Question

Show that 7-root 5 is irrational

Answer 1

Given that,
7-√5 is irrational.
let us assume that 7-√5 is rational.
then, we get,
7-√5=a/b
squaring on both sides.
(7-√5)²=(a/b)²
[a+b]²=a²+2ab+b²

(7)²+2(7)(√5)+(√5)²=a²/b²
square and root should be cancelled....
49+2(7)(√5)+5=a²/b²
49+14√5+5=a²/b²
54+14√5=a²/b²
14√5=a²/b²-54
14√5=a²-54b²/b²
√5=a²-54b²/14b²..................eq---1
there fore LHS≠RHS
a,b are integers, LHS of eq-1 √5 is irrational and RHS of eq-1 a²-54b²/14b²
is rational.
These contradicts the fact that √5 is irrational
Hence 7-√5 is irrational.....
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Answer 2

Let root 5 be rational so that it can be written In the form of p/q and p andq are co prime
Root5=p/q
Squaring both the sides
5=p^2/q^2
5q^2 = p^2
5/p^2 then 5/p
So p is the factor of root5
Let p be 5c
5q^2=25c^2
q^2=5c^2
5/q^2 then 5/q
So q is also the factor of root5
Here it contradicts as root has more than one factor
Hence root5 is irrational
If root5 is irrational then also 7 root is irrational