Question

Show that 5 – √3 is irrational.​

Answer 1

Step-by-step explanation:

Let us assume (5−3–√) is a rational number.

Where a rational number is a number which is represented in the form of ab ,

where b≠0 and a and b has not any common factors except 1.

Then it can be represented as a fraction of two integers.

Let the lowest terms representation be:

(5−3–√)=ab , whereb≠0.

⇒3–√=−ab+5=−a+5bb

Now let −a+5b=p

⇒3–√=pb

Now squaring on both sides we have,

⇒3=p2b2

⇒p2=3b2………………………………. (1)

From above let p2 is even therefore p should also be even.

Let p=2c (c is constant and 2c is an even number)

Squaring both sides we have,

⇒p2=4c2………………… (2)

From equation (1) and (2) we can say that

4c2=3b2

Therefore from above 3b2 is even therefore b2 should also be even and again b is even.

Therefore p and b both are even.

Therefore p and b have some common factors.

But p and b were in lowest form and both cannot be even.

Hence, the assumption was wrong and hence, (5−3–√) is an irrational number.

So, (5−3–√) is an irrational number

Answer 2

Answer:

let us assume that 5 - √3 is irrational.

we can find coprime a and b ( b≠0 ) such that 5-√3=a/b

therefore , 5-a/b=√3

rearranging the equation we get

√3=5-a/b

= (5b-a)/b

Since, A and B are integer

we get :-

5-a/b is a rational

and so √3 is rational.

but, √3 is irrational

so we conclude that 5 -√3 is irrational

Step-by-step explanation:

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