Question

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Answer 1

${\bf{\red{(1)}}}$

Prove that √2is irrational.

SolutiOn:

Let if possible √2 is not irrational,i.e.√2 is rational number.

Then by diffinition of rational number,we can write

\sqrt{2} =\frac{p}{q}

where p and q are integers such that p and q have no common factor expect 1,i.e,

HCF(p,q) = 1

Now, squaring both sides,we get

$( \sqrt{2) {}^{2} } = \frac{p {}^{2} }{q {}^{2} } \\ = > \: 2 = \frac{p {}^{2} }{q {}^{2} } \\ = > \: 2q {}^{2} = {p}^{2} \\ = > \: \frac{2}{p {}^{2} } \: \: \: \:$

_________________(∴ 2/2q² and 2q² = p²)

=>. 2/p ..............(1)

=> p = 2r for some integer r

Therefore,

p² = 4r²

=> 2q² = 4r²

=> q² = 2r²

=> 2/q²

=> 2/q.....................(2)

From (1) & (2) we conclude that 2 is a common factor of p & q which is a contradiction, because we have assumed that p and q have no common factor expect 1.

Therefore ,our assumption is wrong .

Hence,√2 is an irrational number.

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${\bf{\red{(2)}}}$

Prove that 3+2√5 is irrational.

SolutiOn:

Let if possible 3+2√5is not irrational,i.e., it is rational.

Then, according to diffinition:

3+2\sqrt{5} = \frac{p}{q}

$= > \: 2 \sqrt{5} = \frac{p}{q} - 3 \\ i.e \: \sqrt{5} = \frac{p - 3q}{2q}$

=> \sqrt{5} is a rational number, which is a contradiction because\sqrt{5} is an irrational number.

Therefore, our assumption is wrong.

Hence,the given number is irrational.

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${\bf{\red{(3)}}}$

(i) Let 1/√2 is rational.

Then, by diffinition of rational number,we can write.

\frac{1}{ \sqrt{2} } =\frac{p}{q} ,q≠0

Squaring both sides,we get

$\frac{1}{2} = \frac{ {p}^{2} }{q {}^{2} } \\ 2 {p}^{2} = q {}^{2}$

• 2 divides q²
• 2 divides q...........(1)

=> p = 2r, for some r

=> 2p² = 4r²

=> p² = 2r²

• 2 divides p²
• 2divides p...........(2)

From (1) and (2) Nikon conclude that p and q have common factor 2 ,other that which is contradiction so, our assumption is wrong.

Hence, 1/√2 is irrational.

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(ii) Let 7√5 is not irrational.

by diffinition,

7√5 =p/q..........(1)

=> \sqrt{5} =p/7q.........(2)

Since,p and q are integers , therefore p and q are also integers.

=> p/7q is rational number.

i.e., RHS of (2) is rational therefore LHS ,i.e., √5 is also rational , which is a contradiction because √5 is irrational.

Hence, 7√5 is irrational.

(iii) same as question 2nd.

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