Question

prove that (7-5√3) is irrational​

Answer 1

Answer:

-1.66025404

It is a number that cannot be written as a ratio of two integers (or cannot be expressed as a fraction). A number is irrational if it has endless decimals after it. Hence, we know that -1.66025404 (7-5√3) is irrational

Hope this helps! :D

Answer 2

Step-by-step explanation:

Let 7-$\sqrt{5}$ is a rational number

7+$\sqrt{5}$=$\frac{a}{b}$,b=0.....(1)

Where a and b co-prime integer number

Equation (1) can be written as

$\sqrt{5}$=$\frac{a}{b}$-7

or

$\sqrt{5}$=$\frac{a-7b}{b}$.......(2)

Since , a and b are integers. So $\frac{a-7b}{b}$ will be rational number, so from equation (2) we find $\sqrt{5}$ is a rational number. But we know that  is  irrational number

So this result is contradicted .

So our hypothesis is wrong.

hence, $7-\sqrt{5}$ is a rational number.

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