# Criterion for Firing a perceptron

The criterion for firing a perceptron is as follows

Why is it that when the function $w \cdot x + b = 0$ the output is zero as well. Why couldn't it have been set to 1?

If one were to simulate the behavior of the perceptron with a sigmoid function, $w \cdot x + b$ can be multiplied by any arbitrary constant $C>0$ (as $C$ tends to infinity) as long as $w \cdot x+b$ is not equal to 0. This is because at 0 condition, the firing always gives output as 1 instead of zero it should have giving. However, changing the firing rule overcomes this problem. Then, is there any reason why 0 should be the output when $w \cdot x+b= 0$?

• I think it makes no difference because $wx+b$ will almost never be exactly zero. Remember that all these variables are real numbers. Welcome to the site!
– Emre
Aug 28 '17 at 20:15

So, in short Rosenblatt could have chosen for any version of the heaviside step function but he chose for the one where $w \cdot x + b = 0$ leads to a zero output. If you think binary values this makes sense, a zero input (i.e. False) leads to a zero output.