Home / Expert Answers / Advanced Math / use-the-quotient-remainder-theorem-with-divisor-equal-to-2-to-prove-the-following-statement-the-s-pa287

(Solved): Use the quotient-remainder theorem with divisor equal to 2 to prove the following statement. The s ...



Use the quotient-remainder theorem with divisor equal to 2 to prove the following statement.
The square of any integer can beCase 2, n = 2q + 1 for some integer q:
Consider the sentences in the following scrambled list.
Now 4q² + 4g + 1 = 4(q² + g) +

Use the quotient-remainder theorem with divisor equal to 2 to prove the following statement. The square of any integer can be written in one of the two forms 4k or 4k+ 1 for some integer k. Proof: Suppose n is any integer. By the quotient-remainder theorem with divisor equal to 2, n = 2q or n = 2q + 1 for some ---Select--- ?q. Because n could have either one of these two forms, we must consider two separate cases. Case 1, n = 2q for some integer q: Consider the sentences in the following scrambled list. Then, by substitution, k is an integer with the property that n² = 4k. By substitution, n² = (29)² = 4g². By substitution, n² = (2g)² = 2g². By substitution, n² = 2(2g) = 4g. Nm & in Let k = Let k = q². Let k = q. Since n is an even integer and n = 2q, then q is an integer. Since q is an integer, then q2 is also an integer because it is a product of integers. Since q is an integer, then 29² is also an integer because it is a product of integers. By the assumption in Case 1, n = 2q for some integer q. To prove Case 1, select sentences from the list and put them in the correct order below. 1. --Select--- 2. --Select--- 3. --Select--- 2 4. --Select--- 5. -Select--- Case 2, n = 2q + 1 for some integer q: Consider the sentences in the following scrambled list. Now 4q² + 4g + 1 = 4(q² + g) + 1. Now 4q² + 4g + 1 = 4(q + 1) + 1. Now 49² + 4g + 1 = 4g(q² + q) + 1. Since q is an integer, then q² + q is also an integer because products and sums of integers are integers. Since q is an integer, then q(q² + q) is also an integer because products and sums of integers are integers. Since q is an integer, then q + 1 is also an integer because it is a sum of integers. Let k = q + 1. Let k = g(q² + q). Let k = q² + q. By substitution, n² = (2q + 1)² = 4q² + 4g + 1. Then, by substitution, k is an integer with the property that n² = 4k + 1. By the assumption in Case 2, n = 2q + 1 for some integer q. To prove Case 2, select sentences from the list and put them in the correct order below. 1. --Select--- 2. --Select--- 3. ---Select--- 4. --Select--- 5. --Select--- 6. --Select--- Conclusion: Since any integer n can be expressed in one of the ways shown in cases 1 and 2, we conclude that no matter what integer n is chosen, n can be written in one of the two forms specified in the given statement.


We have an Answer from Expert

View Expert Answer

Expert Answer


Kindly upvote my solutio
We have an Answer from Expert

Buy This Answer $5

Place Order

We Provide Services Across The Globe