Question

You’re the grader. To each “Proof”, assign one of the following grades:

• A (correct), if the claim and proof are correct, even if the proof is not the simplest, or the proof you would have given.

• C (partially correct), if the claim is correct and the proof is largely a correct claim, but contains one or two incorrect statements or justications.

• F (failure), if the claim is incorrect, the main idea of the proof is incorrect, or most of the statements in the proof are incorrect.

For grades of C or F, please give a brief justication.

(P1) Claim. If a divides both b and c, then a divides b + c.

“Proof.” Assume that a does not divide b+c. Then there is no integer k such that ak = b+c. However, a divides b, so am = b for some integer m; and a divides c, so an = c for some integer n. Thus, am+an = a(m+n) = b+c. Therefore, k = m+n is an integer satisfying ak = b+c. Then the assumption that a does not divide b+c is false, and a does divide b + c.

(P2) Claim. Let t be a real number. If t is irrational, then 5t is irrational.

“Proof.” Suppose 5t is rational. Then 5t = p/q for integers p and q with q not equal to 0. Therefore, t = p/(5q), where p and 5q are integers and 5q not equal to 0, so t is rational. It follows that if t is irrational, then 5t is irrational.

(P3) Claim. Let x and y be integers. If x and y are even, then x + y is even.

“Proof.” Suppose x and y are even but x + y is odd. Then, for some integer k, x + y = 2k + 1. Therefore, x + y + (−2)k = 1. The left side of the equation is even because the sum of even numbers is even. However, the right side, 1, is odd. Since an odd cannot equal an even, we have a contradiction. Therefore, x+y is even.

(P4) Claim. If x is a positive real number, then the sum of x and its reciprocal is at least 2, i.e., x + 1 x ≥ 2.

“Proof.” Multiplying by x, we get x2 +1 ≥ 2x. By algebra, x2−2x+1 ≥ 0. Thus, (x−1)2 ≥ 0. Any real number squared is greater than or equal to 0, so x + 1 x ≥ 2is true.

(P5) Claim. If A ⊆ B and B ⊆ C, then A ⊆ C.

“Proof.” Suppose A ⊆ B and B ⊆ C. Then x ∈ A and x ∈ B because A ⊆ B. Then x ∈ B and x ∈ C because B ⊆ C. Thus x ∈ A and x ∈ C, so A ⊆ C.

(P6) Claim. If A ⊆ B and B ⊆ C, then A ⊆ C.

“Proof.” Suppose x is any object. If x ∈ A, then x ∈ B because A ⊆ B. But then x ∈ C because B ⊆ C. Therefore, A ⊆ C.

(P7) Claim. If A ⊆ B and B is not⊆ C, then A is not⊆ C.

“Proof.” Suppose A ⊆ B and B is not⊆ C. Then there exists x ∈ B such that x does not ∈ C. Since x ∈ B, x ∈ A by denition of subset. Then x ∈ A and x does not ∈ C. Therefore, A is not⊆ C.

Answer 1

ClaimP1: this is a correct proof(A).

ClaimP2: this is a correct proof(A).

ClaimP3:this is a partially correct (C) .

let x, y be two even integers s. t x+y is odd. Then there is a K s. t x+y=2k+1.

Now for even integers x, y we take x=2a, y =2b for some integers a, b. Then x+y=2k+1 gives us 2(a+b)=2k+1, which is absurd. Hence x+y is even . (proof).

ClaimP4: this is a correct proof (A).

ClaimP5: this is a correct proof(A).

ClaimP6: this is a correct proof(A).

ClaimP7:correct proof (A).