Before going to learn the proof of 3√5 +4 is irrational, we know the following:
Every rational number can be written in the
form of p/q (Where p, q are integers and q ≠ 0)
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
Transposing a term/factor from one side to other
Transpose “+a” to LHS/RHS, it changed to “-a”
Transpose “-a” to LHS/RHS, it changed to “+a”
Transpose “ x a” to LHS/RHS, it changed to “÷ a”
Transpose “÷ a” to LHS/RHS, it changed to “ x a”
Transpose “x a/b” to LHS/RHS, it changed to “x b/a”
Simplifications of fractions
For Example:
2/3 + 5 = (2+5 x 3)/3 = (2 +15)/3 = 17/3
7 – 5/4 = (7 x 4 – 5)/4 = (28 – 5)/4 = 23/4
3/5 + 4/7 = (3x7+4x5)/5x7 =(21+ 20)/35=41/35
And know the "proof of √2 is an irrational"
And also read proof of "5√2 + 7√3 is an irrational number"
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Prove that 3√5 + 4 is an irrational.
(We can prove this by contradiction method)
Proof:
Assume that (3√5 + 4) is not an irrational number
So 3√5 + 4 is a rational number
∴ 3√5 + 4 = p/q
Transpose "+4" to RHS
3√5 = P/q - 4
3√5 = (p - 4q)/q ( Multiplied 4 with q)
Transpose " x 3 " to RHS
√5 =[(p - 4q)/q] ÷ 3
√5 =[(p - 4q)/q] x (1/3)
√5 = (p - 4q)/3q
Since p and q are integers,
“(p - 4q)/3q” is a rational number
But √5 is an irrational number
This is a contradiction
This contradiction is arisen due to our assumption
So our assumption is wrong
“ (3√5 + 4) is not an irrational” is wrong
Therefore (3√5 + 4) is an irrational
Solve the following model problems for Practice
1. Prove that 5√3 + 8 is an irrational number
2. Prove that 2 -6√5 is an irrational number
3. Prove that 5 +√7 is an irrational number
4. Prove that 6√7 - 8 is an irrational number