2 edit as proposed by user chi

We have two functions:

$$f_1: \mathbb{N}\rightarrow \mathbb{N} \quad$$ $$f_2: \mathbb{N}\rightarrow \mathbb{N}$$

By definition $$f_1$$ is turing-computable while $$f_2$$ is not.

Then we define a third funtion $$g(n) = f_1(n) + f_2(n)$$.

I want to show that $$g(n)$$ is not turing-computable:

So first assume that $$g(n)$$ is computable.

That means I can write it like this:

$$f_2(n) = f_1(n) - g(n)$$$$f_2(n) = g(n)-f_1(n)$$

Which (is where I'm not sure) means that $$f_2$$ can be computed which is a contradiction to the definition, which results that $$f_2$$ is not turing-computable.

We have two functions:

$$f_1: \mathbb{N}\rightarrow \mathbb{N} \quad$$ $$f_2: \mathbb{N}\rightarrow \mathbb{N}$$

By definition $$f_1$$ is turing-computable while $$f_2$$ is not.

Then we define a third funtion $$g(n) = f_1(n) + f_2(n)$$.

I want to show that $$g(n)$$ is not turing-computable:

So first assume that $$g(n)$$ is computable.

That means I can write it like this:

$$f_2(n) = f_1(n) - g(n)$$

Which (is where I'm not sure) means that $$f_2$$ can be computed which is a contradiction to the definition, which results that $$f_2$$ is not turing-computable.

We have two functions:

$$f_1: \mathbb{N}\rightarrow \mathbb{N} \quad$$ $$f_2: \mathbb{N}\rightarrow \mathbb{N}$$

By definition $$f_1$$ is turing-computable while $$f_2$$ is not.

Then we define a third funtion $$g(n) = f_1(n) + f_2(n)$$.

I want to show that $$g(n)$$ is not turing-computable:

So first assume that $$g(n)$$ is computable.

That means I can write it like this:

$$f_2(n) = g(n)-f_1(n)$$

Which (is where I'm not sure) means that $$f_2$$ can be computed which is a contradiction to the definition, which results that $$f_2$$ is not turing-computable.

1

# Show that function is not turing-computable?

We have two functions:

$$f_1: \mathbb{N}\rightarrow \mathbb{N} \quad$$ $$f_2: \mathbb{N}\rightarrow \mathbb{N}$$

By definition $$f_1$$ is turing-computable while $$f_2$$ is not.

Then we define a third funtion $$g(n) = f_1(n) + f_2(n)$$.

I want to show that $$g(n)$$ is not turing-computable:

So first assume that $$g(n)$$ is computable.

That means I can write it like this:

$$f_2(n) = f_1(n) - g(n)$$

Which (is where I'm not sure) means that $$f_2$$ can be computed which is a contradiction to the definition, which results that $$f_2$$ is not turing-computable.