(a) Show that there are infinitely many subgroups of the additive group Q of rational numbers. (15 marks) (b) Using contour…

(a) Infinitely many subgroups of (ℚ, +)

For each n ≥ 1 let Hₙ = m/n! : m ∈ ℤ. It contains 0 and is closed under addition and negatives (m/n! + k/n! = (m + k)/n!), so it is a subgroup of ℚ.

Hₙ ⊂ Hₙ₊₁ since m/n! = m(n + 1)/(n + 1)!. The inclusion is strict: 1/(n + 1)! ∈ Hₙ₊₁, but 1/(n + 1)! = m/n! would give m = 1/(n + 1), not an integer. So H₁ ⊊ H₂ ⊊ H₃ ⊊ …, and ℚ has infinitely many distinct subgroups. ∎

(b) ∫_−∞^∞ sin x/[x(x² + a²)] dx, a > 0

Consider f(z) = e^(iz)/[z(z² + a²)] on a contour made of the real axis from −R to −ε, a small upper semicircle γ_ε of radius ε about 0 (to avoid the pole at 0), the real axis from ε to R, and the large upper semicircle Γ_R. The only pole inside is z = ia.

On Γ_R, |f| ≤ 1/[R(R² − a²)] and |e^(iz)| ≤ 1, so the integral over Γ_R tends to 0 as R → ∞.

Residue at ia: e^(−a)/[ia · 2ia] = −e^(−a)/(2a²). Residue at 0: 1/a².

The integral over γ_ε (clockwise) tends to −πi·(1/a²) as ε → 0.

By the residue theorem, P.V.∫ e^(ix)/[x(x² + a²)] dx − πi/a² = 2πi · (−e^(−a)/(2a²)), so

P.V.∫ e^(ix)/[x(x² + a²)] dx = (πi/a²)(1 − e^(−a)).

Taking imaginary parts (the integrand sin x/[x(x² + a²)] has no singularity at 0):

∫_−∞^∞ sin x/[x(x² + a²)] dx = π(1 − e^(−a))/a².

(c) Big-M method

Maximise Z = 4x₁ + 5x₂ + 2x₃ with 2x₁ + x₂ + x₃ ≥ 10, x₁ + 3x₂ + x₃ ≤ 12, x₁ + x₂ + x₃ = 6.

Standard form: 2x₁ + x₂ + x₃ − s₁ + A₁ = 10; x₁ + 3x₂ + x₃ + s₂ = 12; x₁ + x₂ + x₃ + A₂ = 6; objective Z = 4x₁ + 5x₂ + 2x₃ − MA₁ − MA₂, with initial basis A₁ = 10, s₂ = 12, A₂ = 6.

Eliminating A₁, A₂: Z = (4 + 3M)x₁ + (5 + 2M)x₂ + (2 + 2M)x₃ − Ms₁ − 16M. The largest coefficient is that of x₁; ratios 10/2 = 5, 12, 6 are smallest for A₁, so A₁ leaves.

Iteration 1: pivot on x₁. Rows become x₁ + ½x₂ + ½x₃ − ½s₁ + ½A₁ = 5; (5/2)x₂ + ½x₃ + ½s₁ + s₂ − ½A₁ = 7; ½x₂ + ½x₃ + ½s₁ − ½A₁ + A₂ = 1. Now x₂ has the largest coefficient, (6 + M)/2; ratios 10, 14/5, 2 are smallest for A₂, so A₂ leaves.

Iteration 2: pivot on x₂. Rows become x₁ − s₁ + A₁ − A₂ = 4; −2x₃ − 2s₁ + s₂ + 2A₁ − 5A₂ = 2; x₂ + x₃ + s₁ − A₁ + 2A₂ = 2. The objective row is Z = 26 − 3x₃ − s₁ − (M − 1)A₁ − (M + 6)A₂. For M > 1 every coefficient is non-positive, so the solution is optimal.

Solution: x₁ = 4, x₂ = 2, x₃ = 0 (with s₂ = 2 and all artificial variables zero), maximum Z = 26.