Why is $\sqrt{3}$ an irrational number since it can be expressed as ratio of two numbers $(2\sqrt{3}+3)$ and $(2+\sqrt{3})$ ?
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1Where do you not understand? – user1176409 Aug 11 '23 at 03:44
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3A rational number is a ratio of two integers. Neither $2\sqrt{3} + 3$ nor $2 + \sqrt{3}$ are integers. – Theo Bendit Aug 11 '23 at 03:44
2 Answers
I would like to focus on the key issue here (the definition of rationality) as opposed to the actual proof (which is commonplace, straightforward, and available here anyways).
Being rational or irrational is not a matter of being a ratio of two numbers, but instead of two integers.
If it was the former, every number $x$ would be rational: consider $\frac x 1$.
To prove that $x$ is irrational, you need to show that there exists no pair of integers $a,b$ such that $x = \frac a b$.
$2\sqrt 3 + 3$ and $2+\sqrt 3$ are not integers.

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Square root of 3 is irrational because it can not be expressed as a ratio between two integers but what you are trying is just to express it as a ratio between two real numbers. If you are interested you can use the method of contradiction to prove that square root 3 is irrational number.

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