Solution Found!
a. Use proof by contradiction to show that for any integer
Chapter 4, Problem 16E(choose chapter or problem)
a. Use proof by contradiction to show that for any integer n, it is impossible for n to equal both \(3q_{1}+r_{1}\) and \(3q_{2}+r_{2}\), where \(q_{1}, q_{2}, r_{1}\), are integers, 0 \(\leq r_{1}\) < 3, 0 \(\leq r_{2}\) < 3, and \(r_{1} \neq r_{2}\).
b. Use proof by contradiction, the quotient-remainder theorem, division into cases, and the result of part (a) to prove that for all integers n, if \(n^{2}\) is divisible by 3 then n is divisible by 3.
c. Prove that \(\sqrt{3}\) is irrational.
Text Transcription:
3q_1 + r_1
3q_2 + r_2
q_1, q_2, r_1
leq r_1
leq r_2
r_1 neq r_2
n^2
sqrt 3
Questions & Answers
QUESTION:
a. Use proof by contradiction to show that for any integer n, it is impossible for n to equal both \(3q_{1}+r_{1}\) and \(3q_{2}+r_{2}\), where \(q_{1}, q_{2}, r_{1}\), are integers, 0 \(\leq r_{1}\) < 3, 0 \(\leq r_{2}\) < 3, and \(r_{1} \neq r_{2}\).
b. Use proof by contradiction, the quotient-remainder theorem, division into cases, and the result of part (a) to prove that for all integers n, if \(n^{2}\) is divisible by 3 then n is divisible by 3.
c. Prove that \(\sqrt{3}\) is irrational.
Text Transcription:
3q_1 + r_1
3q_2 + r_2
q_1, q_2, r_1
leq r_1
leq r_2
r_1 neq r_2
n^2
sqrt 3
ANSWER:Solution:
Step 1:
(a) In this question we have to prove by contradiction that for any integer , it is impossible for n to equal both and , where, , , and , are integers, 0 ≤ <3, 0 ≤ <3, and ≠ .
(b) We have to prove by contradiction that for all integers n, if is divisible by 3 then n is divisible by 3.
(c) We have to prove that is irrational.