WebbThe sum of any two rational numbers is a rational number The statement is true, but the following proposed proof is incorrect. Proposed proof: 1. Suppose r and s are any rational numbers. 2. By definition of rational, r = and s = for some integers a, b, c, and b d with b = 0 and d = D. с a 3. By substitution, r + s = b d 4. WebbSuppose r and s are any rational numbers. Then r = a/b and s = c/d. for some integers a, b, c, and d with b ≠ 0 and d ≠ 0 (by definition of rational). 2. Then r + s = a/b + c/d. 3. But this is a sum of two fractions, which is a fraction. 4. So r − s is a rational number since a rational number is a fraction.

Prove that the sum of two irrational numbers can be irrational

WebbProve that the sum of any two rational numbers is rational. ! Solution: Begin by mentally or explicitly rewriting the statement to be proved in the form “∀_____, if _____ then _____.” ! … WebbAs 5 divides 5m 2, so 5 divides q 2 but 5 is prime. \Rightarrow 5 \text { divides } q \qquad \text { (Theorem 1)} ⇒ 5 divides q (Theorem 1) Thus, p and q have a common factor 5. … thai e visa sign in

Is the sum of two rationals or two irrationals irrational?

WebbAnswer. Let us assume that 5 + \sqrt {2} 5+ 2 is a rational number, say r. But this contradicts the fact that \sqrt {2} 2 is irrational. Hence, our assumption is wrong. ∴ 5 + \sqrt {2} 5+ 2 is an irrational number. Let us assume that 3 - … WebbFind the mistakes in the “proofs” that the sum of any two rational numbers is a rational number. “Proof: Any two rational numbers produce a rational number when added together. So if r and s are particular but arbitrarily chosen rational numbers, then … WebbAs 5 divides 5m 2, so 5 divides q 2 but 5 is prime. \Rightarrow 5 \text { divides } q \qquad \text { (Theorem 1)} ⇒ 5 divides q (Theorem 1) Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1). Hence, \sqrt {5} 5 is not a rational number. So, we conclude that \sqrt {5} 5 is an irrational number. thai evisa official