Question

prove that 3√3 is a irrational number​

Answer 1

$\large\bf\underline {To \: prove:-}$

• 3√3 is irrational.

$\huge\bf\underline{Solution:-}$

Let us assume that 3√3 is rational

then there co -primes p and q where,

q ≠ 0

such that,

$\mapsto \rm \: 3 \sqrt{3} = \frac{p}{q} \\ \\ \mapsto \rm \: \sqrt{3} = \frac{p}{q} \times \frac{1}{ 3} \\ \\ \mapsto \rm \: \sqrt{ 3} = \frac{p}{3q}$

Since p and q are intigers,

So, p/3q is rational

But we know that√3 is irrational.

So, this contradiction is arissen because of our assumption.

therefore, 3√3 is irrational.

Hence Proved.

Answer 2

Answer:

Answer:

Let us assume that 3√3 is a rational number.

Hence, it can be written in p/q form, where p and q are integers and q is not equal to 0.

So, 3√3 = p/q

⇒ √3 = p/3q

Now, we can see that LHS is totally irrational number whereas the RHS is rational.

So, our assumption was wrong.

Hence, 3√3 is a irrational number.

Number system:

Numbers from 1, 2, 3.... are called natural numbers.

Adding 0 to natural numbers, we call it whole numbers.

Now, adding negative numbers to whole numbers, we call it integers.

Then, the numbers which can be written in the form of p/q, where p and q are integers, and q isn't equal to 0, is said to be a rational number.

And, irrational numbers are numbers which can not be written in the form of p/q, where p and q are integers, and q isn't equal to 0.

All the rational numbers, irrational numbers, also whole numbers, natural numbers and integers are real numbers.

Now, we have unreal numbers too, that is imaginary numbers which includes root of negative numbers.

Also, real numbers and imaginary numbers are subsets of a type of numbers called complex numbers which are numbers written in the form of a + ib. They have real part as well as imaginary part.