Show that if p or q is divisible by 2, so is the other. watch

SamKeene · 18-01-2015 14:31

Here is the question:

I haven't been able to start at all, I am not familiar with this type of question, any pointers? I had tried to assume the one of the variables is divisible and rewrite it as 2*D but that didn't seem to follow through.

A tip or two would be great

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StrangeBanana · 18-01-2015 14:38

Assuming p is divisible by 2 worked out for me

You can prove q is even from that, then assume q is divisible by 2 and do something similar to show that that implies p is even.

TenOfThem · 18-01-2015 14:40

(Original post by SamKeene)
Here is the question:

I haven't been able to start at all, I am not familiar with this type of question, any pointers? I had tried to assume the one of the variables is divisible and rewrite it as 2*D but that didn't seem to follow through.

A tip or two would be great

Setting p=2n and rearranging gives q^2(2n+q) =-8n

Is this possible if q is odd?

Beaten by an edit

StrangeBanana · 18-01-2015 14:43

(Original post by TenOfThem)
Beaten by an edit

That's a nice problem, OP, where's it from?

SamKeene · 18-01-2015 14:55

(Original post by StrangeBanana)
Assuming p is divisible by 2 worked out for me

You can prove q is even from that, then assume q is divisible by 2 and do something similar to show that that implies p is even.

(Original post by TenOfThem)
Setting p=2n and rearranging gives q^2(2n+q) =-8n

Is this possible if q is odd?

Beaten by an edit

Is the next step:

either

therefore divisible, or

$q^2 = -8a^3 q=\sqrt{-8}a^{\frac{3}{2}} q=2i\sqrt{2}a^{\frac{3}{2}}$
therefore divisible?

(Original post by StrangeBanana)

That's a nice problem, OP, where's it from?

From the 2009 Pure Maths entrance exam into Aber.

TenOfThem · 18-01-2015 15:00

(Original post by SamKeene)
Is the next step:

either

therefore divisible, or

$q^2 = -8a^3 q=\sqrt{-8}a^{\frac{3}{2}} q=2i\sqrt{2}a^{\frac{3}{2}}$
therefore divisible?

From the 2009 Pure Maths entrance exam into Aber.

No

ab = c does not mean that a or b is c UNLESS. C is 0

You just have to consider the fact that two odds multiplied give an odd

james22 · 18-01-2015 15:01

(Original post by SamKeene)
Is the next step:

either

therefore divisible, or

$q^2 = -8a^3 q=\sqrt{-8}a^{\frac{3}{2}} q=2i\sqrt{2}a^{\frac{3}{2}}$
therefore divisible?

From the 2009 Pure Maths entrance exam into Aber.

You seem to have lost a q^2 in your second line of working.

SamKeene · 18-01-2015 15:06

(Original post by TenOfThem)
No

ab = c does not mean that a or b is c UNLESS. C is 0

You just have to consider the fact that two odds multiplied give an odd

Of course, I'm getting flustered because I am so unused to this kind of question. Forgetting basic maths...

I still don't quite see it tho, even considering odd*odd = odd.

Could I say:

We know (q + 2a) * q^2 is even.

Therefore q^2 and (q + 2a) cannot be odd.

if q is odd then q^2 is odd and so is q+2a.

Therefore it must be even?

TenOfThem · 18-01-2015 15:07

(Original post by SamKeene)
Of course, I'm getting flustered because I am so unused to this kind of question. Forgetting basic maths...

I still don't quite see it tho, even considering odd*odd = odd.

I think I need another pointer >.<

If q us odd, then q+2n is odd and q^2 is odd

SamKeene · 18-01-2015 15:15

(Original post by TenOfThem)
If q us odd, then q+2n is odd and q^2 is odd

Just figured that out and edited it in the above post, and you just beat me to it :P

For the second part, I think I got it right:

If p and q are odd then:

is even (odd + odd = even)

but the RHS is odd, and odd*even = even.

Therefore p and q cannot be odd and solve (1).

Smaug123 · 18-01-2015 15:17

(Original post by SamKeene)
Just figured that out and edited it in the above post, and you just beat me to it :P

For the second part, I think I got it right:

If p and q are odd then:

is even (odd + odd = even)

but the RHS is odd, and odd*even = even.

Therefore p and q cannot be odd and solve (1).

Yup, quite right. Have you heard of modular arithmetic, by the way? It, along with a couple of simple results in it, makes this problem quite a lot easier conceptually. Don't worry if not.

SamKeene · 18-01-2015 15:19

(Original post by Smaug123)
Yup, quite right. Have you heard of modular arithmetic, by the way? It, along with a couple of simple results in it, makes this problem quite a lot easier conceptually. Don't worry if not.

No, but I know It's useful for STEP, do you have a good resource that I can learn it from?

Smaug123 · 18-01-2015 15:22

(Original post by SamKeene)
No, but I know It's useful for STEP, do you have a good resource that I can learn it from?

I seem to remember it's treated nicely in How to Think Like a Mathematician, which is an excellent book (accessible to people who haven't done any undergraduate maths at all) you should read. My memory might be faulty, though.

SamKeene · 18-01-2015 15:30

(Original post by Smaug123)
I seem to remember it's treated nicely in How to Think Like a Mathematician, which is an excellent book (accessible to people who haven't done any undergraduate maths at all) you should read. My memory might be faulty, though.

Thanks man, I have heard of that. Will have to have a look soon ^.^

Could you assist me a little on the last part?

If x can be written as the quotient of two integers then:

$x = \frac{p}{q}$

$\frac{p^3}{q^3} + \frac{p}{q} +1$

Now I can see we can quote our (b) result to show that the quotient of two integers cannot both be odd.

I see from (a) we know that if one of the integers is even, so is the other.

But how can I show that two even numbers cannot be a solution to the equation?

Smaug123 · 18-01-2015 15:34

(Original post by SamKeene)
Thanks man, I have heard of that. Will have to have a look soon ^.^

Could you assist me a little on the last part?

If x can be written as the quotient of two integers then:

$x = \frac{p}{q}$

$\frac{p^3}{q^3} + \frac{p}{q} +1$

Now I can see we can quote our (b) result to show that the quotient of two integers cannot both be odd.

I see from (a) we know that if one of the integers is even, so is the other.

But how can I show that two even numbers cannot be a solution to the equation?

Think about what that means about $x = \dfrac{p}{q}$ . That's the most oblique hint I can give - don't want to give the idea away, because it's quite pretty and it turns up all over the place

SamKeene · 18-01-2015 15:41

(Original post by Smaug123)
Think about what that means about $x = \dfrac{p}{q}$ . That's the most oblique hint I can give - don't want to give the idea away, because it's quite pretty and it turns up all over the place

Argh it's not enough! Stop teasing me Smaug! Just let me climax with hot loads of mathematical clarity!

When you say 'Think about what that means about $x = \dfrac{p}{q}$ ' are you specially saying to consider that equation when both p and q are even?

Is it something to do with if they are both even, they will cancel down until one of the integers is odd, which is not allowed?

Smaug123 · 18-01-2015 15:46

(Original post by SamKeene)
Argh it's not enough! Stop teasing me Smaug! Just let me climax with hot loads of mathematical clarity!

When you say 'Think about what that means about $x = \dfrac{p}{q}$ ' are you specially saying to consider that equation when both p and q are even?

Is it something to do with if they are both even, they will cancel down until one of the integers is odd, which is not allowed?

Precisely that. More concretely, suppose $x = \dfrac{p}{q}$ expressed in its lowest terms. Then at least one of is odd, as otherwise it wouldn't be in its lowest terms. But then , and we've already shown then that neither of can be odd. Contradiction.

Smaug123 · 18-01-2015 15:47

(Original post by SamKeene)
Argh it's not enough! Stop teasing me Smaug! Just let me climax with hot loads of mathematical clarity!

When you say 'Think about what that means about $x = \dfrac{p}{q}$ ' are you specially saying to consider that equation when both p and q are even?

Is it something to do with if they are both even, they will cancel down until one of the integers is odd, which is not allowed?

It's a pretty idea which can also be used to show that $\sqrt{2}$ is irrational, for instance.

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Show that if p or q is divisible by 2, so is the other. watch

Negatives of studying at Oxbridge

Brother erased my work on deadline day!

How do you hint at guys?

My mum won't let me wear shorts

Grade 9 in GCSE English - AMA

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