio can be simplified into lowest terms. Therefore, we can assume that 3. Cubing both sidesof 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 haveestablished that 7. which 8. Thus, we must conclude that the assumption that 2 is a rational number isfalse.A. we proceed by direct proofB. proof by contradictionC. proof by contrapositiveD. irrationalE. rationalF. there is no integer greater than 1 that divides both n and dG. there is integer greater than 1 that divides both n and dH. a multiple of 3 means that n³ is odd, and therefore n is evenI. a multiple of 3 means that n³ is odd, and therefore n is oddJ. a multiple of 2 means that n³ is even, and therefore n is oddK. a multiple of 2 means that n³ is even, and therefore n is evenL. n is evenM. nis oddN. is a multiple of 3, d³ is odd and therefore d is also oddO. is a multiple of 3, d³ is odd and therefore d is also evenP. is a multiple of 2, d³ is even and therefore d is also oddQ. is a multiple of 2, d³ is even and therefore d is also evenR. n and d are evenS. n and d are oddT. n and d are rationalU. n and d are irrationalV. confirms the assumption that no integer greater than I can divide both n and dW. confirms the assumption that there is an integer greater than I can divide both n and dX. contradicts the assumption that there is an integer greater than 1 can divide both 7 and dY. contradicts the assumption that no integer greater than I can divide both n and dZ. (nothing, we leave this part blank)

Question

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. 4and 5). These missing parts should be selected from the list of statement options below. Note that partsmay be re-used more than once and with correct capitalisation. So, if you use option A after a full stop, itwould be correctly capitalised to &#34;Proof by contradiction&#34; (i.e., do not worry about capitalisation aspects).Proposition: 2 is irrationalProof: 1. Suppose that 2 is 2. Therefore, 2 can be expressed as the ratio of two integers, whered# 0. Any ratio can be simplified into lowest terms. Therefore, we can assume that 3. Cubing both sidesof 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 haveestablished that 7. which 8. Thus, we must conclude that the assumption that 2 is a rational number isfalse.A. we proceed by direct proofB. proof by contradictionC. proof by contrapositiveD. irrationalE. rationalF. there is no integer greater than 1 that divides both n and dG. there is integer greater than 1 that divides both n and dH. a multiple of 3 means that n³ is odd, and therefore n is evenI. a multiple of 3 means that n³ is odd, and therefore n is oddJ. a multiple of 2 means that n³ is even, and therefore n is oddK. a multiple of 2 means that n³ is even, and therefore n is evenL. n is evenM. nis oddN. is a multiple of 3, d³ is odd and therefore d is also oddO. is a multiple of 3, d³ is odd and therefore d is also evenP. is a multiple of 2, d³ is even and therefore d is also oddQ. is a multiple of 2, d³ is even and therefore d is also evenR. n and d are evenS. n and d are oddT. n and d are rationalU. n and d are irrationalV. confirms the assumption that no integer greater than I can divide both n and dW. confirms the assumption that there is an integer greater than I can divide both n and dX. contradicts the assumption that there is an integer greater than 1 can divide both 7 and dY. contradicts the assumption that no integer greater than I can divide both n and dZ. (nothing, we leave this part blank)

Accepted Answer

We need to complete the proof to show 23 is irrational.