Prove that √3+√4 is an Irrational number.
Dear Student,
Let us assume that
(√3 + √4) is a rational number.
√3 + √4 = r , where r is a rational number.
Squaring both sides, we get
[√3 + √4 ]² = r²
⇒3 + 2√12 + 4 = r² [ ∵ (a+b)2 = a2+b2+2ab]
⇒7 + 2√12 = r²
⇒2√12 = r² - 6
⇒√12 = [ r² - 6] / 2
R.H.S is purely rational, whereas, L.H.S is irrational.
This is a contradiction.
This means that our assumption was wrong.
Hence, (√3 + √4) is irrational.
Regards
Let us assume that
(√3 + √4) is a rational number.
√3 + √4 = r , where r is a rational number.
Squaring both sides, we get
[√3 + √4 ]² = r²
⇒3 + 2√12 + 4 = r² [ ∵ (a+b)2 = a2+b2+2ab]
⇒7 + 2√12 = r²
⇒2√12 = r² - 6
⇒√12 = [ r² - 6] / 2
R.H.S is purely rational, whereas, L.H.S is irrational.
This is a contradiction.
This means that our assumption was wrong.
Hence, (√3 + √4) is irrational.
Regards