prove that is an irrational number
Asked by koushalvishal109 | 28th Feb, 2019, 02:16: PM
If 4-√3 is a rational number, then we can write 4-√3 = a/b , where a and b are integers.
4-√3 = a/b
4 = ( a/b) +√3
4b = a + √3b
By squaring both sides, 16b2 = ( a +√3b )2 = a2 + 3b2 + 2√3ab
(13b2 - a2)/(2ab) = p/q = √3
where p = (13b2 - a2) and q = 2ab are integers.
But right side of above equation is a irrartional number that can not be equal to p/q if p and q are integers
we get this contradiction due to the assumption that 4-√3 is a rational number.
Hence 4-√3 is an irrational number
Answered by Thiyagarajan K | 28th Feb, 2019, 03:03: PM
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