Maths Question - Help Needed

Gaverick · 28 Apr 2007 at 18:03

A and B are numbers

Where A = B + 2

A + B = A x B

Show that A and B are not integers

Angus-Higgins · 28 Apr 2007 at 18:07

Well:

B = A - 2

A(A - 2) = A + A - 2

A² - 2A = 2A - 2

A² - 4A + 2 = 0

Then solve to give:

3.414213562373095

Or:

0.585786437626904

(According to an online quadratic equation calculator).

Then simply substitute A into the first equation to give:

B = 1.414213562373095 or -2.585786437626904

Am I right?

Angus Higgins

Gaverick · 28 Apr 2007 at 18:16

Well if

A = 3.414213562373095

and B = 0.585786437626904

then the statement A + B = A x B doesn't hold true

Angus-Higgins · 28 Apr 2007 at 18:19

Gaverick said:
Well if

A = 3.414213562373095

and B = 0.585786437626904

then the statement A + B = A x B doesn't hold true

No, I meant "A" has two possible values (from solving the quadratic equation).

A = 3.414213562373095

Or:

A = 0.585786437626904

And:

B = 1.414213562373095

Or:

B = -2.585786437626904

Angus Higgins

Gaverick · 28 Apr 2007 at 18:28

Thanks for the help, that's cleared it up now

Angus-Higgins · 28 Apr 2007 at 18:30

It's interesting really, we had something similar in our mock examination for GCSE recently.

Almost the same question in fact.

Angus Higgins

Gaverick · 28 Apr 2007 at 18:40

Well it goes to show what happens when you don't study maths for 2 years

A* at GCSE and now 2 years on i can't even solve that

Sleepy · 28 Apr 2007 at 19:17

Angus-Higgins said:
No, I meant "A" has two possible values (from solving the quadratic equation).

A = 3.414213562373095

Or:

A = 0.585786437626904

Thats better written as A = 2 ± √2

But thats not an elegant way of solving the problem.

From AB=A+B then

B = AB-A = A(B-1)

So A = B/(B-1) given that A & B are both integers then then this equation cannot be true as B/(B-1) does not produce an integer solution except for when B = 2. However given that A = B+2 B=2 cannot be true for both equations. So as B is not an integer, then neither is A

There's going to be a better way to express the above.

Angus-Higgins · 28 Apr 2007 at 19:41

Sleepy said:
But thats not an elegant way of solving the problem.

Yes, but it does involve very simple Mathematics that most people understand, and it was the first way of solving it. I saw it as the most simple way to solve the problem and as "help" was "needed", I decided that using a simple method would be easier than a more complex one.

I, personally, like to see nice proofs, but I felt it would be acceptable to simply do something that would solve the problem rather than something that could be less obvious.

Angus Higgins

|Ric| · 28 Apr 2007 at 21:36

Sleepy said:
But thats not an elegant way of solving the problem.

Um I'm not convinced that it does solve the problem, it says show that they can't be integers. All you have done is found a solution that isn't an integer.

I'm not sure how satisfactory it is because you need to make it a bit more obvious that solving the quadratic gives all possible answers.

Sleepy · 28 Apr 2007 at 21:43

|Ric| said:
Um I'm not convinced that it does solve the problem, it says show that they can't be integers. All you have done is found a solution that isn't an integer.

Nope I arrived at an answer that produces one integer solution - which the other initial condition didn't allow, thus ruling out that specific answer. And a general condition which rules out integer answers other than the aforementioned. Thus demonstating that A & B are not integers.

|Ric| · 28 Apr 2007 at 22:00

I agree with your solution, but the one by Angus arguably doesn't answer the question let alone it not being elegant.

Angus-Higgins · 28 Apr 2007 at 22:10

Is this OK then:

A = B + 2

Therefore:

B + 2 + B = B(B + 2)

2B + 2 = B² + 2B

2 = B²

B = 2^(1/2)

?

Or am I not allowed to assume 2^(1/2) is irrational and thus not an integer? (One could prove this, but I don't know the proof).

Angus Higgins

Rebelius · 28 Apr 2007 at 22:11

|Ric| said:
the one by Angus arguably doesn't answer the question let alone it not being elegant.

why not? it's a direct proof - he's turned the given equations into a single quadratic equation in A and shown that the solutions to this are 2 ± √2 with corresponding solutions for B being ±√2.

√2 is not an integer, it's 1.414...

so if you accept that there are only 2 solutions to a quadratic equation (fundamental theorem of algebra) then there can be no solution to this problem with A and B integer.

He's actually done more than what is needed.

edit: in the above post he has done what I would do and just put A=B+2 into the 2nd equation and shown that B cannot be an integer, thus A and B can't both be integers.

Sleepy · 28 Apr 2007 at 22:38

Angus-Higgins said:
Is this OK then

I would say that your answer is the simplest way to go about solving the problem. However depending on the level of exam, it might not be enough for the examiners, in that your answer does not reveal any deeper understanding of integers and how they are defined.

BTW proving the square root of a prime is irrational isn't very hard, a quick google should list a number of different yet simple proofs.

Rebelius · 28 Apr 2007 at 22:51

integers are whole numbers - you don't need to show that root 2 is irrational, the fact that it's between 1 and 2 is sufficient to show it's not an integer.

simple since 1<2<4 so 1<√2<2

what has A-H omitted from his solution that could be essential at any level? a proof of the fundamental theorem of algebra?