Question

prove that root 5 minus root 3 is not a rational number​

Answer 1

First of all, we have to prove that √5 is an irrational number. By similar procedure, it can also be proved that √3 is also an irrational number.

After proving, suppose (let us assume) that √5 - √3 is rational number (say, r)

Then, √5 - √3 = r (r ≠ 0)

Or, √5 = r + √3

As "r" is rational, so (r+√3) is also rational , and hence √5 is also rational, since (r+√3) is equal to √5.

But, this contradicts our assumption since we know √5 is irrational number.

Therefore, √5 - √3 is an irrational number.

One picture is mentioned regarding how to prove that √5 is an irrational number.

Answer 2

Answer:

$\sqrt{5} -\sqrt{3}$ is  not a rational number proved .

Step-by-step explanation:

Explanation :

Given ,$\sqrt{5} -\sqrt{3}$

Let $\sqrt{5} -\sqrt{3}$ be a rational numbers  

Now , let $\sqrt{5} -\sqrt{3} = \frac{a}{b}$

Step1 :

Squaring both side we get ,

$(\frac{a}{b}) ^{2} = (\sqrt{5} -\sqrt{3})^{2}$

$(\frac{a}{b} )^{2} = 5 + 3 -2\sqrt{5} .\sqrt{3}$

⇒$(\frac{a}{b})^{2} = 8 - 2\sqrt{15}$

⇒$\sqrt{15} = \frac{8b^{2} -a^{2} }{2b^{2} }$

Hence , x is a rational number , so $x^{2}$ is also a rational number .

Therefore , $8b^{2} -a^{2}$ is a rational number.......(i)

⇒$\frac{8b^{2} -a^{2} }{2b^{2} }$  = $\sqrt{15}$is also rational number

But , $\sqrt{15}$ is a irrational number .

∴ Therefore , $\frac{8b^{2} -a^{2} }{2b^{2} }$ is also a irrational number . ..............(ii)

Therefor , from (i )and (ii) we arrive at a contradiction .

Final answer :

Hence , $\sqrt{5} -\sqrt{3}$ is a irrational number .

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