I am supposed to "write each expression with a rational denominator"

Question

I am supposed to "write each expression with a rational denominator"

1/thesqrootof 3. I follow the book's explanation, until it suggests dividing the sq. root of 3 by the square root of 3 times the square root of three. I do not understand how that reduces to the square root of three divided by three. What happened to the square root signs? How come this is not the square root of nine?

Thanks for your clarification!

Answer 1

rational denominator is a whole number denomiator.

example:

1/sqrt3 = sqrt3/(sqrt3*sqrt3)= sqrt3 /3

your last question; sqrt 9 is 3, a rational number.

One last example

1/(1-sqrt2) multiply numerator and denominor by the conjugate. Look that term up.

1*(1+sqrt2)/[(1-sqrt2)(1+sqrt2)]=
(1+sqrt2)/(1-2)= - (1+sqrt2)

I bet you see something like this very soon.

Answer 2

To write the expression 1/√3 with a rational denominator, we need to eliminate the square root in the denominator.

To do this, we use a technique called rationalizing the denominator. It involves multiplying both the numerator and denominator by the conjugate of the denominator. In this case, the conjugate of √3 is also √3.

So, we have:

1/√3 * (√3/√3)

Multiplying the numerator and denominator together gives:

√3/(√3 * √3)

Simplifying the denominator, we have:

√3/√(3 * 3)

Since 3 * 3 = 9, we have:

√3/√9

Now, √9 is equal to 3, so we can simplify further:

√3/3

So, the expression 1/√3 with a rational denominator is √3/3.

It's important to note that the square root sign in the denominator disappears after simplifying because the square root of 9 is 3, and the original denominator was √3 * √3, which is equivalent to √9.