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Standard IX

Mathematics

Direct and Indirect Proof

To prove √2 a...

Question

To prove $\sqrt{2}$ as an irrational number-

(i) We assume $\sqrt{2} = \frac{a}{b}, (b \neq 0)$ such that a and b are co-primes.

(ii) On simplifying, we get 2 divides a, which results in contradiction.
Which of the above two statements is/are correct?

Only (i)

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Only (ii)

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Both (i) and (ii)

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None

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Solution

The correct option is A Only (i)

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Similar questions

Q. To prove

\sqrt{2}

as an irrational number-

(i) We assume

\sqrt{2} = \frac{a}{b}, (b \neq 0)

such that a and b are co-primes.

(ii) On simplifying, we get 2 divides a, which results in contradiction.
Which of the above two statements is/are correct?

Q. Fill in the blanks to complete the proof of

2 - \sqrt{5}

as an irrational number, provided

\sqrt{3}

is an irrational number.

Let’s assume

2 - \sqrt{5}

be a rational number

\Rightarrow 2 - \sqrt{5} = \frac{a}{b}, b \neq 0

\Rightarrow \sqrt{5} =

(i)___________ which is a (ii)_________ number.

\Rightarrow \sqrt{5}

is a rational number.
But

\sqrt{5}

is an irrational number.
Hence, we have arrived at a contradiction.

∵ 2 - \sqrt{5}

is an irrational number.

Q. Fill in the blanks to complete the proof of

2 \sqrt{5}

as an irrational number, provided

\sqrt{5}

is an irrational number.

Let’s assume

2 \sqrt{5}

be a rational number

2 \sqrt{5} = \frac{a}{b}, b \neq 0

\Rightarrow \sqrt{5} =

(i)___________ which is a (ii)_________ number.

\Rightarrow \sqrt{5}

is a rational number.
But

\sqrt{5}

is an irrational number.
Hence, we have arrived at a contradiction.

∵ 2 \sqrt{5}

is an irrational number.

Q. Fill in the blanks to complete the proof of

1 + 7 \sqrt{3}

as an irrational number, provided

\sqrt{3}

is an irrational number.

Let’s assume

1 + 7 \sqrt{3}

be a rational number

\Rightarrow 1 + 7 \sqrt{3} = \frac{a}{b}, b \neq 0

\Rightarrow \sqrt{3} =

(i)___________ which is a (ii)_________ number.

\Rightarrow \sqrt{3}

is a rational number.
But

\sqrt{3}

is a/an irrational number.
Hence, we have arrived at a contradiction.

∵ 1 + 7 \sqrt{3}

is an irrational number.

Q. On simplifying

(a + b)^{3} + (a - b)^{3} + 6 a (a^{2} - b^{2})

, we get.

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To prove √2 as an irrational number- (i) We assume √2=ab,(b≠0) such that a and b are co-primes. (ii) On simplifying, we get 2 divides a, which results in contradiction. Which of the above two statements is/are correct?

The correct option is A Only (i)

To prove $\sqrt{2}$ as an irrational number-

(i) We assume $\sqrt{2} = \frac{a}{b}, (b \neq 0)$ such that a and b are co-primes.

(ii) On simplifying, we get 2 divides a, which results in contradiction.
Which of the above two statements is/are correct?