Question

please jaldi answer karo it's important

Answer 1

Assume that 4 - 5√3 is rational.

So that 4 - 5√3 can be written as p/q, where p, q are coprime integers and q ≠ 0.

Here it contradicts our earlier assumption that 4 - 5√3 is rational.

Because, the RHS, p²/q² subtracted from 91, is rational, where p²/q² = (p/q)² is rational as it is square of a rational number p/q, but while the LHS, 40√3, is irrational.

∴ 4 - 5√3 is irrational.

Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Concept}}}}$

Here in this question, we are asked to prove 4-5√3 is an irrational number. Here we are going to prove it by using contradictory method. First we will try to prove it rational and then if our statement is proved wrong that means the given number is an irrational number.

So let's do it!!

Let us assume to the contradiction that 4-5√3 is a rational number. So we can write it in the form of p/q where p and q are integers as well as co-prime.

$\large\tt 4-5\sqrt3=\dfrac{p}{q}$

Now transport 4 from LHS to RHS!

$\large\tt -5\sqrt3=\dfrac{p}{q}-4$

Now divide -5 from both LHS and RHS!

$\large\tt -5\sqrt3\div(-5)=\bigg(\dfrac{p}{q}-4\bigg)\div(-5)$

$\large\tt \sqrt3=\dfrac{p}{-5q}-\dfrac{4}{-5}$

We know that √3 is an irrational number. But value in RHS is rational.

As we know that irrational number can't be equal to rational number.

This contradicts the fact that√3 is an irrational number. This contradiction has arisen because of our incorrect assumption that 4-5√3 is a rational number.

Hence 4-5√3 is an irrational number.

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