Simplifying rational numbers help

[nobahar]

Express
$$\frac{1+\sqrt{2}}{3-\sqrt{2}}$$ as $$a+b\sqrt{2}$$ where a and b are rational numbers.

I started by
$$\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{3+\sqrt{2}}{3+\sqrt{2}}$$

But, I obtain

$$\frac{5}{7}-\frac{4}{7}\sqrt{2}$$

I believe that, here, a and b are rational, but is there a more tidy version? I tried playing with the square root so that it is a multiple of 2, so that I could 'split' it into two square roots, on of the square root of two so that I could use it to cancel out the square root of two on the bottom, then I could also remove the other square root if it was a rational square root. For example, I tried:

$$\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{x+\sqrt{8}}{x+\sqrt{8}}$$
because
$$\sqrt{8}= \sqrt{2*4} = \sqrt{2} * \sqrt{4} = 2\sqrt{2}$$

I haven't specified x, since its just an example. I actually tried making a relationship between x and the square root I introduced, since I could represent it algebraically as
$$\sqrt{g}$$
where g is a multiple of two, would give
$$\sqrt{2}*\sqrt{\frac{1}{2}g}$$.

I'm guessing its more straight forward than this.
Thanks in advance.

 

[Mark44]

Express
$$\frac{1+\sqrt{2}}{3-\sqrt{2}}$$ as $$a+b\sqrt{2}$$ where a and b are rational numbers.

I started by
$$\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{3+\sqrt{2}}{3+\sqrt{2}}$$

But, I obtain

$$\frac{5}{7}-\frac{4}{7}\sqrt{2}$$

The sign in the middle is wrong -- should be +.

I believe that, here, a and b are rational, but is there a more tidy version?

I don't believe there is. Assuming you make the correction I mentioned, you will have written the original expression in the form a + b*sqrt(2), where a and be are rational. You can't get any tidier than that.

I tried playing with the square root so that it is a multiple of 2, so that I could 'split' it into two square roots, on of the square root of two so that I could use it to cancel out the square root of two on the bottom, then I could also remove the other square root if it was a rational square root. For example, I tried:

$$\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{x+\sqrt{8}}{x+\sqrt{8}}$$
because
$$\sqrt{8}= \sqrt{2*4} = \sqrt{2} * \sqrt{4} = 2\sqrt{2}$$

I haven't specified x, since its just an example. I actually tried making a relationship between x and the square root I introduced, since I could represent it algebraically as
$$\sqrt{g}$$
where g is a multiple of two, would give
$$\sqrt{2}*\sqrt{\frac{1}{2}g}$$.

I'm guessing its more straight forward than this.
Thanks in advance.

 

[Mentallic]

I believe that, here, a and b are rational, but is there a more tidy version? I tried playing with the square root so that it is a multiple of 2, so that I could 'split' it into two square roots, on of the square root of two so that I could use it to cancel out the square root of two on the bottom, then I could also remove the other square root if it was a rational square root. For example, I tried:

$$\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{x+\sqrt{8}}{x+\sqrt{8}}$$
because
$$\sqrt{8}= \sqrt{2*4} = \sqrt{2} * \sqrt{4} = 2\sqrt{2}$$

I haven't specified x, since its just an example. I actually tried making a relationship between x and the square root I introduced, since I could represent it algebraically as
$$\sqrt{g}$$
where g is a multiple of two, would give
$$\sqrt{2}*\sqrt{\frac{1}{2}g}$$.

I'm guessing its more straight forward than this.
Thanks in advance.

Quite ambitious, but no. Since the original expression is irrational, if you're going to simplify it, it will always still be irrational. That is why the $\sqrt{2}$ must be there. You can of course "simplify" it into many different ways, but there wil always be an irrational portion of the expression.

 

[nobahar]

Many thanks Mark44 and Mentallic.