Math Q (1 Viewer)

axwe7 · Dec 23, 2015

Just do the last question - (f)

What I wanted to know, was that do I have to prove that root x is rational, and thereby prove a contradiction to find the values of a and b or do I just correspond the pronumerals with their respective values?

Cheers,
Axwe7

InteGrand · Dec 23, 2015

axwe7 said:
View attachment 32748

Just do the last question - (f)

What I wanted to know, was that do I have to prove that root x is rational, and thereby prove a contradiction to find the values of a and b or do I just correspond the pronumerals with their respective values?

Cheers,
Axwe7

$\noindent I think they want you to assume that $\sqrt{x}$ is irrational. Then the fact that $a$ and $b$ are rational means that $a$ has to be $\frac{2}{3}$ and $b$ has to be $3$. This is based on the fact that if $\alpha$ is an irrational number and $a,a^\prime ,b,b^\prime$ are rational numbers, and $a+b\alpha = a^\prime + b^\prime \alpha$, then $a=a^\prime$ and $b=b^\prime$. This can be proved as an exercise for the reader (you may wish to use proof by contradiction). If $\sqrt{x}$ is rational, then $a$ and $b$ are not uniquely determined.$

axwe7 · Dec 23, 2015

InteGrand said:
$\noindent I think they want you to assume that $\sqrt{x}$ is irrational. Then the fact that $a$ and $b$ are rational means that $a$ has to be $\frac{2}{3}$ and $b$ has to be $3$. This is based on the fact that if $\alpha$ is an irrational number and $a,a^\prime ,b,b^\prime$ are rational numbers, and $a+b\alpha = a^\prime + b^\prime \alpha$, then $a=a^\prime$ and $b=b^\prime$. This can be proved as an exercise for the reader (you may wish to use proof by contradiction). If $\sqrt{x}$ is rational, then $a$ and $b$ are not uniquely determined.$

So I actually don't need any working out, I could just state that a= A and b=B?

I did the question with working out, and I also stated the contradiction, however there is not contradiction as the question doesn't say that root x is irrational,

leehuan · Dec 23, 2015

If you look at the previous parts, I think the assumption to be made here is that sqrt(x) is irrational here. However the case when x is a perfect square is interesting:

To visualise, consider the simplest case of x=1:

Then a+b=2+2/3
a+b=8/3

Well, we can literally pick any two rational numbers a and b that will satisfy a simple statement such as this.

We then move up to x=4:

a+2b = 2 + 4/3
a+2b=10/3

We can consider the next case of x=9:

a+3b=2+2
a+3b=4

3 is secure enough, but x=16 just for the sake of even more reliability:
a+4b=2+8/3
a+4b=14/3

No matter which pair of simultaneous equations we solve, we arrive at THE SAME VALUES FOR A AND B so it doesn't matter

axwe7 · Dec 23, 2015

leehuan said:
If you look at the previous parts, I think the assumption to be made here is that sqrt(x) is irrational here. Like InteGrand stated, if sqrt(x) was rational or by equivalent x is a perfect square, then there are an infinite number of solutions for a and b.

Understood, thanks.

leehuan · Dec 23, 2015

axwe7 said:
Understood, thanks.

Actually, please check my edit. I tried doing an investigation.

axwe7 · Dec 23, 2015

leehuan said:
Actually, please check my edit. I tried doing an investigation.

Hahah, yeah, just did, my message was posted a bit late, idk why.

Just one last question, why does this relate to how a number can be written in a fractional form or if it can't. How is that the determining factor of the values of a and b?

leehuan · Dec 23, 2015

axwe7 said:
Hahah, yeah, just did, my message was posted a bit late, idk why.

Just one last question, why does this relate to how a number can be written in a fractional form or if it can't. How is that the determining factor as to the values of a and b?

I'm not sure what you mean here. Given a and b are rational it just means we can express it as a fraction; doesn't mean it has to be written as one (e.g. integers such as 2). It just means the answer isn't something like pi.

Math Q (1 Viewer)

axwe7

Member

InteGrand

Well-Known Member

axwe7

Member

leehuan

Well-Known Member

axwe7

Member

leehuan

Well-Known Member

axwe7

Member

leehuan

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)