Discussion

Question:

If x = (9 + 4√5)⁴⁸ = [x] + f

where [x] is defined as integral part of x and f is a faction, then x(1 – f) equals

Explanation:

x = (9 + 4√5)⁴⁸

Let

y = (9 - 4√5)⁴⁸

Now,

(9 + 4√5)⁴⁸ × (9 - 4√5)⁴⁸ = (81 - 80)⁴⁸ = 1 ...(i)

Also, (9 + 4√5)⁴⁸ + (9 - 4√5)⁴⁸

= [⁴⁸C₀ 9⁴⁸ + ⁴⁸C₁9⁴⁷ (4√5) + ⁴⁸C₂9⁴⁶ (4√5)² + .... + ⁴⁸C₄₇ (9)(4√5)⁴⁷ + ⁴⁸C₄₈ (4√5)⁴⁸]

+ [ ⁴⁸C₀ 9⁴⁸ - ⁴⁸C₁ 9⁴⁷ (4√5) + ⁴⁸C₂ 9⁴⁶ (4√5)₂ - ... - ⁴⁸C₄₇ (9)(4√5)⁴⁷ + ⁴⁸C₄₈ (4√5)⁴⁸]

= 2[⁴⁸C₀ 9⁴⁸ + ⁴⁸C₂ 9⁴⁶ (4√5)² + ... ⁴⁸C₄₈ (4√5)⁴⁸]

∴ x + y = 2(k) = even

Now,

0 < 9 - 4√5 < 1

∴ 0 < (9 - 4√5)⁴⁸ < 1

∴ 0 < y < 1 ...(ii)

Also, x = [x] + f, 0 < f < 1 … (iii)

∴ [x] + f + y is even

As [x] is an integer, f + y is an integer.

From (ii) and (iii)

0 < f + y < 2

∴ f + y = 1

Now, x (1 – f) = xy

But from (i), xy = 1

∴ x(1 – f) = 1

Hence, option (a).

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