Can someone please help?

To count the points (x, y) with positive integer coordinates that lie below the hyperbola xy = 16, we can consider the following:

 

The hyperbola xy = 16: This is a downward-sloping curve that passes through the points (1, 16), (2, 8), (4, 4), (8, 2), and (16, 1). Any point below this curve will have a product of coordinates less than 16.

 

Restricting to positive integers: Since we only want points with positive integer coordinates, we can focus on the region in the first quadrant where both x and y are positive.

 

Counting the points: We can systematically count the points in this region that satisfy xy < 16:

 

For x = 1, we can have y = 1, 2, ..., 15 (15 points).

 

For x = 2, we can have y = 1, 2, ..., 7 (7 points).

 

For x = 3, we can have y = 1, 2, ..., 5 (5 points).

 

For x = 4, we can have y = 1, 2, 3 (3 points).

 

For x = 5, we can have y = 1, 2, 3 (3 points).

 

For x = 6, we can have y = 1, 2 (2 points).

 

For x = 7, we can have y = 1, 2 (2 points).

 

For x = 8, we can have y = 1 (1 point).

 

Summing these up, we get 15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 = 43 points.

 

Therefore, there are 43 points of the form (x, y), where both coordinates are positive integers, that lie below the graph of the hyperbola xy = 16.

To find how many points of the form $(x, y)$ lie below the graph of the hyperbola defined by the equation $xy = 16$, we start by expressing $y$ in terms of $x$:

$$y = \frac{16}{x}$$

We want to find integer points $(x, y)$ where both $x$ and $y$ are positive integers, and $y < \frac{16}{x}$. Hence, we can reformulate our problem to determine the constraints on $x$:

1. $y$ must be a positive integer:

To ensure $y$ is an integer, $x$ must be a divisor of $16$.

2. The divisors of $16$ can be calculated:
The positive divisors of $16$ are $1, 2, 4, 8, 16$.

Now we will determine $y$ for each divisor of $16$ and check how many pairs $(x, y)$ we can find such that $y$ remains below $\frac{16}{x}$:

$$\begin{align*} x = 1 & \Rightarrow y = \frac{16}{1} = 16 \quad (\text{valid since } y < 16)\\ x = 2 & \Rightarrow y = \frac{16}{2} = 8 \quad (\text{valid since } y < 8)\\ x = 4 & \Rightarrow y = \frac{16}{4} = 4 \quad (\text{valid since } y < 4)\\ x = 8 & \Rightarrow y = \frac{16}{8} = 2 \quad (\text{valid since } y < 2)\\ x = 16 & \Rightarrow y = \frac{16}{16} = 1 \quad (\text{valid since } y < 1)\\ \end{align*}$$

Next, we need to determine how many integer values of $y$ are valid (i.e., $y$ must be positive and lie below the computed value):

- For $x = 1$, $y < 16$ gives valid values: $1, 2, \ldots, 15$ (15 values).

- For $x = 2$, $y < 8$ gives valid values: $1, 2, \ldots, 7$ (7 values).

- For $x = 4$, $y < 4$ gives valid values: $1, 2, 3$ (3 values).

- For $x = 8$, $y < 2$ gives valid values: $1$ (1 value).

- For $x = 16$, $y < 1$ yields no positive integers.

Now we add up the number of valid $y$ values:

$$15 + 7 + 3 + 1 + 0 = 26$$

Thus, the total number of points $(x, y)$ where both $x$ and $y$ are positive integers that lie below the hyperbola defined by $xy = 16$ is:

$$\boxed{26}$$