Alphabets and Numbers? : Another Grandpa Mystery

Question

Now we all know how crazy Grandpa is when it comes to Math. He says to me:

"Use your imagination and tell me,

If
Seven + L = Zero
And
Zero - C = One
Then
Three + I = ?

I have no idea!”

Answer 1

Based on DEEM’s hint, the answer is

Eight.

I think this is because

When you look at the digital number for 7, and add an L to the left of it, you get L7 which looks like a digital Zero. When you look at a digital Zero and remove a C from the left of it, you get a digital One. When you take a digital 3 and add an I to the left of it, you get a digital Eight.

Answer 2

I think it should be

One, also.

The function is

$f(x) = (x+1) \pmod 2$, where $x$ is the result of the addition or subtraction when you convert the letters to Roman numerals.

So:

For $x = \mbox{Seven}+L = 7+50=57$, $f(x) = (57+1) \pmod 2 = 0 = \rm Zero$.
For $x = \text{Zero}-C = 0-100=-100$, $f(x) = (-100+1) \pmod 2 = 1 = \rm One$.
For $x= \mathrm{Three} + I = 3+1 = 4$, $f(x) = (4+1) \pmod 2 = 1 = \rm One$.

Of course, the answer could also be

Zero.

If the function was

$f(x) = H(-x)$, where $H$ is the Heaviside function, equal to 1 for $x\geq 0$ and 0 for $x<0$.

Then

For $x = \mbox{Seven}+L = 7+50=57$, $f(x) = H(-57) = 0 = \rm Zero$.
For $x = \text{Zero}-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = \rm One$.
For $x= \mathrm{Three} + I = 3+1 = 4$, $f(x) = H(-4) = 0 = \rm Zero$.