Question

prove that ti is irrational 5-2√3

Answer 1

let us assume on the contrary that 5-2√3 is a rational no
Then It can be expressed in p/q form where q ≠0

p/q = 5 - 2√3
2√3 = 5 - p/q
√3 = 5 - p/q × 1/2

Here p/q is rational , 1/2 is also rational and 5 is an integer
Therefore √3 is a rational
But we that √3 is an irrational no
Hence our assumption is wrong and 5 - 2√3 is an irrational number...

Answer 2

Let us assume that 5-2√3 is a rational...
5-2√3=a/b
squaring on both sides...
(5-2√3)²7 =(a/b)²
{(a-b)²=a²-2ab+b²}
(5)²-2(5)(2√3)+(2√3)²=(a/b)²
25-10(2√3)+(2)²(√3)²=a²/b²
(((((√3))))² .......here square and root should be cancelled..
there fore we get...
25-10(2√3)+4(3)=a²/b²
25-20√3+12=a²/b²
37-20√3=a²/b²
-20√3=a²/b²-37
-20√3=a²-37b²/b²
√3=a²-37b²/-20b²...............eq-1
LHS≠RHS
there fore a, b are integers, LHS of eq-1 √3 is irrational and RHS of eq-1 is a²-37b²/-20b².
Hence this contradicts the fact that √3 =a²-37b²/-20b² is irrational.
there fore 5-2√3 is also irrational..
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