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# Maths Café

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# Maths in pubs

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stevencarrwork 17 January 2010 00:15
Posts680
Joined28/07/09
I gave some Foundation level maths problems to my friends in the pub tonight.

They found them very interesting.

One example, prove that the 1st April and 1st July are always on the same day of the week as each other in every year.

This sort of problem is very useful in real life, but the important thing is that my friends were intrigued by it, and came up with 2 different ways of solving it - both different from the way that I would do it.

Another one was two identical planks are put together so that they overlap by 8 feet. The total length of the arrangement is 30 feet. How long is each plank? (Answer - 19 feet)

If people are happy to do maths problems in pubs, how can we get students to be motivated by them? Perhaps giving them a couple of lagers first?
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baius 17 January 2010 13:11
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Joined03/02/09
That's a new twist on giving students "incentives".
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Rebecca_Hanson 17 January 2010 23:44
Posts1105
Joined28/01/08

Sometimes intake of prohibited substances can shut the disruptive students up to allow you to properly engage the others.

Bit unpredictable though.  And best to run with whatever they sort out for themselves - for the sake of your career.

Let them buy you a drink when you meet them in the pub in a year or so and they are ready to apologise for and regret being so disaffected.  Then you could try the maths problems....... or maybe not!

On the slightly less dark side, remember by year 11 star stamps, smiley stamps and so on all start to work again.

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Christine_Jones 18 January 2010 21:03
Posts770
Joined09/10/09
Steven, I remember working with those exact problems with 2 of my GCSE students. They did not show any real kind of interest in the problems. I don't know if lager would work as I think they all drink alco-pops now don't they?

The real difference is in the pub it doesn't matter if you get the question wrong and you can argue about how you would work the answer out and no one judges you. In school they are constantly under scrutiny for every little action they make.

One example, prove that the 1st April and 1st July are always on the same day of the week as each other in every year.

This sort of problem is very useful in real life, but the important thing is that my friends were intrigued by it, and came up with 2 different ways of solving it - both different from the way that I would do it.

My first question is why do you say this information is useful in real life? In what way is that useful to know? I must be missing something because I'm pretty sure that I could manage in life without knowing that those two dates fall on the same day of the week.

The other question is what are the three different ways that you and your mates came up with to solve this problem? I can't come up with two different methods let alone three!
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stevencarrwork 18 January 2010 21:52
Posts680
Joined28/07/09
When I worked at Nissan, and we worked out delivery dates as being so many weeks and days from a build date on a certain day, it was important to know exactly when that delivery date would be.

If you know what day of the week a certain date is on, and can calculate what day of the week a later date is on, then you can avoid making plans for Saturdays when you already have something on etc.

So those sorts of calculations are useful.

One person took 1/4 as a Monday, and calculated 8,15 ,22, 29, 6 May , 13 May etc until getting to 1 July.

Another took 30 days for April as being remainder 2 when divided by 7, 31 is remainder 3, 30 is remainder 2 . The three remainders add up to 7.

I added 30, 31 and 30 and got 91 which is 13 times 7.

You should be able to argue about how you would work things out. When I do 1 on 2 tuition, I try to get my students to talk to each other about how to solve problems. That is really difficult.

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Christine_Jones 19 January 2010 08:41
Posts770
Joined09/10/09

When I worked at Nissan, and we worked out delivery dates as being so many weeks and days from a build date on a certain day, it was important to know exactly when that delivery date would be.

If you know what day of the week a certain date is on, and can calculate what day of the week a later date is on, then you can avoid making plans for Saturdays when you already have something on etc.

So those sorts of calculations are useful.

I stand corrected. I guess it was then important to know about the other matching dates of September and December. I guess because I would just check on a calendar/diary I just couldn't conceive of this information being that useful to people. However I presume that it's only useful during the months of April and September and for activities projected to happen 3 months in the future.

One person took 1/4 as a Monday, and calculated 8,15 ,22, 29, 6 May , 13 May etc until getting to 1 July.

Another took 30 days for April as being remainder 2 when divided by 7, 31 is remainder 3, 30 is remainder 2 . The three remainders add up to 7.

I added 30, 31 and 30 and got 91 which is 13 times 7.

The problem for me is I consider all these answers to be essentially arithmetic so I don't really see them as that different. The first and last one are exactly the same. i would have usefd the same method as you but all we are really doing is exploiting our knowledge of arithmetic in a more efficeint way than the first suggestion.

The second suggestion above is interesting though. Who ever suggested this has to have some mathematical background. People very rarely use modulus arithmetic unless they have been shown it in the past.

You should be able to argue about how you would work things out. When I do 1 on 2 tuition, I try to get my students to talk to each other about how to solve problems. That is really difficult.

I agree that you should be able to argue about how you would work things out. I can justify my own methods and can understand other peoples methods but my biggest problem is finding multiple methods to attack a single problem.

For instance in the OU course I'm doing we had the following task to look at,(sorry it hasn't formatted that well):

$1\frac{1} {2}$x3=$1\frac{1} {2}$+3

$1\frac{1} {3}$x4=$1\frac{1} {3}$+4

$1\frac{1} {4}$x5=$1\frac{1} {4}$+5

What is the pattern here? Does this pattern continue?

This is a no-brainer, using algebra the structure of the pattern is revealed, problem finished. However one student had thought of this problem in terms of pictures. Firstly 3 lots of $1\frac{1} {2}$ strips then a single $1\frac{1} {2}$ and 3 lots of 1 strips. If you draw this it is very obvious how you can get from one image to the other.

This is absolutely brilliant as far as teaching students is concerned. I think even KS2 pupils could understand what was happening with this problem and how it generalises.

That's the real difficulty I have, I would never have thought in terms of pictures like that. In my defence the pictures don't contain a recognised proof but the point is to make mathematics understandable to pupils and the picture idea wins.

It's a bit like solving simultaneous equations. You can show it graphically so that students get the hang of what you are trying to do but in the end they must use algebra to solve them.

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stevencarrwork 19 January 2010 11:32
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Joined28/07/09
You can just look up these dates on diaries, but if you are programming computers to calculate delivery dates, you have to know the maths behind it.

I tend to turn straight to algebra to solve problems, because it is a tool I know I can use to solve problems.

But pictures are often more useful for teaching. Maths is more than just symbol processing, even if processing symbols is the best way to get the answer to a maths problem.
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MarkDawes 19 January 2010 13:51
Posts318
Joined09/04/07

Thank you, Christine, for posting such an interesting problem.  I hadn't seen it before.

I realised a while ago that I learnt maths from algebraists and that I don't readily or easy think like a geometer.  In many situations, though, while the algebra will provide a watertight proof, it doesn't explain why.  My favourite proofs are those that explain as well as proving, and I think the graphical proof you mention here does just that.

Christine_Jones wrote:
What is the pattern here? Does this pattern continue?
This is a no-brainer, using algebra the structure of the pattern is revealed, problem finished. However one student had thought of this problem in terms of pictures.

It may well be the case that the structure of the pattern is revealed by using algebra, but I don't think this explains the structure of the mathematics

I would encourage others to think about what a diagrammatical proof might look like for this.

I have created my own diagrams, but have posted them in a Word document (so I don't spoil it for everyone else!).  I would welcome thoughts on this.

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tombutton 19 January 2010 14:03
Posts89
Joined21/07/06
These are nice diagrams Mark, and I think there are often situations like these where a diagramatical proof gives more insight than the algebra.

If you haven't seen Roger B Nelson's "Proof Without Words" I highly recommend it (it can be found on Amazon).  I've used some these with A level students/teachers and they work really well.

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Christine_Jones 19 January 2010 14:23
Posts770
Joined09/10/09
Mark, I like your diagrams. The ones I discussed above only looked at one case whereas yours cover the general case.

I have always turn to algebra when I come across problems like this. No help to my students but there you go.

You may be interested in these similar problem, from the OU course:

1) 2 - $\frac{2} {3}$ = 2 x $\frac{2} {3}$

3 - $\frac{3} {4}$ = 3 x $\frac{3} {4}$

4 - $\frac{4} {5}$ = 4 x $\frac{4} {5}$ ...

2) 1 x $\frac{1} {2}$ = 1 - $\frac{1} {2}$

2 x $\frac{1} {3}$ = 1 - $\frac{1} {3}$

3 x $\frac{1} {4}$ = 1 - $\frac{1} {4}$ ...

The last problem is causing me difficulties since I need the division symbol and I can't find it in the special characters box. (I've used forward slash instead.)

3) 1 $\frac{1} {3}$ / $\frac{2} {3}$ = 1 $\frac{1} {3}$ + $\frac{2} {3}$

2 $\frac{1} {4}$ / $\frac{3} {4}$ = 2 $\frac{1} {4}$ + $\frac{3} {4}$

3 $\frac{1} {5}$ / $\frac{4} {5}$ = 3 $\frac{1} {5}$ + $\frac{4} {5}$ ...

I think the really clever thing is thinking these questions up in the first place! These are all simple to sort out algebraically. The second one seems a non-starter to me but I suppose it follows the pattern of the others. There must be an underlying pattern that enables these kinds of examples to be generated. I haven't spent any time looking at that yet.
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