Question

(1) Show that 4 √2 is an irrational number.

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Answer 1

Solution :

• In this question we have to prove that 4√2 is and irrational number.

• Let 4√2 be an rational number.

Then, 4√2 = p/q

Shifting 4 from LHS to RHS :

→ √2 = p/4q

Now,since p and q are integers. So, p/4q is a rational number and thus √2 also be rational number. But it is the fact that √2 is an irrational number.

⛬ This is a contradiction our assumption is wrong

⛬ 4√2 is and irrational number .

Related concept :-

• An irrational number is a real number that can't be written as a fraction.

• A rational number is a real number such as a whole number, fraction, decimal, or integer.

• The set of counting numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ...} is called the set of natural numbers.

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large\underline{\red{\bf To \: Prove :-}}$

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• 4 √2 is an irrational number

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$\large\underline{\orange{\bf Solution :-}}$

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• Let us assume 4√2 to be a rational number

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We know that any rational number can be expressed in the form of $\sf \dfrac { p } { q }$ where q ≠ 0

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So,

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➜ $\sf 4 \sqrt 2 = \dfrac { p } { q }$

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➨ $\sf \sqrt 2 = \dfrac { p } { 4q }$

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Here as we assumed "p" & "q" are integers and "4" too is an integer . Thus $\sf \dfrac { p } { 4q }$ is a rational number

So we expressed $\sf \sqrt 2$ as a rational . But it contradicts the fact that √2 is an irrational number.

This contradiction has been arisen due to our wrong assumption that 4√2 is a rational number hence 4√2 is an irrational number

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Proved

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