Mathematics: Prove that root 7 is irrational by the method of contradiction

Prove that root 7 is irrational by the method

Ответ:

RickandMorty420710

15.03.2022

Let assume that $\sqrt7$ is a rational number. Therefore it can be expressed as a fraction $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$ and $\text{gcd}(a,b)=1$ .

$\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2$

This means that a^2 is divisible by 7, and therefore also is divisible by 7.

So, a=7k where $k\in\mathbb{Z}$

$(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2$

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is .

But if both numbers and are divisible by 7, then $\text{gcd}(a,b)=7$ which contradicts our earlier assumption that $\text{gcd}(a,b)=1$ .

Therefore $\sqrt7$ is an irrational number.

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Ответ:

keripressley148

27.05.2020

Mathematics

Step-by-step explanation:

First you should measure the with

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Prove that root 7 is irrational by the method of contradiction

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