Question

prove that the following is irrational
$3 + \sqrt{2}$
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Answer 1

Solution:-

=> Let us assume, to the contrary that ( 3 + √2 ) is rational

Then , there exist co - prime a and b ( b is not equal to zero ) Such that

=> ( 3 + √2 ) = a/b

=> √2 = a/ b - 3

=> √2 = ( a - 3b )/b

Since a and b are integers, so (a - 3b)/b is rational

thus √2 is also rational

But this contradicts the fact that √2 is irrational, So our assumptions is incorrect

Hence ( 3 + √2 ) is irrational

More information about irrational number

=> The number which when expressed in decimal from are expressible as non terminating and non repeating decimals are know as irrational number

Example => Note that every non terminating and non repeating decimals is irrational

Type 1 => Clearly, 0.1010010001...... is a non terminating and non repeating decimals so it is irrational

Type 2 => if m is a positive integer which is not a prefect square then √m is irrational

Type 3 => π is irrational, while 22/7 is rational