Deductive proof and contradiction Test

{"questions":[{"content":"Which is a complete proof that the product of an even integer and an integer is even?\u003Cbr\u003E[[choice-proof-methods-test-1]]","widgets":{"choice-proof-methods-test-1":{"type":"choice","options":["Write a = 2k; then ab = 2(kb), where kb is an integer","For a = 2 and b = 3, ab = 6","Assume ab is even, so the claim holds"],"answer":[0],"choice_feedback":["This covers arbitrary inputs and uses the definition.","One example is not a universal proof.","This assumes the desired result."]}},"hints":[],"explanation":"This covers arbitrary inputs and uses the definition.","id":"proof-methods-test-1","coverage_topics":["aa-hl-deductive-proof"],"correct_feedback":"This covers arbitrary inputs and uses the definition.","incorrect_feedback":"This covers arbitrary inputs and uses the definition.","evidence_kind":"objective-concept-test"},{"content":"For an integer k, expand $(2k+1)^2$ in a form proving it is odd.\u003Cbr\u003E[[choice-proof-methods-test-2]]","widgets":{"choice-proof-methods-test-2":{"type":"choice","options":["$2(k^2+k)+1$","$4k^2+1$","$2(2k^2+2k)+1$"],"answer":[2],"choice_feedback":["This has incorrect coefficients.","The cross term 4k has been lost.","The expression is twice an integer plus 1."]}},"hints":[],"explanation":"The expression is twice an integer plus 1.","id":"proof-methods-test-2","coverage_topics":["aa-hl-deductive-proof"],"correct_feedback":"The expression is twice an integer plus 1.","incorrect_feedback":"The expression is twice an integer plus 1.","evidence_kind":"objective-concept-test"},{"content":"Give the smallest positive integer disproving “n² + n + 2 is always odd”. [[input-proof-methods-test-3]]","widgets":{"input-proof-methods-test-3":{"type":"input","answer":"1"}},"hints":[],"explanation":"At n = 1 the result is 4, which is even.","id":"proof-methods-test-3","coverage_topics":["aa-hl-deductive-proof"],"correct_feedback":"At n = 1 the result is 4, which is even.","incorrect_feedback":"At n = 1 the result is 4, which is even.","evidence_kind":"objective-concept-test"},{"content":"A student claims that if a+b is even, then both integers a,b are even. Which response is sufficient?\u003Cbr\u003E[[choice-proof-methods-test-4]]","widgets":{"choice-proof-methods-test-4":{"type":"choice","options":["a = 1 and b = 3 give an even sum but neither is even","a = 2 and b = 4 give an even sum, so the claim is proved","a = 1 and b = 2 are not both even, so the claim is false"],"answer":[0],"choice_feedback":["One allowed pair with a true premise and false conclusion refutes the implication.","That is a supporting example; it cannot establish all possible pairs.","The premise fails because the sum is odd. A counterexample must satisfy the premise."]}},"hints":[],"explanation":"One allowed pair with a true premise and false conclusion refutes the implication.","id":"proof-methods-test-4","coverage_topics":["aa-hl-deductive-proof"],"correct_feedback":"One allowed pair with a true premise and false conclusion refutes the implication.","incorrect_feedback":"One allowed pair with a true premise and false conclusion refutes the implication.","evidence_kind":"objective-concept-test"},{"content":"Assume there is a smallest rational number r greater than 2. Which value contradicts its minimality?\u003Cbr\u003E[[choice-proof-methods-test-5]]","widgets":{"choice-proof-methods-test-5":{"type":"choice","options":["$r/2$","$r-1$","$(r+2)/2$"],"answer":[2],"choice_feedback":["It is smaller but may be at most 2, outside the specified set.","Subtracting 1 may also leave the set; the argument must work for every possible r \u0026gt; 2.","The average of r and 2 is rational and lies strictly between 2 and r, so it belongs to the set and is smaller."]}},"hints":[],"explanation":"The average of r and 2 is rational and lies strictly between 2 and r, so it belongs to the set and is smaller.","id":"proof-methods-test-5","coverage_topics":["aa-hl-proof-by-contradiction"],"correct_feedback":"The average of r and 2 is rational and lies strictly between 2 and r, so it belongs to the set and is smaller.","incorrect_feedback":"The average of r and 2 is rational and lies strictly between 2 and r, so it belongs to the set and is smaller.","evidence_kind":"objective-concept-test"},{"content":"To prove there are infinitely many primes, assume finitely many p₁,…,pₙ. Why consider N = p₁⋯pₙ + 1?\u003Cbr\u003E[[choice-proof-methods-test-6]]","widgets":{"choice-proof-methods-test-6":{"type":"choice","options":["Every listed prime divides N","No listed prime divides N, but N \u003E 1 has a prime divisor","N must itself always be prime"],"answer":[1],"choice_feedback":["Each leaves remainder 1.","That divisor is absent from the allegedly complete list.","N need not be prime; a new prime divisor suffices."]}},"hints":[],"explanation":"That divisor is absent from the allegedly complete list.","id":"proof-methods-test-6","coverage_topics":["aa-hl-proof-by-contradiction"],"correct_feedback":"That divisor is absent from the allegedly complete list.","incorrect_feedback":"That divisor is absent from the allegedly complete list.","evidence_kind":"objective-concept-test"},{"content":"An irrationality proof obtains p² = 2q² and hence p is even. What is still needed for a lowest-terms contradiction?\u003Cbr\u003E[[choice-proof-methods-test-7]]","widgets":{"choice-proof-methods-test-7":{"type":"choice","options":["Write p = 2k, substitute, and deduce q is even too","Stop: an even numerator contradicts lowest terms","Cancel 2 and conclude p = q"],"answer":[0],"choice_feedback":["Substitution gives q² = 2k², so q is even. Only then has a common factor been forced.","A reduced fraction can have an even numerator and an odd denominator.","Taking square roots or cancelling does not turn p² = 2q² into p = q."]}},"hints":[],"explanation":"Substitution gives q² = 2k², so q is even. Only then has a common factor been forced.","id":"proof-methods-test-7","coverage_topics":["aa-hl-proof-by-contradiction"],"correct_feedback":"Substitution gives q² = 2k², so q is even. Only then has a common factor been forced.","incorrect_feedback":"Substitution gives q² = 2k², so q is even. Only then has a common factor been forced.","evidence_kind":"objective-concept-test"},{"content":"Which assumption negates “every element of set S has property P”?\u003Cbr\u003E[[choice-proof-methods-test-8]]","widgets":{"choice-proof-methods-test-8":{"type":"choice","options":["No element of S has P","Every element of S has P","At least one element of S does not have P"],"answer":[2],"choice_feedback":["That is stronger than the negation.","That repeats the original claim.","Negating a universal statement gives an existential counterexample."]}},"hints":[],"explanation":"Negating a universal statement gives an existential counterexample.","id":"proof-methods-test-8","coverage_topics":["aa-hl-proof-by-contradiction"],"correct_feedback":"Negating a universal statement gives an existential counterexample.","incorrect_feedback":"Negating a universal statement gives an existential counterexample.","evidence_kind":"objective-concept-test"}],"mix":0,"tinkumo_module_mode":"test"}