# How do I prove 0 1 1

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“The two kinds of proofs of existence
“Examples of proofs of existence

##### The two kinds of proofs of existence

Some mathematical theorems say that an object with a certain property exists. There are two fundamentally different methods of proving these theorems:

1. Constructive proof: You specify an object and show that it has the required properties; or one gives a method to find such an object. This adds the additional information to the record as to how an object with the required properties can look.

2. Non-constructive proof: One shows with the help of logical inferences (and possibly other propositions) that such an object must exist, but does not indicate how it can be found. This proof can also be used as evidence of contradiction. (Assuming there was no object with these properties, then ...)

##### Examples of proofs of existence

Let \ (f: \ mathbb {R} \ rightarrow \ mathbb {R} \) be defined by \ (f (x) = -x ^ 2 + 1 \). We show: \ (f \) has (at least) one zero.

###### 1st variant: constructive proof

To show that \ (f \) has a zero, we can simply specify a zero. Such a zero is the number 1. We can prove this by inserting it: \ (f (1) = -1 ^ 2 + 1 = -1 + 1 = 0 \). Therefore \ (f \) has at least one zero.

###### 2nd variant: non-constructive proof

For this proof we need the intermediate value theorem. This is as follows:

If a function \ (f \) is defined and continuous on the interval \ ([a, b] \), then \ (f \) takes all values ​​between \ (f (a) \) and \ (f (b) \ ) in the interval \ ([a, b] \) at least once.

We are looking for a way to apply the intermediate value theorem. To do this, try different values ​​in \ (f \): \ begin {align *} f (0) & = 0 ^ 2 + 1 = 1> 0 \ f (2) & = -2 ^ 2 + 1 = - 4 +1 = -3 <0 \ end {align *}

In order to be able to apply the intermediate value theorem, we must first check the requirements. We put \ (a = 0 \) and \ (b = 2 \). The function \ (f \) is defined over the whole of \ (\ mathbb {R} \), thus also in the interval \ ([0,2] \). As a polynomial function, \ (f \) is continuous in the entire domain, i.e. also in the interval \ ([0,2] \). Therefore we can apply the intermediate value theorem and get: \ (f \) takes every value between \ (f (0) = 1 \) and \ (f (2) = - 3 \) in the interval \ ([0,2] \) ) at least once. In particular, \ (f \) has a zero in this interval.