# Is 7 0 7 0 or infinite

### What are irrational numbers?

Like rational numbers, you cannot represent irrational numbers as fractions, periodic or terminating numbers.

you are **non-periodic** and **infinite**.

**Examples:**

$$ sqrt (2) = 1.414213562 ... $$

$$1,41441444144441444441…$$

Non-square roots are always irrational numbers.

You can already pull some roots

- $$ sqrt (9) = 3 $$
- $$ sqrt (0.16) = 0.4 $$, since $$ 0.4 * 0.4 = 0.16 $$
$$ sqrt (4/9) = 2/3 $$, since

$$ 2 * 2 = 4 $$ and $$ 3 * 3 = 9 $$

The square numbers $$ 1, 4, 9, 16, 25,… $$ will help you with this

Note: Square numbers are always natural numbers.

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### Nest irrational numbers in an interval

With interval nesting, you can save irrational numbers as **decimal number** without using the root key of your calculator.

**Example:** $$ sqrt (2) $$

### Step 1: Find the first interval.

Between which natural numbers is $$ sqrt (2) $$?

- Try the square numbers $$ 1 $$, $$ 4 $$, $$ 9 $$ and $$ sqrt (2) ^ 2 $$.
- Since $$ 1 ^ 2 = 1le2le2 ^ 2 = 4 $$, $$ sqrt (2) $$ lies between $$ 1 $$ and $$ 2 $$.
- Always choose that
**smallest interval**, in which the value $$ 2 $$ is also present. So not $$ [1; 9] $$, but just $$ [1; 2] $$.

**interval**

An interval is a set of numbers between two numbers.

The closed interval $$ [2; 5] = {x in QQ | -2lexle5} $$ contains the $$ - 2 $$ and the $$ 5 $$ and all rational numbers in between.

### Select the interval nesting more closely

Note: Calculation steps marked in blue are calculated with the calculator.

### Step 2: Box the interval further.

- Add a digit after the decimal point.
- Use the calculator to find out between which of the numbers $$ (1,1) ^ 2, (1,2) ^ 2, (1,3) ^ 2,…, (1,9) ^ 2 $$ the number $$ 2 $$ lies.
- $$ 1,4lesqrt (2) le1,5 $$, because $$ (1,4) ^ 2 = 1.96 $$$$ le2le $$$$ (1.5) ^ 2 = 2.25 $$

### 3rd step: Two decimal places

- Use the calculator to calculate between which of the numbers $$ (1.41) ^ 2, (1.42) ^ 2, (1.43) ^ 2,…, (1.49) ^ 2 $$ the number $$ 2 $$ lies.
- $$ 1.41lesqrt (2) le1.42 $$,

because $$ (1.41) ^ 2 = 1.9881 $$$$ le2le $$$$ (1.42) ^ 2 = 2.0164 $$

### 4th step: three decimal places

- Use the calculator to calculate between which of the numbers $$ (1.411) ^ 2, (1.412) ^ 2, (1.413) ^ 2,…, (1.419) ^ 2 $$ the number $$ 2 $$ lies.
- $$ 1,414lesqrt (2) le1,415 $$,

because $$ (1.414) ^ 2 = 1.999396 $$$$ le2le $$$$ (1.415) ^ 2 = 2.002225 $$

So you can nest $$ sqrt (2) $$ more and more precisely and get one **Approximate value**.

### Proof by contradiction: $$ sqrt (2) $$ is irrational

### I. Claim: $$ sqrt (2) $$ is irrational

### II. Assumption: $$ sqrt (2) $$ is rational (is a shortened fraction)

To show: There is a contradiction.

### Preliminary considerations:

- If you multiply a number $$ n $$ by $$ 2 $$, the result is an even number $$ (2 * n) $$.
- If the square of a number is even, so is the number itself.
**Example:**64 is straight and 8 is too. - Fractions can be shortened if the numerator and denominator have a common factor.

**Proof of contradiction**

In this evidence process, you show an assertion by assuming the opposite of the assertion and that leads to a contradiction.

**Procedure:**

I. Claim

II. Assumption to the contrary of assertion

III. Contradiction

IV. Assumption wrong, assertion holds

Already around 300 BC The mathematician showed **Euclid**that $$ sqrt (2) $$ is an irrational number. He also carried out a contradiction proof.

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### Proof by contradiction: $$ sqrt (2) $$ is irrational

Proof step | Explanations | |
---|---|---|

1) | $$ sqrt (2) = p / q $$ | According to the claim, $$ sqrt (2) $$ can be represented as an abbreviated fraction (Have $$ p $$ and $$ q $$ none common divisor). |

2) | $$ 2 = p ^ 2 / q ^ 2 $$ | Squar both sides of the equation. |

3) | $$ 2 * q ^ 2 = p ^ 2 $$ | Transform the equation according to $$ p $$. |

4) | $$ p ^ 2 $$ is just | This follows from the representation of $$ p $$. |

5) | $$ p $$ is just | This follows from the second preliminary consideration. |

6) | $$ p = 2 * n $$ | $$ p $$ is even, i.e. double of any number $$ n $$. |

7) | $$ p ^ 2 = 4 * n ^ 2 $$ | Squar both sides of the equation. |

### Proof by contradiction: $$ sqrt (2) $$ is irrational

Proof step | Explanation | |
---|---|---|

8) | $$ 4 * n ^ 2 = 2 * q ^ 2 $$ | Equalization of $$ p ^ 2 = 4 * n ^ 2 $$ and $$ p ^ 2 = 2 * q ^ 2 $$. |

9) | $$ 2 * n ^ 2 = q ^ 2 $$ | Division by 2. |

10) | $$ q ^ 2 $$ is just | This follows from the representation of $$ q ^ 2 $$. |

11) | $$ q $$ is even | This follows from the second preliminary consideration. |

12) | $$ q = 2 * m $$ | $$ q $$ is even, i.e. double of any number $$ m $$. |

13) | $$ sqrt (2) = p / q = (2 * n) / (2 * m) $$ | $$ p $$ and $$ q $$ are even and both divisible by $$ 2 $$. |

### III. That is a contradiction to assumption.

$$ p $$ and $$ q $$ have a common factor. So $$ sqrt (2) $$ is not an abbreviated fraction after all.

### IV. The assumption is wrong, the claim is valid.

This proves: $$ sqrt (2) $$ is irrational.

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