Prove that 5 , \sqrt{3} is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that

5, \sqrt{3}

is an irrational number.

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Step 1: Use the definition of rational number and prove p is divisible by 3.

Let 5 \sqrt{3} is a rational number.

∴ 5 \sqrt{3} = \frac{p}{q},

where p and q are some integer and HCF(p,q)=1

- - (1)

\Rightarrow 5 \sqrt{3} . q = p

[Squaring on both sides.]

\Rightarrow (5 \sqrt{3} . q)^{2} = p^{2}

\Rightarrow 3 (25 q^{2}) = p^{2}

\Rightarrow p^{2}

is divisible by 3

\Rightarrow p

is divisible by 3

- - (2)

Step 2: Prove q is divisible by 3.

Let p = 3 m,

where m is some integer.

\Rightarrow 5 \sqrt{3} . q = 3 m

[Squaring on both sides.]

\Rightarrow (5 \sqrt{3} . q)^{2} = (3 m)^{2}

\Rightarrow 3 (25 q^{2}) = 9 m^{2}

\Rightarrow 25 q^{2} = 3 m^{2}

\Rightarrow q^{2}

is divisible by 3

\Rightarrow q

is divisible by 3

- - (3)

From (2) and (3), there is a common factor for p and q, which contradicts (1).

Hence,

5 \sqrt{3}

is an irrational number proved.