solve root 5 is a rational number
Answers
Suppose that √5 is rational, and express it in lowest possible terms (i.e., as a fully reduced fraction) as
m /n for natural numbers m and n. Then √5 can be expressed in lower terms as 5n − 2m / m − 2n , which is a contradiction.[3] (The two fractional expressions are equal because equating them, cross-multiplying, and canceling like additive terms gives 5n2 = m2 and m / n = √5, which is true by the premise. The second fractional expression for √5 is in lower terms since, comparing denominators, m − 2n < n since m < 3n since m / n < 3 since √5 < 3. And both the numerator and the denominator of the second fractional expression are positive since 2 < √5 < 5 / 2 and m / n = √5.)
2. This irrationality proof is also a proof by contradiction:
Suppose that √5 = a / b where a / b is in reduced form. Thus 5 = a² / b² and 5b² = a². If b were even, b², a², and a would be even making the fraction a / b
not in reduced form. Thus b is odd, and by following a similar process, a is odd.
Now, let a = 2m + 1 and b = 2n + 1 where m and n are integers.
Substituting into 5b² = a² we get:
5(2n+1)²=(2m+1)²
which simplifies to:
5(4n²+4n+1)=4m²+4m+1
making:
20n²+20n+5=4m²+4m+1
By subtracting 1 from both sides, we get:
20n²+20n+4=4m²+4m
which reduces to:
5n²+5n+1=m²+m
In other words:
5n(n+1)+1=m(m+1)
The expression x(x + 1) is even for any integer x (since either x or x + 1 is even). So this says that 5 × even + 1 = even, or odd = even. Since there is no integer that is both even and odd, we have reached a contradiction and √5 is irrational.
√5
=√2.2×2.2(approx value)
=2.2
=22/10
therefore√5 is a rational no.