Case Analysis On Integers And Real Numbers

1) A case analysis that splits on “n is divisible by 3” and “n is not divisible by 3” is:

Exhaustive

Not exhaustive

Overlapping

Invalid

This split is exhaustive because for any integer n, either n is divisible by 3 or it is not. There are no integers that are neither divisible by 3 nor not divisible by 3. Therefore, the cases cover all possibilities, making the split exhaustive. It is not overlapping because the cases are mutually exclusive: if n is divisible by 3, it cannot be not divisible by 3, and vice versa. It is not invalid as a split.

Explanation

This split is exhaustive because for any integer n, either n is divisible by 3 or it is not. There are no integers that are neither divisible by 3 nor not divisible by 3. Therefore, the cases cover all possibilities, making the split exhaustive. It is not overlapping because the cases are mutually exclusive: if n is divisible by 3, it cannot be not divisible by 3, and vice versa. It is not invalid as a split.

2) Which case split is NOT exhaustive for integers?

Even or odd

Prime or composite

Positive, negative, zero

Less than 0, equal to 0, greater than 0

The split into prime or composite is not exhaustive for integers because it misses several types of integers. Prime numbers are greater than 1 and have no positive divisors other than 1 and themselves. Composite numbers are greater than 1 and have other divisors. However, this split misses negative integers, zero, and 1, which is neither prime nor composite. Therefore, it is not exhaustive. The other splits are exhaustive: even or odd covers all integers, positive/negative/zero covers all integers, and less than 0/equal to 0/greater than 0 covers all integers.

Explanation

The split into prime or composite is not exhaustive for integers because it misses several types of integers. Prime numbers are greater than 1 and have no positive divisors other than 1 and themselves. Composite numbers are greater than 1 and have other divisors. However, this split misses negative integers, zero, and 1, which is neither prime nor composite. Therefore, it is not exhaustive. The other splits are exhaustive: even or odd covers all integers, positive/negative/zero covers all integers, and less than 0/equal to 0/greater than 0 covers all integers.

3) A proof splitting on “x > 0” and “x < 0” without handling x = 0 is:

Complete

Elegant

Incomplete

Invalid for reals

For real numbers, the case x = 0 is not covered by the splits x > 0 and x < 0. Therefore, if the proof does not handle x = 0, it is incomplete because there might be properties that need to be verified at x = 0. It is not complete because a case is missing. It may or may not be elegant, but elegance doesn't compensate for incompleteness. It is invalid for reals if the conclusion fails at x=0, but strictly speaking, the proof is incomplete due to the missing case.

Explanation

For real numbers, the case x = 0 is not covered by the splits x > 0 and x < 0. Therefore, if the proof does not handle x = 0, it is incomplete because there might be properties that need to be verified at x = 0. It is not complete because a case is missing. It may or may not be elegant, but elegance doesn't compensate for incompleteness. It is invalid for reals if the conclusion fails at x=0, but strictly speaking, the proof is incomplete due to the missing case.

4) Which case split is exhaustive for real numbers?

Integer only

Less than 0 only

Rational or irrational

Positive only

The split into rational and irrational numbers is exhaustive for real numbers because every real number is either rational or irrational. There are no real numbers that are neither. The other splits are not exhaustive: "integer only" misses non-integer reals, "less than 0 only" misses non-negative reals, "positive only" misses non-positive reals. Therefore, rational or irrational is exhaustive.

Explanation

The split into rational and irrational numbers is exhaustive for real numbers because every real number is either rational or irrational. There are no real numbers that are neither. The other splits are not exhaustive: "integer only" misses non-integer reals, "less than 0 only" misses non-negative reals, "positive only" misses non-positive reals. Therefore, rational or irrational is exhaustive.

5) When splitting on “n is even” vs “n is odd,” you are using:

Overlapping cases

Complementary exhaustive cases

Arbitrary cases

Redundant cases

The cases "n is even" and "n is odd" are complementary because they are mutually exclusive and together they cover all integers. Every integer is either even or odd, and no integer is both. Therefore, they are complementary and exhaustive. They are not overlapping, arbitrary, or redundant.

Explanation

The cases "n is even" and "n is odd" are complementary because they are mutually exclusive and together they cover all integers. Every integer is either even or odd, and no integer is both. Therefore, they are complementary and exhaustive. They are not overlapping, arbitrary, or redundant.

6) When proving a statement about all integers n, which case analysis is most appropriate?

Prime vs Composite

Positive vs Negative

Greater than, equal to, or less than zero

None of the above

For statements about all integers, the most comprehensive case analysis is n > 0, n = 0, or n < 0. This covers all possible integers and the cases are mutually exclusive and exhaustive.

Explanation

For statements about all integers, the most comprehensive case analysis is n > 0, n = 0, or n < 0. This covers all possible integers and the cases are mutually exclusive and exhaustive.

7) A proof that uses cases “x > 0” and “x ≥ 0” is:

Exhaustive

Overlapping

Invalid

Non-exhaustive

The cases “x > 0” and “x ≥ 0” are overlapping because x > 0 is included in x ≥ 0. Therefore, these cases are not mutually exclusive; they overlap for all x > 0. This means that for x > 0, it is covered in both cases, which is redundant. The cases are exhaustive if the domain is x ≥ 0, but if the domain is all reals, it misses x < 0. However, the primary issue here is the overlap, so overlapping is the correct description.

Explanation

The cases “x > 0” and “x ≥ 0” are overlapping because x > 0 is included in x ≥ 0. Therefore, these cases are not mutually exclusive; they overlap for all x > 0. This means that for x > 0, it is covered in both cases, which is redundant. The cases are exhaustive if the domain is x ≥ 0, but if the domain is all reals, it misses x < 0. However, the primary issue here is the overlap, so overlapping is the correct description.

8) Which case split is exhaustive for integers modulo 7?

{0,1,2,3,4,5}

{1,2,3,4,5,6}

{0,2,4,6}

{0,1,2,3,4,5,6}

For integers modulo 7, the possible remainders are 0,1,2,3,4,5,6. Therefore, the set {0,1,2,3,4,5,6} includes all cases. The set {0,1,2,3,4,5} misses remainder 6, {1,2,3,4,5,6} misses remainder 0, and {0,2,4,6} misses 1,3,5. Thus, {0,1,2,3,4,5,6} is exhaustive.

Explanation

For integers modulo 7, the possible remainders are 0,1,2,3,4,5,6. Therefore, the set {0,1,2,3,4,5,6} includes all cases. The set {0,1,2,3,4,5} misses remainder 6, {1,2,3,4,5,6} misses remainder 0, and {0,2,4,6} misses 1,3,5. Thus, {0,1,2,3,4,5,6} is exhaustive.

9) To prove “If n is integer, then n² + n is even,” you should split on:

N mod 2

N mod 3

N is prime

N > 0

As previously explained, n² + n =n(n+1) is always even for integers n. The standard proof uses case analysis on n mod 2, i.e., even or odd. This split is exhaustive and straightforward. Other splits are not necessary or exhaustive for this property.

Explanation

As previously explained, n² + n =n(n+1) is always even for integers n. The standard proof uses case analysis on n mod 2, i.e., even or odd. This split is exhaustive and straightforward. Other splits are not necessary or exhaustive for this property.

10) Which case split for real numbers is NOT mutually exclusive?

X < 0, x = 0, x > 0

X ≤ 0, x > 0

X < 0, x ≤ 0

X ∈ ℚ, x ∉ ℚ

The cases x < 0 and x ≤ 0 are not mutually exclusive because x < 0 is included in x ≤ 0. Therefore, there is overlap for all x < 0. The other splits are mutually exclusive: in (a), the cases are disjoint; in (b), x≤0 and x>0 are disjoint; in (d), rational and irrational are disjoint. Thus, x < 0 and x ≤ 0 are not mutually exclusive.

Explanation

The cases x < 0 and x ≤ 0 are not mutually exclusive because x < 0 is included in x ≤ 0. Therefore, there is overlap for all x < 0. The other splits are mutually exclusive: in (a), the cases are disjoint; in (b), x≤0 and x>0 are disjoint; in (d), rational and irrational are disjoint. Thus, x < 0 and x ≤ 0 are not mutually exclusive.

11) In case analysis, if you miss the boundary case x=0, the proof is:

Still valid

Invalid

Stronger

Shorter

If a boundary case like x=0 is missed in the case analysis, the proof is invalid because it fails to establish the conclusion for all values. There might be properties that only fail at x=0, so without checking that case, the proof is incomplete and cannot be considered valid. It is not stronger or shorter; it is flawed.

Explanation

If a boundary case like x=0 is missed in the case analysis, the proof is invalid because it fails to establish the conclusion for all values. There might be properties that only fail at x=0, so without checking that case, the proof is incomplete and cannot be considered valid. It is not stronger or shorter; it is flawed.

12) A case analysis that splits on “n is divisible by 2” and “n is divisible by 3” is:

Exhaustive for integers

Not exhaustive (e.g., n=5)

Mutually exclusive

Both a and c

This split is not exhaustive because there are integers that are neither divisible by 2 nor by 3, such as n=5. Therefore, it does not cover all integers. It is also not mutually exclusive because some integers are divisible by both 2 and 3 (e.g., n=6). So, it is not exhaustive and not mutually exclusive. The correct choice is that it is not exhaustive.

Explanation

This split is not exhaustive because there are integers that are neither divisible by 2 nor by 3, such as n=5. Therefore, it does not cover all integers. It is also not mutually exclusive because some integers are divisible by both 2 and 3 (e.g., n=6). So, it is not exhaustive and not mutually exclusive. The correct choice is that it is not exhaustive.

13) Which is the most common case split for parity proofs?

N mod 3

N mod 2 (even/odd)

N mod 4

N mod 5

For parity proofs, which involve evenness or oddness, the most common case split is on n mod 2, i.e., even or odd. This is because parity is directly related to divisibility by 2. Other modulo splits are used for other properties, but for parity, even/odd is standard.

Explanation

For parity proofs, which involve evenness or oddness, the most common case split is on n mod 2, i.e., even or odd. This is because parity is directly related to divisibility by 2. Other modulo splits are used for other properties, but for parity, even/odd is standard.

14) To prove “∀x∈ℝ, |x| ≥ 0,” you might split on:

X > 0, x = 0, x < 0

X > 0 only

X < 0 only

X = 0 only

To prove that the absolute value |x| is always non-negative for all real x, we can use case analysis on the sign of x. If x > 0, then |x| = x > 0. If x = 0, then |x| = 0 ≥ 0. If x < 0, then |x| = -x > 0. Thus, in all cases, |x| ≥ 0. This split is exhaustive for real numbers. Other splits are not exhaustive.

Explanation

To prove that the absolute value |x| is always non-negative for all real x, we can use case analysis on the sign of x. If x > 0, then |x| = x > 0. If x = 0, then |x| = 0 ≥ 0. If x < 0, then |x| = -x > 0. Thus, in all cases, |x| ≥ 0. This split is exhaustive for real numbers. Other splits are not exhaustive.

15) A complete case analysis requires:

At least one case proven

Every possible case proven

Interesting cases only

Two cases minimum

For case analysis to be complete, every possible case must be proven. This means that for each case, the conclusion must be established. It is not sufficient to prove only some cases or interesting cases. There is no minimum number of cases; it depends on the problem. Therefore, every possible case must be proven.

Explanation

For case analysis to be complete, every possible case must be proven. This means that for each case, the conclusion must be established. It is not sufficient to prove only some cases or interesting cases. There is no minimum number of cases; it depends on the problem. Therefore, every possible case must be proven.

Case Analysis on Integers and Real Numbers

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