Math for class 10
Before going to learn the proof of 5√2+7√3 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
-------------------------------------------------------------------------
Prove that 5√2 +7√3 is an irrational.
(We can
prove this by contradiction method)
Proof:
Assume that 5√2 + 7√3 is not an
irrational number
So 5√2 + 7√3 is a
rational number
∴ 5√2 + 7√3 = p/q (p, q are integers and q ≠ 0)
Transpose "+7√3" is to
RHS
5√2 = P/q - 7√3
Squaring
on both sides, we get
(5√2)2 = (P/q – 7√3 )2
[Apply (a - b)2 = a2 - 2ab + b2]
52(√2)2= (P/q)2 –
2 x (p/q) x ((7√3) + (7√3)2
[Here a= p/q and b=7√3 ]
25(2) = (P2/q2)
– (14√3)(p/q) + 72(√3)2
50 = (P2/q2) – (14p/q)(√3)+ 49(3)
Transpose “– (14p/q)(√3)” to
LHS
50 + (14p/q)(√3)= P2/q2 + 147
Transpose “+50” to RHS
(14p/q)(√3)= P2/q2 + 147 -
50
(14p/q)(√3)= P2/q2 + 97
Multiply 97 by q2
(14p/q)(√3) = (p2 + 97q2)/q2
Transpose “ x (14p/q)” to RHS
(√3) = (p2 + 97q2)/q2 x (q/14p)
(√3) = (p2 + 97q2)/14pq
(“q” cancelled)
Since p and q are integers,
“(p2 + 97q2)/14pq”
is a rational number
But √3 is an
irrational number
This is a contradiction
This contradiction is arisen due to our assumption
So our assumption is wrong
“ 5√2 + 7√3 is not an irrational” is wrong
Therefore 5√2 + 7√3 is an irrational
Hence proved
Solve the following model problems for Practice
1. Prove that 2√3 + 3√7 is an irrational number
2. Prove that 2√7 - √5 is an irrational number
3. Prove that √5- √2 is an irrational number