How to generate a "1" when the input voltage is above 5 V and a "0" when it's below 5 V? [closed]

The usual way to accomplish this is with a comparator. I use the classic LM393 here:

^{simulate this circuit – Schematic created using CircuitLab}

It compares two potentials, \$V_P\$ and \$V_Q\$, outputting high (+5V here, since that's the supply I've used) when \$V_P>V_Q\$, and low (0V) otherwise.

Potential \$V_P\$ is a fraction of the input \$V_{IN}\$:

$$ V_P = V_{IN}\frac{R_2}{R_1+R_2} $$

Potential \$V_Q\$ is a fraction of the supply:

$$ V_Q = 5 \times \frac{R_4}{R_3+R_4} $$

I have arranged things such that when \$V_{IN}=+5.1\rm V\$, then \$V_P = V_Q = 0.65\rm V\$. I have also included about 0.1V of hysteresis, using positive feedback via R6, to prevent oscillations at the threshold. It means that for \$V_{OUT}\$ to go high, \$V_{IN}\$ must exceed +5.1V, but to return low again it must fall under +5V.

It's important to respect the acceptable input voltage range of the comparator (see the datasheet), so R1 and R2 are chosen to ensure that even a 24V input will not cause P to exceed that.

Here's a simulation showing \$V_{IN}\$ (blue) rising and falling slowly, and the resulting \$V_{OUT}\$ (orange):

what is the best way to approach this problem statement

I would suggest to you that you consider what I'm saying below.

With an uncertainty of 100 mV between 5 volts and 5.1 volts, it's not clear what you want to happen but, if you accept that this 100 mV uncertainty can be regarded as hysteresis (can be very useful) then you will be best served by using a comparator that can handle the 24 volt input range and produce the relevant (but unspecified) logic levels at the output.

^{simulate this circuit – Schematic created using CircuitLab}

Here's a basic sketch of a comparator being used to work out if a voltage is more or less than 5V, given a 24V supply. The voltage dividor gives 5V.

I have made no attempt to add filtering, overvoltage protection, etc.