Posts Tagged ‘Proof’

A lovely sequence proof problem

May 28, 2020 Leave a comment

Another puzzle I found in Chris Smith’s (@aap03102) newsletter, and had fun doing is this one:

It’s a nice little puzzle that relies on ones ability to use algebra, reasoning and to understand what an arithmetic sequence is.

I considered the problem and thought that as I know the difference between each consecutive term in an arithmetic sequence is always the same I should first start by considering the differences. I looked at each one and expressed it as a single fraction:

When I had the two defences I knew that I could then equate them:

At this point I simplified the equation:

And was surprised it had fallen out so nicely . This obviously show that the differences between x^2, y^2 and z^2 are also equal so this is also and arithmetic sequence. I think this is a lovely result.

It occurred to me later that I could have set the mean of the 1st and 3rd term equal to the second and the algebra may have been a little easier. I’d love to hear your approach.

Which is heavier? Prove it!

May 22, 2020 2 comments

Today’s puzzle comes from Chris Smith (@aap03102) again, he has a weekly newsletter that if you dont subscribe to you should. It’s full of great maths stuff and features a weekly puzzle. Often I do them when I get the newsletter, but a lot of the time I save them for later if I don’t have time there and then. The pandemic has allowed me to find the time to catch up with some of them. This one is a nice little puzzle:

I looked at it, I didn’t really know where to start. I could see the right one (sack 2) was easily simplified, so I thought that might be a route. I wondered if I could simplify the left one (sack 1), but nothing obviously jumped out. I tried some numbers and found sack 2 heavier and given the way the question was phrased reasoned that it would always be so under the conditions, so that was infact an answer to the question.

However, I didn’t like that as a solution so decided to rephrase the question for me as “prove sack 2 is always heavier”.

I started with sack 2, as I could simplify it:

I then figured I could express that as a fraction with the same denominator as sack 1:

With some expanding and rearranging I could write it as sack 1 add a fraction. And that the fraction was always positive:

Which proves that sack 2 is always heavier:

I think this is a nice little proof problem and one that should be accessible to GCSE students, so I hope to use in lessons.

Proving sums and differences

May 20, 2020 Leave a comment

Today I want to look at a proof puzzle that I got from a previous addition of Chris Smith’s (@aap03102) brilliant maths newsletter. The puzzle is this:

It’s one I saw a few weeks ago, had a think about, got stuck and promptly forgot all about. But today I found the screenshot on my camera roll and thought I’d another go.

I tried a few things out that became fruitless endeavours, then I considers different ways to express the integers. As we were thinking I thought about modulo 100, what are the remainders we could have, as we are dealing with sums and differences we need to get the remainders to equal 0 for the sum or difference to be divisible.

Using mod 100 we would have 100 possible remainders. I only had 52 integers though, I I needed more thought. I considered this:

Using negatives I could express b as:

So I could get the magnitude of b to be one of 51 options. That meant that if I had 52 integers, they could be expressed this way and at least 2 of them would have the same value.

If the b values had the same sign, then their difference would be 0 hence the difference of the integers would be divisible by 100. Likewise if the signs were different then their sum would be divisible by 100.

I then considered how else I could write that.

I could express the integers as 100a + b (argh, why did I use a gain but for something different. Grrr.)

Then it follows the same way that at least 2 must have the same magnitude of b either the sum or the difference is divisible by 100. As shown here with the example b = 49:

A nice puzzle that I enjoyed exploring. I think there are probably nice proofs, certainly ones laid out better. I’d love to see yours.

Hippocrates’s First Theorem

April 11, 2016 6 comments

Over the half term I was doing some reading for my MA and I happened across Hippocrates’s First Theorem. (Not THAT Hippocrates, THIS Hippocrates!)

Here is the mention in the book I was reading (Simmons 1993):



It’s not a theorem I’d ever come across before, and it doesn’t seem to have any real applications, however it is still a nice theorem and it made me wonder why it worked, so I set about trying to prove it.

First I drew a diagram and assigned an arbitrary value to the hypotenuse of triangle A.


I selected 2x, as I figured it would be easier than x later when looking at sectors.

I then decided to work out the area of half of A.


A nice start – splitting A into two smaller right angled isosceles triangles made it nice and easy.

I then considered the area b. And that to find it I’d need to work out the area the book had shaded, I called this C.


Then the area of B was just the area of a semi circle with the area of C subtracted from it:


Which worked out as the area of the triangle (ie half the area of A) as required.


This made me wonder if it worked for all triangles that are inscribed in semi circles this way – ie the areas of the semicircles on the short legs that fall outside the semicircle on the longest side equal the area of the triangle.

My first thought was that for all three vertices to sit on the edge of a semi circle in this was then the triangle must be right angled (via Thales’s Theorem).


I called the length eg  (ie the diameter of the large semi circle and the hypotenuse of efg) x and used right angled triangle trigonometry to get expressions for the two shorter sides ef and fg. Then I found the area of the triangle:


Then the semi circles:



I then considered the diagram, to see where to go next:


I could see that the shaded area needed to be found next, and that this was the area left when you subtract the triangle from the semicircle.

I could now subtract this from the two semi circles to see if it did equal the triangle.


Which it did. A lovely theorem that I enjoyed playing around with and proving.

I think there could be a use for this when discussing proof with classes, it’s obviously not on the curriculum, but it could add a nice bit of enrichment.

Have you come across the theorem before? Do you like it? Can you see a benefit of using it to enrich the curriculum?


Simmons M, 1993, The Effective Teaching of Mathematics, Longman: Harlow

Tests of divisibility

December 11, 2014 Leave a comment

Today I was marking my year 9 classes books and came across some work on prime factor decomposition and tests of divisibility. Yesterday I had been arguing with Colin Beveridge (@icecolbeveridge) about the use of formula triangles, (see the comment section on this post) and any other method of anything that was presented in the fashion “do it this way and don’t worry about the why.” I felt a wave of hypocrisy flow over me. I had never explained the tests of divisibility to the class, and furthermore I didn’t even know myself why they worked! I thought I’d explore them and see what I came up with.

If the last digit is divisible by 2, the number is divisible by 2.

This test of divisibility is obvious, the definition of an even number is that it has 2 as a prime factor, and all even numbers have a last number divisible by 2. No further exploration needed. I know why this one works and I’m certain the class do too.

If the sum if the digits is a multiple of 3, then the number is divisible by 3.

This is a fact I’ve known since I was at primary school back in the 1980s, but I can honestly say that I’ve never thought about why it works. I wrote a couple if equations out, realised I should be working in modulo 3 and came up with this:


Which basically boils down to the fact any number can be split into the digits multiplied by a powerful of ten.

A= (x0)10^0 + (x1)10^1 +… + (xn)10^n

As (10^n) mod 3 = 1 for all natural numbers (and 0) then it follows that:

(A)mod 3 = (x0 +x1+…+xn) mod 3

Which implies our test of divisibility. This also implies the test if divisibility for 9 (ie is the sum of digits is divisible by 9 then so is the number) as (10^n)mod9 = 1 for all natural numbers and 0. To prove you just follow the exact working but in mod 9.

A number is divisible by 4 if it’s last two digits are divisible by 4.

This one uses the fact that if two numbers are both divisible by a number then so is their sum.


We know any number bigger than 100 can be expressed 100a + b where a and b are natural numberso and b<100. We also know that as 4 is a factor of 100, 100a is divisible by 4 if a is a natural number. Hence the whole number is divisible by 4 iff the best is divisible by 4, and b is the last two digits.

From this we can deduce the test of divisibility for 8 (ie a number larger than 1000 is divisible by 8 iff the last three digits are divisible by 8) as 1000 is divisible by 8. The proof is the same, but you split the number into 1000a + b where a and b are natural numbers and b is less than 1000.

Testing for 5 and 6

Testing for 5 (the last number is 5 or 0) is another obvious one that needs no further exploration. And the test for 6 is simple, if 2 and 3 are prime factors 6 must be a factor, as all numbers can be expressed as a product of their prime factors and 6 is the product of 2 and 3.

Testing divisibility for 10

Again, this is obvious as we are talking about a base ten system! That leaves just one more test.

To test for divisibility by 7, take the last digit, double it. Take this away from the number you are left with if you subtract the last number and divide by ten. If the answer is 0 or divisible by 7 (ie if the answer mod 7 = 0) then the original number us divisible by 7.

What a ridiculously complicated divisibility test. It looks much nicer algebraically:

(10a + b) mod 7 = 0 (a and b are natural numbers, b < 10)


(a -2b)mod7 = 0

Once I expressed it algebraically I had two simultaneous equations and solved like this:


I enjoyed working through them and now I’m confident I know why these tests work, but I’m fairly sure that the ones that aren’t obvious would be too complicated to explain to year 9. This made me think about their use, and I certainly think it’s fine to use them even without understanding. Which brought me back to the formula triangles debate. This is one that will rage, but as I explained in my post on it, I don’t have a great issue with people who understand the algebra using them. I also don’t have a great issue with people who struggle at maths and aren’t going to pursue it past GCSE using them. My issue is with higher attainers who could and should understand the algebra being taught then with no deeper conceptual understanding. I guess it’s a topic I will need to think more on.

Vedic Multiplication

November 30, 2014 2 comments

On Friday, timehop told me it was a year since I wrote this post on multiplication methods. I’d forgotten about the post, and tweeted a link to it. A number of people commented and tweeted about it and a nice discussion ensued. Part of the discussion moved to Vedic Multiplication.

I know a few Vedic Multiplication methods, and believe there to be many more. Most of the ones I know link to the most common algorithms but there is this curious one used for numbers close to one hundred.

Start with a multiplication problem:


First, take bother numbers away from 100:


Multiply those numbers and make them your last two digits:


Take one of the differences from 100 away from the other number (it should be the same):


That becomes your first two digits:


It’s an interesting little trick. I don’t see it as something that there is any reason to teach, and I don’t think it promotes understanding at all, but it’s interesting nonetheless. I think it may have a use in lessons, as an interesting introduction to algebraic proof.

How would you prove it?

First, consider the product ab, and apply the same steps:


The last two digits are (100-a)(100-b). The first two are a-(100-b) which equals a+b-100. Or b-(100-a) which also equals a+b-100. To make these the first two digits of a four digit number we need to multiply the expression by 100.

This gives:


Which expands to:


Which cancels to:

ab As required.


A nice, accessible, algebraic proof that proves this works. It works for all numbers, not just those close to 100, but if your product (100-a)(100-b) > 99 (ie more than 2 digits) you need to carry the digits.

A nice little thought puzzle

October 21, 2014 3 comments

Earlier today I listened to episode 19 of my favourite podcast, “Wrong, But Useful (if you haven’t listened, and like maths, then go listen quick!). Every month the hosts Colin Beveridge (@icecolbeveridge) and Dave Gale (@reflectivemaths) set a maths puzzle for listeners to solve.

This month it was Colin’s turn and he set a nice algebra puzzle:

Show that n^4 + 4 is not prime for any integer greater than or equal to two.

Have a little go if you haven’t already….

My thought process went like this:

For even n then the whole things even, but what about the odds?

Could it be solved by induction?

N=3 gives 85. N=5 gives 629. Nothing jumping out there.

I’ll factorise it, see what happens:


After a brief false start where I wrote the square root of 2 is 1 I came up with the solution, which is quite neat.

n^4+4=(n^2-2n +2)(n^2+2n+2)

n^2 – 2n +2 is greater than 1 for all integers greater than 1.

n^2 + 2n +2 is greater than 1 for all positive integers.

Hence n^4 + 4 is a product of two numbers greater than 1 for all integers n greater than or equal to two.

A lovely little midweek puzzle.

More triangles!

June 5, 2014 5 comments

Earlier this evening, Jane Moreton (@PGCE_Maths) tweeted this:


I looked at it for a moment and started pondering. In this case c was clearly always going to be 45. I wondered whether the others would change and then realised they wouldn’t. I decided to draw it out and play around with some angle reasoning.

When I drew it out something else occured to me, there would be 3 right angled triangles, all with the same opposite side (from the given angle) and differing adjacent sides that were multiples of each other. It occurred that the sidelengths didn’t matter, and I could reason it out with tangent ratios instead (and everyone knows trigonometry is more fun).

Tan (a) = 1/3
Tan (b) = 1/2
Tan (c) = 1

I realise that as the angles will always be the same I could evaluate each one and show that a+b = c. But a) I didn’t have a calculator on me and b) that’s no fun!

Instead, I used the addition formula for Tan(a+b):

Tan (a+b) = (Tan (a) + Tan (b))/(1-Tan(a)Tan(b))

Using our values for Tan (a) and Tan (b) we get a numerator of 1/3 + 1/2 which equals 5/6. We get a denominator of 1-(1/3)(1/2) which also equals 5/6. So we get Tan (a+b) = 1, Tan (c) = 1 too so Tan (a+b) = Tan (c). As a,b and c are all in right angled triangles they all fall between 0 and 90 degrees, so a + b must equal c which equals 45 degrees.

A nice little mental workout. I will show some of my classes tomorrow and next week.

Here’s my back of an envelope workings:


Pythagorean Triples

April 30, 2014 6 comments

So, pythagorean triples. For those who don’t know, pythagorean triples are right angled triangles with integer (whole number) side lengths. The most famous pythagorean triple is the 3, 4, 5 triangle.

These triangles are great for maths teachers, as they can be used to make the arithmetic easier when introducing higher mathematical concepts. They are often favoured by exam boards for the same reason. Recently I have been working on the Edexcel M3 module and it seems that EVERY triangle they use is a 3,4,5 or a multiple there of. Noticing this fact has help my students and I save lots of time, we lost 20 minutes to this by not spotting it:


Yesterday we were revising FP1 and one of my students was finding the modulus of -7 + 24i, as she was typing the required sum into her calculator I said “25” to the general amazement of the class. They were intrigued as to how I did it so quickly, I explained I knew that the 7,24,25 triangle was a pythagorean triple and this led onto a fascinating discussion.

They thought it was an impressive trick, and I told them that yes, it was, and another good trick (courtesy of @icecolbeveridge) was knowing that the angles if a 3,4,5 triangle are roughly 53.13 and 36.87 degrees!

We discussed this further, and they asked how many triples I knew “off the top of my head”. I explained that I could write them forever if we weren’t limited to primitive triples, because I could just add 3,4 and 5 to the respective sides. This lead to a nice discussion about primitive triples. Which are those where the side lengths are co-prime (have no common factors). And then I wrote the primitive ones I could remember:


This prompted the question “is it some sort of sequence? I can see the smallest side is the sequence of odd numbers.” I was impressed how quickly they had spotted than, but I then remembered the 8,15,17 triangle which has an even numbered smallest side. I also told them that I have a vague recollection that there is  a way to use one primitive triple to generate another but I can’t remember it. (A quick Google search didn’t help, so perhaps I imagined it. If you know it, I’d love you to tell me!)

I then discussed Euclid’s formula for generating triples and then moved on to my favourite way:

Take two unit fractions with denominators one apart, add them. The answer in its simplest form will be a fraction in which the numerator and the denominator form the short legs of a pythagorean triple.

I’m not sure where I first heard this method, but I do love it and I often use it when working out questions to set my classes. I explained that this method shows there are infinitely many primitive triples, as it will generate infinitely many, but that I didn’t think it would generate them all. (Again, I don’t know if it will or not, I just imagine it won’t. Do send me a link to a proof either way if you have it!)

The class, interested and enthused as ever, wanted to see proof that this always worked. They had tried lots of numbers and they had all worked, but they knew this didn’t prove it. I then said we would prove it together.

I asked how we might start, had the word induction thrown at me, then one said,

“could we start with 1/m + 1/(m+2), then square the top and bottom of the answer and see if it’s a square?

I shouted “excellent” and so we did:


Obviously this isn’t the whiteboard… I forgot to photograph it so have recreated it on paper.

The start, adding the fractions:


Then squaring the numerator and denominator, manipulating the expression and factorising:


And finally, the all important statement:


Do you have a preferred method of generating triples? Or do you know any interesting facts about them? I’d love to hear them.

Resilience in the classroom

April 14, 2014 3 comments

Recently I’ve written a lot of posts around puzzles and problems that people have set me to solve. This is something I find fun to do and I have enjoyed solving them. I have also enjoyed sharing my thought processes and proofs here, including the dead ends I went down.

The ideas involved in doing this have led me to think a lot about which, if any, of my students would have been able to solve each of the problems.

I started thinking about my A level classed, I know that the vast majority would have given them a go. One or two might have not started because they didn’t know where to start, but the rest would have given it a try. I’m fairly certain that at least 5 of them, given enough time, would have solved most of the puzzles. I intend to test this hypothesis after the holidays.

I then thought about my year 11 class. They are a top set and I have been trying to build a resilience in them this year. When I first took the class over in September one of the students complained to my colleague, her science teacher, that I had “given a worksheet and not even told us how to do it.” It had been a starter task designed to double check the class were familiar with and able to use Pythagoras’s Theorem. My colleague then asked her “could you not do it then?” to which she had replied “I could actually, it was Pythagoras’s Theorem.” He then asked, “so why did you want him to tell you what to do?” She had no answer.

This is quite common in schools, it’s an idea a lot of pupils have. That they should be explictly told what to do for each type if question. But I think that if we are to create the mathematicians of the future we need to be building a resilience into pupils. We need to equip them with the skills and knowledge required to solve the problems, and allow them to select whichever bit of it they want/need to solve it.

I think the puzzles I’ve discussed recently are good examples of tasks that do this. Some of them have the added bonus of being solvable multiple ways, often given rise to a “low barrier, high ceiling” task that can be set to 11-18 year olds and be solvable, yet challenging, to all.

This one which started the chain is a lovely puzzle based around algebraic fractions. It is solvable to a clever yr7 student who just has a basic knowledge of fractions, or via simultaneous equations and complex numbers, which is how most people with an advanced knowledge went about it.

This triangle puzzle was particularly nice, it had a lovely solution which requires a knowledge of the sine rule for the area of a triangle and knowledge of the sine curve, or a much simpler visual one which only requires a primary school level understanding of triangles!

This triangle puzzle is the best example of one with multiple solutions. I used Heron’s Formula (which no one else seems to have heard of!) But it is equally solvable using accurate drawing, similar and congruent triangles rules and/or Trigonometry (including Pythagoras’s Theorem).

These problems are great, and will build resilience, but the two most recent ones are the ones which illustrate this best.

While discussing constructing a proof with my brother he was asking how algebraic proof worked. He has two A levels in maths (at A no less) but he stopped studying maths a decade ago so is a little rusty. His questions, though, made me think about my classes. I know my year 13 pupils can construct proofs, but I’m not sure about all of those pupils who are younger. I am going to ensure I build more opportunities for this into my lessons.

In solving this problem I noticed a pattern in the numbers, I expressed this pattern algebraically and manipulated the algebra to prove the pattern was true for all numbers. This is what mathematicians have done for centuries and how theorems are born. And this is a skill I need to instil in my students.

In the root of the problem I discussed solving a problem which involved searching for integer solutions of an equation in two variables. It was fun, and again I was asked, “how on earth did you work that out? I wouldn’t know where to start.” Well I didn’t know where to start either, I just tried things until I got something that was right. This is what the best of my students do. This post gives some great examples of this. It needs to be more though. As maths teachers we need to make sure our students are willing to do this, if they don’t know what to do to apply things they know until they get an answer. I have been using this approach with my pupils. I won’t help them unless they have tried something first.

The way the maths exams have been previously has made this spoonfeeding possible and far too common. In the last couple if years the exam boards have thrown a few curveballs, which has meant that students have had to apply their knowledge in different ways to the past papers. I think this is the way forward, and hope the new GCSE and A Level exams address this.

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