Let K0 s t Squareroot 2 s t a Show that Forall x y K0 x y

Let K_0 = {s + t Squareroot 2: s, t}, a) Show that Forall x, y K_0, (x + y) K_0, (x - y) K_0 and xy K_0 b) Let f: 2 times 2 rightarrow R, be defined by f(s, t) = s + t Squareroot 2. Show that f is injective (one to one). c) Show that K_0 (0, 1) is infinite. By Bolzano-Weierstrass, K_0 has an accumulation point in [0, 1]. d) Show that 0 is an accumulation point K_0.

Solution

ANSWER TO Q 3 PART (a)

x K₀ => x = s₁ + t₁2; y K₀ => y = s₂ + t₂2;

Now, x + y = (s₁ + s₂) + (t₁ +  t₂)2 ....... (1)

If s₁, t₁; s₂, t₂ Z, (s₁ + s₂ ) and (t₁ +  t₂) also Z. So, we may write

s =(s₁ + s₂ ) and t = (t₁ +  t₂) where s and t Z ......... (2)

(1) and (2) =>  x + y = s + t2 => x + y K_0.

Similarly,

x - y = (s₁ - s₂) + (t₁ - t₂)2 ....... (3)

If s₁, t₁; s₂, t₂ Z, (s₁ - s₂ ) and (t₁ - t₂) also Z. So, we may write

s =(s₁ - s₂ ) and t = (t₁ - t₂) where s and t Z ......... (4)

(3) and (4) =>  x - y = s + t2 => x - y K_0.

Also,

xy = (s₁s₂) + 2t₁t₂ + (s₁t₂ + t₁s₂)2 ....... (5)

If s₁, t₁; s₂, t₂ Z, (s₁s₂ + 2t₁t₂ ) and (s₁t₂ + t₁s₂) also Z. So, we may write s = (s₁s₂ + 2t₁t₂ ) and

t = (s₁t₂ + t₁s₂) where s and t Z ......... (6)

(5) and (6) =>  xy = s + t2 => xy K_0.

All three subparts of Q1 (a) proved

ANSWER TO Q 3 PART (b)

Let f(s₁, t₁) = s₁ + t₁2 and f(s2, t2) = s₂ + t₂2.

Let, if posible, s₁ + t₁2 = s₂ + t₂2. Then, s₁ = s₂ and ,t₁ =  t₂ [Because, both s₁ + t₁2 and s₂ + t₂2 are surds and two surds are equal only if rational parts (s₁ and s₂ in this case) are equal and irrational parts (t₁2and t₂2 in this case) are equal separately.

Thus, we have shown that if the two functional parts are equal, the respetive variables are also equal.

=> f is one to one. PROVED