SL 1.6 Question Bank

Solution:

Proof: Let the two odd integers be represented as general forms.

Since the first integer is odd, we can write it as 2a + 1 for some integer a.

Since the second integer is odd, we can write it as 2b + 1 for some integer b.

Sum = (2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1)

Since (a + b + 1) is an integer, 2(a + b + 1) is even by definition.

Therefore, the sum of two odd integers is always even. ∎

✅ Complete proof with algebraic justification

Solution:

Proof by contradiction:

Step 1: Assume the opposite – that there IS a largest integer.

Let N be the largest integer.

Step 2: Consider the integer N + 1.

Since N is an integer, N + 1 is also an integer.

Step 3: Note that N + 1 > N.

This contradicts our assumption that N is the largest integer.

Step 4: Since our assumption leads to a contradiction, it must be false.

Therefore, there is no largest integer. ∎

✅ Proof by contradiction complete

Solution:

Let me test this statement with odd values of n:

For n = 1: n² + n = 1² + 1 = 2 (even)

For n = 3: n² + n = 3² + 3 = 9 + 3 = 12 (even)

For n = 5: n² + n = 5² + 5 = 25 + 5 = 30 (even)

Actually, let me prove this algebraically:

If n is odd, then n = 2k + 1 for some integer k.

n² + n = n(n + 1) = (2k + 1)(2k + 2) = (2k + 1) × 2(k + 1) = 2(2k + 1)(k + 1)

This is always even (divisible by 2).

❌ No counterexample exists – the statement is actually TRUE!

Solution (Proof by Contradiction):

Step 1: Assume the opposite.

Assume n² is odd but n is even.

Step 2: Since n is even, we can write n = 2k for some integer k.

Step 3: Calculate n².

n² = (2k)² = 4k² = 2(2k²)

Step 4: Identify the contradiction.

Since 2k² is an integer, n² = 2(2k²) is even.

But we assumed n² is odd. This is a contradiction.

Step 5: Conclude.

Our assumption must be false. Therefore, if n² is odd, then n is odd. ∎

✅ Proof by contradiction complete

Solution:

Proof by contradiction:

Step 1: Assume √3 is rational.

Then √3 = p/q where p, q are integers with no common factors and q ≠ 0.

Step 2: Square both sides.

3 = p²/q², so 3q² = p²

Step 3: Analyze divisibility.

Since 3q² = p², we know 3 divides p². This means 3 divides p.

So p = 3r for some integer r.

Step 4: Substitute back.

3q² = (3r)² = 9r², so q² = 3r²

Step 5: Find contradiction.

Since q² = 3r², we know 3 divides q² and hence 3 divides q.

Both p and q are divisible by 3, contradicting our assumption that they have no common factors.

Therefore, √3 is irrational. ∎

✅ √3 is irrational

Solution:

(a) Direct Proof:

Given: x > 0 and y > 0

To prove: xy > 0

Proof: Since x > 0, we know x is positive.

Since y > 0, we know y is positive.

By the properties of real numbers, the product of two positive numbers is positive.

Therefore, xy > 0. ∎

(b) Contrapositive:

Original: “If (x > 0 and y > 0), then xy > 0”

Contrapositive: “If xy ≤ 0, then (x ≤ 0 or y ≤ 0)”

✅ Direct proof and contrapositive complete

Geometric Proof:

Construction: Draw a square with side length (a + b).

Inside this square, place four identical right triangles with legs a and b.

Step 1: Calculate the area of the large square.

Area = (a + b)² = a² + 2ab + b²

Step 2: Calculate the areas of the components.

Four triangles: 4 × (½ab) = 2ab

Inner square (formed by hypotenuses): c²

Step 3: Set up the equation.

Total area = Area of triangles + Area of inner square

a² + 2ab + b² = 2ab + c²

Step 4: Solve for c².

a² + 2ab + b² – 2ab = c²

a² + b² = c² ∎

✅ Pythagorean theorem proven

Solution:

Given: Two distinct rational numbers r₁ and r₂ where r₁ < r₂

To prove: There exists a rational number r such that r₁ < r < r₂

Proof: Consider the average of r₁ and r₂:

r = (r₁ + r₂)/2

Step 1: Show r is rational.

Since r₁ and r₂ are rational, r₁ = p₁/q₁ and r₂ = p₂/q₂ for integers p₁, p₂, q₁, q₂

r = (p₁/q₁ + p₂/q₂)/2 = (p₁q₂ + p₂q₁)/(2q₁q₂)

Since the numerator and denominator are integers, r is rational.

Step 2: Show r₁ < r < r₂.

r – r₁ = (r₁ + r₂)/2 – r₁ = (r₂ – r₁)/2 > 0 (since r₂ > r₁)

r₂ – r = r₂ – (r₁ + r₂)/2 = (r₂ – r₁)/2 > 0

Therefore, r₁ < r < r₂. ∎

✅ Density of rationals proven

Solution:

Let the four consecutive integers be n, n+1, n+2, n+3

Product P = n(n+1)(n+2)(n+3)

We need to show P is divisible by 24 = 2³ × 3

Step 1: Divisibility by 8 (2³)

Among any 4 consecutive integers:

• At least 2 are even (divisible by 2)

• At least 1 is divisible by 4

• The product contains factors 2 × 4 × (another even) ≥ 2³ = 8

Step 2: Divisibility by 3

Among any 4 consecutive integers, at least one is divisible by 3.

This can be shown by considering n ≡ 0, 1, or 2 (mod 3)

Step 3: Combine results

Since P is divisible by both 8 and 3, and gcd(8,3) = 1,

P is divisible by 8 × 3 = 24. ∎

✅ Product divisible by 24

Solution:

(a) Truth value:

The statement is FALSE.

(b) Counterexample:

Consider x = -2.

Verification:

x² = (-2)² = 4 ✓

But x = -2 ≠ 2 ✗

Explanation:

The equation x² = 4 has two solutions: x = 2 and x = -2.

The statement ignores the negative solution.

Correct statement would be: “If x² = 4, then x = 2 or x = -2”

❌ Statement false; counterexample: x = -2