## Proof by markscheme

While marking my Y11 mocks this week I came across this nice algebraic proof question:

The first student had not attempted it. While looking at it I ran through it quickly in my head. Here is the method i used jotted down:

I thought, “what a nice simple proof”. Then I looked at the markscheme:

There seemed no provision made in the markscheme for what I had done. *(Edit: It is there, my brain obviously just skipped past it)* *How did you approach this question? Please let me know via the comm*ents *or social media.*

Anyway, some of my students gave some great answers. None of them took my approach, but some used the same as the markscheme:

And one daredevil even attempted a geometric proof…….

## Proving Products

Just now one of the great maths based pages I follow shared this:

So naturally I figured I would have a go. I thoughts just get stuck in with the algebra and see what happens, normally a good approach to these things.

My first thought was that if I use 2n – 3, 2n -1, 2n + 1 and 2n +3 then tgere would be less to simplify later. I know that (2n + 1)(2n – 1) = 4n^2 – 1 and (2n – 3)(2n + 3) = 4n^2 – 9 so I multiples these together.

(4n^2 – 1)(4n^2 -9) = 16n^4 – 40n^2 + 9

I thought the best next move would be to complete the square:

(4n^2 – 5)^2 – 16

This shows me that the product of 4 consecutive odd numbers is always 16 less than a perfect square and as such that the product of 4 consecutive odd numbers plus 16 is always a square.

(4n^2 – 5)^2 – 16 + 16 = (4n^2 – 5)^2

A nice little proof to try next time you teach it to your year 11s.

## Vedic Multiplication

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:

**(100-a)(100-b)+100(a+b-100)**

Which expands to:

**10000-100a-100b+ab+100a+100b-10000**

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.

## Constructing a proof

Yesterday, while discussing this problem, my friend Steve posed this problem that he’d been thinking about:

*The sum of two squares doubled is the sum of two squares. Prove it algebraically and geometrically.*

Being the sort of person that can’t let a puzzle go unsolved I picked up a pen and set to work. I started with 2(x^2+y^2) and thought about where to go next. As I was doing this my brother (@andycav_25) posted

*“1 seems to be always involved as one of the squares in the doubled bit” *

then

*“With 5 and 6 you get 122. 11^2 +1*

*2 and 3 give 26. 5^2+1*

*So it’s the sum of the first 2 squared plus 1″*

I could see his error straight away, he had only tried consecutive numbers, I tried 2 and 6. 4+36 = 40 doubled is 80. 2+6=8 and 8^2 is 64. 80-64=16. And there you have it. It’s not the sum of the numbers squared add one, it’s the sum of the numbers squared add their difference squared. All I had to do now was construct a rigorous algebraic proof that would work for all numbers:

Nice isn’t it. The two squares are x^2 and y^2, their sum is (x+y) their difference is (x-y). Set up your equations, show they are equal and there you are, a proof that works for whatever numbers you like. The one case where it doesn’t fully work is where x=y. This relies on the second square being zero. However, the question relies on the same thing, as 2^2 + 2^2 doubled is 16, which is the sum of 4^2 + 0^2 but not any ‘other’ two squares. If 0^2 doesn’t count then this counter example disproves the statement. Perhaps the question should read:

*Prove that the sum of two squares doubled is either a square, or the sum of two squares*

That works better, and is how I’m going to pose it!

I like algebraic proofs, I’m comfortable with them, I’ve worked with them alot, I know how to do them and have been known to do them for fun. This is a good example of how they come about, you noticed a pattern, like Andy did, test some other numbers, like I did, hypothesise the general case and try to use algebra to prove it.

Geometric proofs, however, are a trickier beast. I didn’t opt for many geometry based courses at uni and I’m not as experienced at constructing the proofs. I’m never fully sure what constituents a geometric proof. Is a visualisation of an algebraic proof good enough, or is that just an algebraic proof expressed a different way?

I don’t shirk a challenge, however, so I wanted to have a go. It was even harder than I thought. Here are some early attempts:

Then I had thought I had had break through of sorts:

I thought I had cracked it, but now I’m not sure. I’ve lost a bit of confidence in it. Does it proof what it is supposed to? Area is definitely conserved. The total area of the top is definitely the sum of two squares doubled:

and the bottom is clearly two squares.

It works whatever size the squares are. However the implication in the question is that we are looking for integer side lengths, so I’m fairly sure it isn’t correct. Back to the drawing board!

Steve, who originally brought the puzzle to my attention, constructed this lovely visual proof:

We have 2x^2’s and 2y^2’s which make an (x+y)^2, and with an extra (the overlap) (x-y)^2 therefore 2(x^2 +y^2) = (x+y)^2 + (x-y)^2.

It’s definitely a proof, and a lovely neat one at that. Does it count as a geometric proof, or is it just a visual proof?

I’d love to hear your opinions. I’d also live clarification on what a geometric proof is. Do tell me if you can come up with a geometric proof that works.

## Hippocrates’s First Theorem

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?Reference:Simmons M, 1993,

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