show that 3√3 is an irrational
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Answered by
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Hey mate..
========
Let us assume to the contrary that
![3 \sqrt{3} 3 \sqrt{3}](https://tex.z-dn.net/?f=3+%5Csqrt%7B3%7D+)
is a rational number.
So,
( where p,q are co-prime numbers and q is not equal to zero )
![= > \sqrt{3} = \frac{p}{3q} = > \sqrt{3} = \frac{p}{3q}](https://tex.z-dn.net/?f=+%3D+%26gt%3B+%5Csqrt%7B3%7D+%3D+%5Cfrac%7Bp%7D%7B3q%7D+)
Since, p and q are integers,
is a rational number .
So,
is a rational number
But it is impossible since,
![3 \sqrt{3} 3 \sqrt{3}](https://tex.z-dn.net/?f=3+%5Csqrt%7B3%7D+)
is irrational .
So, We came to conclusion that ,
is an irrational number.
( Hence, Proved )
Hope it helps !!
========
Let us assume to the contrary that
is a rational number.
So,
Since, p and q are integers,
So,
But it is impossible since,
is irrational .
So, We came to conclusion that ,
( Hence, Proved )
Hope it helps !!
rajdeep17:
awesome thank you
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