We need to solve two problems. First, we need to find an irrational number between 2 and 3. Second, we need to solve two equations for $x$ and $y$, where $x$ and $y$ are real numbers. i) $(x + iy)(4i) = 8 + 4i$ ii) $\sqrt{2+i} = x - iy$

1. Problem Description

We need to solve two problems.

First, we need to find an irrational number between 2 and

3. Second, we need to solve two equations for $x$ and $y$, where $x$ and $y$ are real numbers.

i) $(x + iy)(4i) = 8 + 4i$

ii) $\sqrt{2+i} = x - iy$

2. Solution Steps

c) An irrational number between 2 and

3.

The square root of any non-perfect square integer is an irrational number.

Let's consider $\sqrt{5}$. Since $2^2 = 4$ and $3^2 = 9$, we have $2 < \sqrt{5} < 3$.

Therefore, $\sqrt{5}$ is an irrational number between 2 and

3.

d) Solve for $x$ and $y$.

i) $(x + iy)(4i) = 8 + 4i$

$4xi + 4i^2y = 8 + 4i$

$4xi - 4y = 8 + 4i$

Equating the real and imaginary parts:

$-4y = 8$ and $4x = 4$

$y = -2$ and $x = 1$

ii) $\sqrt{2+i} = x - iy$

Square both sides:

$2+i = (x - iy)^2$

$2+i = x^2 - 2ixy + (iy)^2$

$2+i = x^2 - 2ixy - y^2$

$2+i = (x^2 - y^2) - 2ixy$

Equating the real and imaginary parts:

$x^2 - y^2 = 2$ and $-2xy = 1$ which means $xy = -1/2$

From $xy = -1/2$, we get $y = -1/(2x)$. Substituting this into the first equation:

$x^2 - \left(-\frac{1}{2x}\right)^2 = 2$

$x^2 - \frac{1}{4x^2} = 2$

Multiplying by $4x^2$:

$4x^4 - 1 = 8x^2$

$4x^4 - 8x^2 - 1 = 0$

Let $z = x^2$. Then $4z^2 - 8z - 1 = 0$. Using the quadratic formula:

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 - 4(4)(-1)}}{8} = \frac{8 \pm \sqrt{64 + 16}}{8} = \frac{8 \pm \sqrt{80}}{8} = \frac{8 \pm 4\sqrt{5}}{8} = \frac{2 \pm \sqrt{5}}{2}$

Since $z = x^2$, $z$ must be positive. Thus, we take the positive root: $z = \frac{2 + \sqrt{5}}{2}$

$x^2 = \frac{2 + \sqrt{5}}{2}$

$x = \pm \sqrt{\frac{2 + \sqrt{5}}{2}}$

If $x = \sqrt{\frac{2 + \sqrt{5}}{2}}$, then $y = \frac{-1}{2\sqrt{\frac{2 + \sqrt{5}}{2}}} = \frac{-1}{\sqrt{2(2+\sqrt{5})}} = - \sqrt{\frac{1}{2(2+\sqrt{5})}} = -\sqrt{\frac{2-\sqrt{5}}{2(4-5)}} = \sqrt{\frac{2-\sqrt{5}}{2}}$

However, $\sqrt{5} > 2$ so $2 - \sqrt{5}$ is negative which would result in y being imaginary. Thus, $x$ must be negative:

$x = - \sqrt{\frac{2 + \sqrt{5}}{2}}$

$y = - \frac{1}{2x} = - \frac{1}{-2\sqrt{\frac{2 + \sqrt{5}}{2}}} = \frac{1}{2\sqrt{\frac{2 + \sqrt{5}}{2}}} = \frac{1}{\sqrt{2(2+\sqrt{5})}} = \sqrt{\frac{1}{2(2+\sqrt{5})}} = \sqrt{\frac{2-\sqrt{5}}{2(4-5)}} = -\sqrt{\frac{2-\sqrt{5}}{-2}} = \sqrt{\frac{\sqrt{5}-2}{2}}$

$x = - \sqrt{\frac{2 + \sqrt{5}}{2}}$

$y = \sqrt{\frac{\sqrt{5} - 2}{2}}$

3. Final Answer

c) $\sqrt{5}$

d) i) $x = 1$, $y = -2$

ii) $x = - \sqrt{\frac{2 + \sqrt{5}}{2}}$, $y = \sqrt{\frac{\sqrt{5} - 2}{2}}$