1. ## Multiply Two Factors

I am just looking through my course notes right now.

I notice the prof went to multiply two terms and I am kind of lost on the method he used here.

A problem here is that I don't know how to show root of with my keyboard LOL. I am sure someone will

Is this correct or did I copy something down wrong?

2. ## re: Multiply Two Factors

Originally Posted by ehpoc
I am just looking through my course notes right now.

I notice the prof went to multiply two terms and I am kind of lost on the method he used here.

A problem here is that I don't know how to show root of with my keyboard LOL. I am sure someone will

Is this correct or did I copy something down wrong?
Do you know how to expand brackets using FOIL?

3. ## re: Multiply Two Factors

$$\sqrt{2}$$ produces $\sqrt{2}$.

Originally Posted by ehpoc

Is this correct or did I copy something down wrong?
This is correct. Just use distributivity and note that $\sqrt{2}\sqrt{2}=2$.

4. ## re: Multiply Two Factors

Ok there were two reasons why this confused me.

1. the 2bd which should be clear to me.....ooops

2.In my notes where there should have been a "c" where there was an "a" so I was getting a^2 haha ...math in the morning.

But now I have another similar related question about some simple algebra

I have

(a-b√2)/(a^2-2b^2) =( a/a-2b^2)-(b/a^2+2b^2) where did the √2 go?
EDIT: For this one I should have added context. Because from a similar proof it removes the √2 because a and b are rational. I still dont get the nest part though...

From that he went to a^2-2b^2 Not sure what happened here.

5. ## re: Multiply Two Factors

Originally Posted by ehpoc
(a-b√2)/(a^2-2b^2) =( a/a-2b^2)-(b/a^2+2b^2) where did the √2 go?

$\frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a-b\sqrt{2}}{(a-b\sqrt{2})(a+b\sqrt{2})}=\frac{1}{a+b\sqrt{2}}$

6. ## Re: Multiply Two Factors

You can see it in contest in this link.

Numbers of Type Rational a plus b root 2 Form a Field - ProofWiki

7. ## Re: Multiply Two Factors

Originally Posted by ehpoc
(a-b√2)/(a^2-2b^2) =( a/a-2b^2)-(b/a^2+2b^2) where did the √2 go?
I don't see this equation in the link you provided. The link says that the inverse of $a+b\sqrt{2}$, which is $\left({\dfrac {a - b \sqrt 2} {a^2 - 2b^2}}\right)$, equals $\dfrac a {a^2 - 2b^2} - \dfrac {b \sqrt 2} {a^2 - 2b^2}$, with $\sqrt{2}$ in the nominator of the second term. The coefficients $\dfrac a {a^2 - 2b^2}$ and $\dfrac {b} {a^2 - 2b^2}$ are rational number when $a$ and $b$ are.

By the way, according to the standard arithmetic operation precedence, a/a-2b^2 means (a/a)-2b^2 = 1 - 2b^2. It should probably be a/(a^2-2b^2).

8. ## Re: Multiply Two Factors

The equation in my notes is doing the same proof same proof as the link I provided. In that one too the √2 seems to disappear. Whatever though I will just go see my prof about this.