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**A geometric proof for the Arithmetic and Geometric Mean**

There is more than one way to define the mean of a number. The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x_{1}.x_{2}.x_{3}…x_{n})^{1/n}

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)^{1/3}.

With 2 numbers, a and b, the geometric mean is (ab)^{1/2}.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)^{1/2} for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems). Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle. Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras: (MN)^{2} = a^{2} +x^{2}

We find NQ using Pythagoras: (NQ)^{2} = b^{2} +x^{2}

Therefore the length MQ can also be found by Pythagoras:

(MQ)^{2} = (MN)^{2 } + (NQ)^{2}

(MQ)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

But MQ = a + b. Therefore:

(a + b)^{2 } = a^{2} +x^{2} + b^{2} +x^{2}

a^{2}+ b^{2} + 2ab = a^{2} +x^{2} + b^{2} +x^{2}

2ab = x^{2} +x^{2}

ab = x^{2}

x = (ab)^{1/2}

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)^{1/2}. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.

**Non Euclidean Geometry – An Introduction**

It wouldn’t be an exaggeration to describe the development of non-Euclidean geometry in the 19th Century as one of the most profound mathematical achievements of the last 2000 years. Ever since Euclid (c. 330-275BC) included in his geometrical proofs an assumption (postulate) about parallel lines, mathematicians had been trying to prove that this assumption was true. In the 1800s however, mathematicians including Gauss started to wonder what would happen if this assumption was false – and along the way they discovered a whole new branch of mathematics. A mathematics where there is an absolute measure of distance, where straight lines can be curved and where angles in triangles don’t add up to 180 degrees. They discovered non-Euclidean geometry.

**Euclid’s parallel postulate (5th postulate)**

Euclid was a Greek mathematician – and one of the most influential men ever to live. Through his collection of books, *Elements, *he created the foundations of geometry as a mathematical subject. Anyone who studies geometry at secondary school will still be using results that directly stem from Euclid’s *Elements* – that angles in triangles add up to 180 degrees, that alternate angles are equal, the circle theorems, how to construct line and angle bisectors. Indeed you might find it slightly depressing that you were doing nothing more than re-learn mathematics well understood over 2000 years ago!

All of Euclid’s results were based on rigorous deductive mathematical proof – if A was true, and A implied B, then B was also true. However Euclid did need to make use of a small number of definitions (such as the definition of a line, point, parallel, right angle) before he could begin his first book He also needed a small number of postulates (assumptions given without proof) – such as: * “(It is possible) to draw a line between 2 points”* and “*All right angles are equal”*

Now the first 4 of these postulates are relatively uncontroversial in being assumed as true. The 5th however drew the attention of mathematicians for centuries – as they struggled in vain to *prove* it. It is:

*If a line crossing two other lines makes the interior angles on the same side less than two right angles, then these two lines will meet on that side when extended far enough. *

This might look a little complicated, but is made a little easier with the help of the sketch above. We have the line L crossing lines L1 and L2, and we have the angles A and B such that A + B is less than 180 degrees. Therefore we have the lines L1 and L2 intersecting. Lines which are not parallel will therefore intersect.

Euclid’s postulate can be restated in simpler (though not quite logically equivalent language) as:

*At most one line can be drawn through any point not on a given line parallel to the given line in a plane.*

In other words, if you have a given line (l) and a point (P), then there is only 1 line you can draw which is parallel to the given line and through the point (m).

Both of these versions do seem pretty self-evident, but equally there seems no reason why they should simply be assumed to be true. Surely they can actually be proved? Well, mathematicians spent the best part of 2000 years trying without success to do so.

**Why is the 5th postulate so important? **

Because Euclid’s proofs in *Elements *were deductive in nature, that means that if the 5th postulate was false, then all the subsequent “proofs” based on this assumption would have to be thrown out. Most mathematicians working on the problem did in fact believe it was true – but were keen to actually prove it.

As an example, the 5th postulate can be used to prove that the angles in a triangle add up to 180 degrees.

The sketch above shows that if A + B are less than 180 degrees the lines will intersect. Therefore because of symmetry (if one pair is more than 180 degrees, then other side will have a pair less than 180 degrees), a pair of parallel lines will have A + B = 180. This gives us:

This is the familiar diagram you learn at school – with alternate and corresponding angles. If we accept the diagram above as true, we can proceed with proving that the angles in a triangle add up to 180 degrees.

Once, we know that the two red angles are equal and the two green angles are equal, then we can use the fact that angles on a straight line add to 180 degrees to conclude that the angles in a triangle add to 180 degrees. But it needs the parallel postulate to be true!

In fact there are geometries in which the parallel postulate is not true – and so we can indeed have triangles whose angles don’t add to 180 degrees. More on this in the next post.

If you enjoyed this you might also like:

Non-Euclidean Geometry II – Attempts to Prove Euclid – The second part in the non-Euclidean Geometry series.

The Riemann Sphere – The Riemann Sphere is a way of mapping the entire complex plane onto the surface of a 3 dimensional sphere.

Circular Inversion – Reflecting in a Circle The hidden geometry of circular inversion allows us to begin to understand non-Euclidean geometry.

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

**Unbelievable: 1+2+3+4…. = -1/12 ?**

The above video by the excellent team at Numberphile has caused a bit of an internet stir – by providing a proof that 1+2+3+4+5+… = -1/12

It’s well worth watching as an example of what proof means – if something is proved which we “know” is wrong, then should we accept it as true? The particular proof as offered in the video is certainly open to question – even if the end result: 1+2+3+4+5+… = -1/12 can actually be proved under certain definitions, using the Riemann Zeta function.

**Grandi’s Series**

The proof in the video requires that firstly we accept that the infinite summation, 1-1+1-1+1-1… = 1/2. This series is known as the Grandi’s Series – and has been the cause of arguments in the mathematical community for centuries as to what the infinite summation should actually be. One method (called Cesaro Summation) gives an answer of 1/2 – which is the answer accepted in the video.

Alternative interpretations of Grandi’s series would be to group the numbers as 1 + (-1+1) + (-1+1) +(-1+1)…. which you would expect to equal 1. Or, we could group the numbers as (1-1) + (1-1) + (1-1) … which you would expect to equal 0. Therefore it would be also mathematically valid to say that the infinite summation 1-1+1-1… has no sum.

**Divergent Series are the invention of the Devil**

For the proof in the video to be valid we have to therefore accept that the sum of Grandi’s series is 1/2. We also need accept that it is possible to manipulate infinite series by “shifting them along by 1” or by factorising.

However as we have already seen in the case of Grandi’s series, infinite series don’t always follow normal arithmetic rules. Indeed, the 19th century Norwegian mathematician Niels Abel, warned that that, “Divergent series are the invention of the devil, and it is shameful to base on them any demonstration whatsoever!”

Nevertheless it is an interesting method. First they define 3 different infinite series:

S = 1 + 2 + 3 + 4 + 5 …..

S_{1} = 1 – 1 + 1 – 1 + 1 – 1 ….

S_{2} = 1 – 2 + 3 – 4 + 5…

**Step 1:**

The first step is to state that S_{1} = 1 – 1 + 1 – 1 + 1 – 1 …. = 1/2.

**Step 2: **

if S_{2} = 1-2+3-4+5…

then 2S_{2} = 1 -2+3-4+5…

+1-2+3-4…

Here we have “shifted along by one space” the second S_{2}. This means that when we add the two sequences together we end up with:

2S_{2} = 1-1+1-1+1… = 1/2

which gives S_{2} = 1/4.

**Step 3:**

Do S – S_{2} = 1 + 2 + 3 + 4 + 5 …..

-(1 – 2 + 3 – 4 + 5…)

= 4 + 8 + 12 + ….

= 4(1 + 2 + 3….)

= 4(S)

Now if S – S_{2} = 4S we can simply rearrange this equation and substitute the value of S_{2} = 1/4 which we found before to give: S = -1/12

As mentioned above this is not a very rigorous proof. There is a more rigorous (and complicated) method of proving this – which is the method used by Euler, and which employs the Riemann Zeta function. You can watch this method here:

You might notice when watching this proof that at the start of the video they use the infinite summation of a geometric sequence formula – which is only valid for absolute x less than 1. Then later on they substitute x = -1 into a result derived from it. This is OK because of analytical continuation (which is a method of extending the domain of a function beyond its usual domain). This idea starts to get really complicated – but if you’re interested in the basic idea look at the post on the Riemann Sphere below. The Riemann Sphere allows infinity to be included in the domain of the complex numbers.

If you enjoyed this post you might also like:

Mathematical Proof and Paradox: How you can “prove” things like 1 = 2. Can you spot the flaws in the logic?

The Riemann Hypothesis: How the Riemann Zeta function is fundamental to understanding the prime numbers – and how solving the Riemann Hypothesis is one of the greatest puzzles in mathematics.

The Riemann Sphere – an introduction to isomorphic mappings, which is a lot more interesting than it sounds!

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

**Fermat’s Theorem on the sum of two squares**

Not as famous as Fermat’s Last Theorem (which baffled mathematicians for centuries), Fermat’s Theorem on the sum of two squares is another of the French mathematician’s theorems.

Fermat asserted that all odd prime numbers p of the form 4n + 1 can be expressed as:

where x and y are both integers. No prime numbers of the form 4n+3 can be expressed this way.

This is quite a surprising theorem – why would we expect only some prime numbers to be expressed as the sum of 2 squares? To give some examples:

13 is a prime number of the form 4n+1 and can be written as 3^{2} + 2^{2}.

17 is also of the form 4n + 1 and can be written as 4^{2} + 1^{2}.

29 = 5^{2} + 2^{2}.

37 = 6^{2} + 1^{2}.

Prime numbers of the form 4n + 3 such as 7, 11, 19 can’t be written in this way.

The proof of this theorem is a little difficult. It is however easier to prove a similar (though not logically equivalent!) theorem:

All sums of x^{2} + y^{2} (x and y integers) are either of the form 4n + 1 or even.

In other words, for some n:

x^{2} + y^{2} = 4n + 1 or

x^{2} + y^{2} = 2n

We can prove this by looking at the possible scenarios for the choices of x and y.

**Case 1:**

x and y are both even (i.e. x = 2n and y = 2m for some n and m). Then

x^{2} + y^{2} = (2n)^{2} + (2m)^{2 }

x^{2} + y^{2} = 4n^{2} + 4m^{2}

x^{2} + y^{2} = 2(2n^{2} + 2m^{2})

which is even.

**Case 2:**

x and y are both odd (i.e. x = 2n+1 and y = 2m+1 for some n and m).

Then x^{2} + y^{2} = (2n+1)^{2} + (2m+1)^{2 }

x^{2} + y^{2} = 4n^{2}+ 4n + 1 + 4m^{2} + 4m + 1

x^{2} + y^{2} = 4n^{2}+ 4n + 4m^{2} + 4m + 2

x^{2} + y^{2} = 2(2n^{2} + 2m^{2} + 2m + 2n + 1).

which is even.

**Case 3:**

One of x and y is odd, one is even. Let’s say x is odd and y is even. (i.e. x = 2n+1 and y = 2m for some n and m).

Then x^{2} + y^{2} = (2n+1)^{2} + (2m)^{2 }

x^{2} + y^{2} = 4n^{2}+ 4n + 1 + 4m^{2}

x^{2} + y^{2} = 4(n^{2}+m^{2}+n) + 1

which is in the form 4k+1 (with k = (n^{2}+m^{2}+n) )

Therefore, the sum of any 2 integer squares will either be even or of the form 4n+1. Unfortunately this does not necessarily imply the reverse: that all numbers of the form 4n+1 are the sum of 2 squares (which would then prove Fermat’s Theorem). This is because,

A implies B

Does not necessarily mean that

B implies A

For example,

If A is “cats” and B is “have 4 legs”

A implies B (All cats have 4 legs)

B implies A (All things with 4 legs are cats).

A is logically sound, whereas B is clearly false.

This is a nice example of some basic number theory – such investigations into expressing numbers as the composition of 2 other numbers have led to some of the most enduring and famous mathematical puzzles.

The Goldbach Conjecture suggests that every even number greater than 2 can be expressed as the sum of 2 primes and has remained unsolved for over 250 years. Fermat’s Last Theorem lasted over 350 years before finally someone proved that a^{n} + b^{n}=c^{2 } has no positive integers a, b, and c which solve the equation for n greater than 2.

If you liked this post you might also like:

The Goldbach Conjecture – The Goldbach Conjecture states that every even integer greater than 2 can be expressed as the sum of 2 primes. No one has ever managed to prove this.

Mathematical Proof and Paradox – how we can “prove” the impossible

Divisibility tests allow us to calculate whether a number can be divided by another number. For example, can 354 be divided by 3? Can 247,742 be divided by 11? So what are the rules behind divisibility tests, and more interestingly, how can we prove them?

**Divisibility rule for 3**

The most well known divisibility rule is that for dividing by 3. All you need to do is add the digits of the number and if you get a number that is itself a multiple of 3, then the original number is divisible by 3. For example, 354 is divisible by 3 because 3+5+4 = 12 and 12 can be divided by 3.

We can prove this using the modulo function. This allows us to calculate the remainder when any number is divided by another. For example, 21 ≡ 3 (mod 6). This means that the remainder when 21 is divided by 6 is 3.

So we first start with our number that we want to divide by 3:

n = a + 10b + 100c + 1000d + …….

Now, dividing by 3 is the same as working in mod 3, so we can rewrite this n in terms of mod 3:

n = a + 1b + 1c + 1d + ……. (mod 3)

(this is because 10 ≡ 1 mod 3, 100 ≡ 1 mod 3, 1000 ≡ 1 mod 3 etc)

Now, for a number to be divisible by 3 this sum needs to add to a multiple of 3. Therefore if a + b + c + d + …. ≡ 0 (mod 3) then the original number is also divisible by 3.

**Divisibility rule for 11**

This rule is much less well known, but it’s quite a nice one. Basically you take the digits of any number and alternately subtract and add them. If the answer is a multiple of 11 (or 0) then the original number is divisible by 11. For example, 121 is a multiple of 11 because 1-2+1 = 0. 247,742 is because 2-4+7-7+4-2 = 0.

Once again we can prove this using the modulo operator.

n = a + 10b +100c + 1000d + …..

and this time work in mod 11:

n = a + -1b +1c + -1d + ….. (mod 11)

this is because for ease of calculation we can write 10 ≡ -1 (mod 11). This is because -1 ≡ 10 ≡ 21 ≡ 32 ≡ 43 (mod 11). All numbers 11 apart are the same in mod 11. Meanwhile 100 ≡ 1 (mod 11). This alternating pattern will continue.

Therefore if we alternately subtract and add digits, then if the answer is divisible by 11, then the original number will be as well.

**Palindromic Numbers**

Palindromic numbers are numbers which can be read the same forwards as backwards. For example, 247,742 is a palindromic number, as is 123,321. Any palindromic number which is an even number of digits is also divisible by 11. We can see this by considering (for example) the number:

n = a + 10b + 100c + 1000c + 10,000b + 100,000a

Working in mod 11 we will then get the same pattern as previously:

n = a – b +c – c +b – a (mod 11)

so n = 0 (mod 11). Therefore n is divisible by 11. This only works for even palindromic numbers as when the numbers are symmetric they cancel out.

**IB Maths Revision**

I’d strongly recommend starting your revision of topics from Y12 – certainly if you want to target a top grade in Y13. My favourite revision site is Revision Village – which has a huge amount of great resources – questions graded by level, full video solutions, practice tests, and even exam predictions. Standard Level students and Higher Level students have their own revision areas. Have a look!

This classic clip “proves” how 25/5 = 14, and does it three different ways. Maths is a powerful method for providing proof – but we need to be careful that each step is based on correct assumptions.

One of the most well known fake proofs is as follows:

let a = b

Then a^{2} = ab

a^{2} – b^{2} = ab – b^{2}

(a-b)(a+b) = b(a-b)

a+b = b (divide by a-b )

b+b = b (as a = b)

2b = b

2 = 1

Can you spot the step that causes the proof to be incorrect?

Another well known maths problem that appears to prove the impossible is the following:

This was created by magician Paul Curry – and is called Curry’s Paradox. You can work out the areas of all the 4 different coloured shapes on both triangles, and yet by simply rearranging them you created a different area.

A third “proof” shows that -1 = 1:

Let a = b = -1

a^{2} = b^{2}

2a^{2} = 2b^{2}

a^{2} = 2b^{2} – a^{2}

a = √(2b^{2} – a^{2})

a = √(2(-1)^{2} – (-1)^{2})

a = √(1)

-1 = 1

And finally a proof that 1= 0. This last proof was used by Italian mathematician Guido Ubaldus as an example of a proof of God because it showed how something could appear from nothing.

0 = 0 + 0 + 0 + 0 ……

0 = (1-1) + (1-1) + (1-1) + (1-1) ……

0 = 1-1+1-1+1….

0 = 1 + (-1+1 ) + (-1+1) + ….

0 = 1

So, maths is a powerful tool for convincing people of an argument – but you always need to make sure that the maths is accurate! If you want to see the problems in the above proofs, highlight below (explanation in white text):

1) We divide by (a-b) in the 5th line. As a = b, then (a-b) = 0. We can’t divide by zero!

2) Neither of the “triangles” are in fact triangles – the hypotenuse is not actually straight. This discrepancy allows for the apparent paradox.

3) In the second to last line we square root 1, but this has 2 possible answers, 1 or -1. As a is already defined as a = -1 then there is no contradiction.

4) This is very similar to the Cesaro Summation problem which exercised mathematicians for centuries. The infinite summation of 0 + 0 + 0 + 0 … is not the same as the infinite summation 1 – 1 + 1 – 1 + 1 ….