Question

if √3 is an irrational no. prove that 2√3-7 ia an irrational no.
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Answer 1

GIVEN:

• 2√3–7

TO FIND:

• Prove that 2√3–7 is an irrational number.

SOLUTION:

Let 2√3–7 is an irrational number, which can be written in the form of p/q, Where p and q are coprimes and q ≠0

According to question:-

$\sf{\longmapsto 2 \sqrt{3} -7 = \dfrac{p}{q}}$

$\sf{\longmapsto 2 \sqrt{3} = \dfrac{p}{q} + 7 }$

$\sf{\longmapsto 2 \sqrt{3} = \dfrac{p +7q}{q} }$

$\sf{\longmapsto \sqrt{3} = \dfrac{p+7q}{2q}}$

Since, p and q are integers so p+7q/2q is Rational, and √3 is rational.

But this contradicts the fact that √3 is Irrational.

So, we conclude that, √3 is an irrational number.

❝ Hence, 2√3–7 is an Irrational number ❞

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Answer 2

Question:

if √3 is an irrational no. prove that 2√3-7 ia an irrational no.

Given:

★ √3 is a irrational number

★2√3-7

Solution:

Let us assume that 2√3-7 is a rational number, we can written as a/b , a and b are co prime numbers, where b≠0

Now,

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→→→2√3-7=a/b

→→→2√3=a-7b/b

→→→√3=a-7b/2b

we know that

a-7b/2b is a rational number, but √3 is a irrational number.

so, our assumption is wrong.

2√3-7 is a irrational number.

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