[SOLVED] Find the values of p and q

• April 27th 2008, 01:43 AM
anne.
[SOLVED] Find the values of p and q
(2√3 - 1)² = p + q√3

Does anyone know how to solve this?
So far I've only got:

13 - 4√3 = p + q√3

However, as with a similar problem I have done, I am pretty sure there are a range of answers, not just p = 13 and q = -4.

Can someone help me with this?
• April 27th 2008, 01:49 AM
Moo
Hello,

It may depend on how you define p and q. Do they belong to $\mathbb{N}$, or $\mathbb{R}$ ?
• April 27th 2008, 01:57 AM
Isomorphism
Quote:

Originally Posted by Moo
Hello,

It may depend on how you define p and q. Do they belong to $\mathbb{N}$, or $\mathbb{R}$ ?

Or $\mathbb{Q}$, ya know, the rationals :)
• April 27th 2008, 01:58 AM
Moo
If it belongs to $\mathbb{Q}$ there can't be several solutions, just one :)

By the way, I meant $\mathbb{Z}$, certainly not $\mathbb{N}$
• April 27th 2008, 01:59 AM
anne.
The question did not contain any further detail, also, I do not know what $\mathbb{N}$ or $\mathbb{R}$ are...we haven't learned it, so I don't think they belong to....those things. (Itwasntme)

Erm....could someone explain what those large, nice looking letters mean? (Worried)

Thanks. (Bow)(Bow)(Bow)
• April 27th 2008, 02:03 AM
Moo
$\mathbb{N}$ : set of natural integers (means the positive ones)

$\mathbb{Z}$ : set of all integers (positive and negative)

$\mathbb{Q}$ : rational numbers, which means $\frac pq$ with p and q in $\mathbb{Z}$ (and q different from 0)

$\mathbb{R}$ : set of all numbers, including irrational ones.

$\mathbb{N}$ is included in $\mathbb{Z}$, which is included in $\mathbb{Q}$, which is included in $\mathbb{R}$

-----------------------------

Why did I ask ? Because if we're working on $\mathbb{Z}$, there is only one solution.

If we're working on $\mathbb{R}$, then $p=10-5\sqrt{3}$ and $q=1+\sqrt{3}$ are solutions, for example, but there is an infinity of possibilities. (note that $\sqrt{3}$ is an irrational and only belongs to $\mathbb{R}$).
• April 27th 2008, 02:08 AM
Isomorphism
Quote:

Originally Posted by Moo
If it belongs to $\mathbb{Q}$ there can't be several solutions, just one :)

By the way, I meant $\mathbb{Z}$, certainly not $\mathbb{N}$

In the most general sense, shouldnt you be saying $\mathbb{Q}$, since they include $\mathbb{Z}$(this is where uniqueness of the representation holds)?
• April 27th 2008, 02:09 AM
anne.
Well the question did not state anything, so I'd assume the solution would belong to $\mathbb{R}$.
• April 27th 2008, 02:17 AM
Moo
So the solutions will be

$p=a+b\sqrt{3}$ and $q=c+d\sqrt{3}$

$p+q\sqrt{3}=a+b\sqrt{3}+c\sqrt{3}+3d$

So we want $a+3d=13$ and $b+c=-4$

Hence a=13-3d and b=-4-c.

-------> general solutions are $p=13-3d-(4+c)\sqrt{3}$ and $q=c+d \sqrt{3}$, this time, c and d belonging to $\mathbb{Q}$ (yep Isomorphism, it can be done this way indeed :))

This is only for the ones involving $\sqrt{3}$ the basic way. Otherwise, there is an infinity of them.
• April 27th 2008, 02:19 AM
Moo
Since they ask you the "values of p and q", I'd follow the idea that the solutions are simply p=13 and q=-4
• April 27th 2008, 02:20 AM
Isomorphism
Quote:

Originally Posted by anne.
Well the question did not state anything, so I'd assume the solution would belong to $\mathbb{R}$.

Then you have $\mathbb{R}$ solutions :)

Pick any real number. Let it be p. Substitute in your equation and solve for q, another real number!

From my experience, I think you are asked to find rational values(that is the numbers from $\mathbb{Q}$ and not $\mathbb{R}$).

Doing it for $\mathbb{R}$ looks uneducational (Thinking)
• April 27th 2008, 02:30 AM
anne.
O_O;;

wow. I didn't get a word of the second-last post that you posted (the one with all the equations...)

I'll be sticking with the simple solution of p = 13 and q = -4 then.