0

Is it true that a computable number or function is always definable, while the other way around is not?

It seems so based on the following link: just want to confirm

https://math.stackexchange.com/questions/1266587/example-of-uncomputable-but-definable-number#:~:text=The%20point%20here%20is%20that,1%20definable%20functions%20over%20N

Mike
  • 33
  • If a number is computable , then there is a turing machine (let us call it $T$) computing it and then we can define the number as "the result of the computation of $T$". For the other direction , there are counterexamples as pointed out below. – Peter Sep 10 '23 at 10:05

1 Answers1

1

Yes, all computable numbers are definable but not all definable numbers are computable.

Informally, a computable number is one for which we can write a computer program which calculates it to any desired accuracy. So, even though $\pi$ is irrational (and transcendental), it is computable since we know several formulae which can easily be programmed. In the real world, a computer would have limited memory and we have a limited lifetime so we might not really be able to calculate it to any desired accuracy. However, the concept of computable does not require that. It can use an imaginary computable with unlimited memory and runtime. More specifically a Turing Machine.

Computable Number at Wolfram

Here is a well-known formula for $\pi$

$$\pi = 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} ...)$$

This is terribly inefficient and would not be a good way to practically calculate $\pi$ but it is good enough to say that it is computable since this could be programmed easily.

A definable number requires a statement which uniquely identifies it. Any computable number is definable since its program can serve as its definition.

A number which is definable but not computable is necessarily fairly weird. One example is: Chaitin's Constant. This relies on something called the Halting Problem. Surprisingly, a program which can read another program and say whether it will halt (as opposed to an infinite loop) is not possible. So, although any program will either halt or not, this cannot be determined in a finite time. Some more detail is required to make this precise e.g. agree the computing language (usually a Turing Machine) and a way to order the programs but this is easy to do. With that agreement, the constant has a defined value but we will never calculate it. Again, the computers involved have unlimited memory and runtime.

Chaitin's Constant at Wolfram

badjohn
  • 8,204