Square Roots and Prime Numbers

[donnagirl]

For \(\displaystyle \sqrt{2 \cdot 3} \cdot \sqrt{a \cdot b} \cdot \sqrt{3 \cdot 5}\) to be a rational number, with a and b representing positive prime numbers, the sum a + b must equal 7.
Why is this the case? How can I show/prove this? I know everything reduces to \(\displaystyle \sqrt{90ab}\) but not sure how to show the rest of it..

 

[pka]

For \(\displaystyle \sqrt{2 \cdot 3} \cdot \sqrt{a \cdot b} \cdot \sqrt{3 \cdot 5}\) to be a rational number, with a and b representing positive prime numbers, the sum a + b must equal 7. Why is this the case? How can I show/prove this? I know everything reduces to \(\displaystyle \sqrt{90ab}\) but not sure how to show the rest of it..

We need squares in the radical. \(\displaystyle \sqrt{2\cdot3^2\cdot 5\cdot a\cdot b}\).

Thus \(\displaystyle a=2~\&~b=5\) gives \(\displaystyle \sqrt{2^2\cdot3^2\cdot 5^2}=30\)

 

[donnagirl]

But how do we know to choose those squares? Why must it be 2 and 5? I mean yes they work but is there a way to do this algebraically without guess/check?

 

[pka]

But how do we know to choose those squares? Why must it be 2 and 5? I mean yes they work but is there a way to do this algebraically without guess/check?

The is no. There is no strictly algebraic way of doing this.
One must understand the number theory behind the problem.

First, \(\displaystyle \sqrt{n}\) is irrational if \(\displaystyle n\) is not a perfect square.

Next, one needs the know the prime factorization theorem.

Here is an example: Suppose we know that \(\displaystyle \sqrt{3^3\cdot 7^2\cdot 11^7\cdot a\cdot b } \) is a natural number and \(\displaystyle a+b=14\), where \(\displaystyle a~\&~b\) are natural numbers.
Then \(\displaystyle a=3\text{ and }b=11\) OR \(\displaystyle b=3\text{ and }a=11\) because that is the only that \(\displaystyle 3^3\cdot 7^2\cdot 11^7\cdot a\cdot b\) is a perfect square with \(\displaystyle a+b=14.\)