prove that √2 + √3 is irrational
Answers
SOLUTION__✍️
Let √2 + √3 = (a/b) is a rational no.
= On squaring both sides , we get
= 2 + 3 + 2√6 = (a2/b2)
= So,5 + 2√6 = (a2/b2) a rational no.
= So, 2√6 = (a2/b2) – 5 Since, 2√6 is an =irrational no. and (a2/b2) – 5 is a rational no.
= So, my contradiction is wrong. So, (√2 + √3) is an irrational no.
MARK BRAINLIEST
Answer:
Let us presume that √2 + √3 is rational .
Then it can be expressed in the form of a / b where a and b are integers and a and b are co primes such that their H.C.F = 1 .
√2 + √3 = a / b
Squaring both sides we get :
⇒ ( √2 + √3 )² = a² / b²
Using the formula of expansion :
( a + b )² = a² + b² + 2 ab we get :
⇒ 2 + 3 + 2 × √2 × √3 = a² / b²
⇒ 5 + 2 √6 = a² / b²
⇒ 2√6 = a²/b² - 5
⇒ 2√6 = ( a² - 5 b² ) / b²
Since 2√6 is an irrational number and the right hand side is a rational number , we see that there is a contradiction .
Hence √2 + √3 is an irrational number .
Step-by-step explanation:
An irrational number can never be expressed in the a/b form where a and b are integers .
So we prove the above by contradiction method where we assume that the statement is correct but prove that our assumptions are wrong .