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Posts Tagged ‘Algebra’

Proving Products

June 26, 2017 1 comment

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.

A lovely angle puzzle

March 24, 2016 4 comments

I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.

Earlier this week this question popped up:

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What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.

The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:

7x + 2y + 6z – 20 = 360

7x + 2y + 6z = 380 (1)

I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:

2x – 20 = 2y + 2z (2)

And

3x = 2x + 4z (3)

I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:

x = 4z (4)

Subbing into 2 we get:

8z – 20 = 2y + 2z

6z = 2y + 20 (5)

Subbing into 1

28z + 2y + 6z = 380

34z = 380 – 2y (6)

Add equation  (5) to (6)

40z = 400

z = 10 (7)

Then equation 4 gives:

x = 40

And equation 2 gives:

60 = 2y + 20

40 = 2y

y = 20.

From here you can find the solution x + y + z = 40 + 20 + 10 = 70.

A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.

Cross-posted to Betterqs here.

A bizarre solution

October 9, 2015 5 comments

On twitter earlier I saw this picture:

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It was tweeted by “Mathster”, an app which claims to be “a complete solution for UK and IB curricula”. The picture had me scratching my head. Why would you add the unnecessary step one into preceding?  Why would you refer to “moving terms”? Why? Why? Why?!

By saying “move terms” there is no explanation for changing sign, so learners may just move the term as is, which makes it perhaps even worse than the fabled “magic bridge”! There’s no mention of what’s actually going on! The final step is explained well, and I don’t know if that makes it better or worse!

Area Puzzle 21

September 23, 2015 2 comments

I came across the following area puzzle on Ed’s (@solvemymaths) site.

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I found it quite an interesting idea and had a little go at solving it.

First I considered an equilateral triangle, all the angles are 60 degrees so we can see the area would be (x^2 sin (60))/2. As sin 60 = 3^(1/2)/2 it was easy enough to solve.

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I then realised I could have used Heron’s Formula, so did it that way too. Luckily I got the same answer:

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The square was a tough one:

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Then I considered a pentagon, I wasn’t sure how to approach it at first,  so I reverted to my favourite shape, the triangle. Split the pentagon into 5 congruent isosceles triangles and solved with trig.

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Wolfram alpha gave me the lovely exact answer:

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I then considered the hexagonal case, which is really just 6 equilateral triangles.

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I approached the decagon in the same manner I had the Pentagon.

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And got an equally lovely exact answer:

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I enjoyed working through these, and thought it would make a nice lesson to build resilience and cognitive activation. I also thought about what else could be done. What does the sequence of side lengths generated look like? Could an equation be formed to describe it? What if we were looking at the areas of regular polygons with side length six, what would the sequence look like then? All these would make nice investigations.

Bicycle Puzzle

September 12, 2015 5 comments

This week’s puzzle from Chris Smith is a nice contextualised simultaneous equations puzzle I intend to use next week.

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It boils down to:

2X + B + T = 135
2X + 2B + 3T = 269
X + B + T = 118

Via elimination using 1 and 3 we can see that X (number of tandems) is 17.

That gives:

34 + B + T = 135
34 + 2B + 3T = 269
17 + B + T = 118

1 and 3 rearrange to the same leaving:

B + T = 101
2B + 3T = 235

Elimination gives T (number of Tricycles) to be 33 which leave B (number of Bicycles) to be 68.

A nice little problem.

Indices Help

February 16, 2015 Leave a comment

This is the second post in a series of posts which I’m writing to appear on a website for our A level students. You can see the first one, on fractions, here. After sharing the first one I got some great feedback and have some areas to improve it, if you see anything missing from this, or feel I need to clarify / remove anything, please let me know.

Indices, or index numbers. That’s those little numbers that appear raised just after a number or a letter. You know, the “2” that means squared, the “3” that means cubed, etc. They are quite often referred to as “powers”. They are sometimes written like this 3^2 means 3 squared. The power is called the index number (in this case 2) and the number being raised is the base number (in this case 3). Indices are pretty important in mathematics, and you will need to be very well versed in them to be successful on the A level maths course.

What are they all about?

In the first instance, they are the numbers of times a number is multiplied by itself. 2^2 means 2 x 2, 2^3 means 2 x 2 x 2, etc. This means that 2^1 is just 2. Which is something some people forget, don’t be one of those people. Repeat after me: “anything to the power 1 is itself, anything to the power 1 is itself….” 

But what do we do with them?

We need to be able to handle indices. We will save ourselves a lot of time later if we can simplify expressions using them, and it is essential for calculus. So here are “The Rules”, but remember, they only work with the same base number (or letter)….

Rule 1

The first rule of indices is, you don’t talk about indices…. Sorry, I couldn’t resist. Is Fight Club even a film that people still watch?

Rule 1 is: “When multiplying, add the powers“. (I say rule 1, that’s what I call it, not its official name…)

This rule is fairly intuitive. If you have 2^2 x 2^3 then you have (2 x 2) x (2 x 2 x 2) which is the same as 2 x 2 x 2 x 2 x 2, which is by definition 2^5. Try some yourself and see.

Rule 2

Again, this is fairly intuitive. The rule is: “When dividing with indices, subtract powers.

So 3^3/3^2 = 3

This is because we are “cancelling” common factors. Taking this as a fraction we’d have a numerator of 3 x 3 x 3 and a denominator or 3 x 3, so we divide top and bottom to give 3/1 which is just 3.

Rule 3

This rule involves raising a power to a power. Consider the problem (x^3)^2,  this means x^3 multiplied by itself, which gives x^6 (using rule 1). The “shortcut” here is to notice that because of the way multiplication works, “when raising a power to a power you multiply“.

A real common mistake on this type of problem occurs when you get something of the form (2y^4)^3. Often people will evaluate that as 2y^12, but that’s wrong. Don’t be one of those people.

(2y^4)^3 = 2y^4 x 2y^4 x 2y^4

So you get 8y^12.

DON’T FORGET TO APPLY THE POWER OUTSIDE THE BRACKETS TO EACH AND EVERY TERM.

Rule 4

Negative powers are the reciprocal of positive powers

This follows from rule 2, 3^4/3^6 gives 3^(-2) but if we cancel common factors it gives 1/3^2, hence they are the same. This one is extremely important when we get to calculus.

Rule 5

Fractional powers are roots

Think of it like this, 9^(1/2) x 9^(1/2) = 9^1 (using rule 1). Well 9^1 = 9 so 9^(1/2) multiplied by itself is 9. And we know 3 multiplied by itself is 9. This follows for all square roots. By the same logic we can see that a power of 1/3 is a cube root, a power of 1/n is an nth root, etc. This is quite important for calculus.

Another thing you need to be able to do with fractional roots is evaluate them, and I don’t mean just unit fractions. You need to be able to evaluate stuff like 32^(3/5).

This isn’t as hard as it looks, you just need to tackle it in stages. Split the fraction up using rule 3:

32^(3/5) = (32^(1/5))^3 always do the root first, it makes the number easier to deal with.

In this case 32^(1/5) = 2 and 2^3 = 8 so 32^(3/5) = 8. Questions like this do come up in a level papers, and they come up in the non calculator c1 paper, so it’s handy to know the first 10 powers of 2, 3 and 5. If you are struggling to remember them in an exam, you can work them out and write them down.

The final note on fractional indices is that when they are involved in problems using the other rules you deal with then in the same was as any fraction problem. See this page for help in fractions.

Rule 6Anything to the power 1 is itself” – as mentioned before.

Rule 71 to any power is 1

This is straightforward. 1×1=1 no matter how many times you repeat it.

Rule 8 – “Anything, except 0, to the power zero is 1

This is one people sometimes struggle to get their heads around. An nice way to think of it is this: 2^2/2^2 = 2^0 but 4/4=1. You can try with any base and any power, this always works. The only exception comes when the base is 0. 0^0 is undefined (or indeterminate), in the same way that dividing by zero is.

Fractions Help

February 15, 2015 2 comments

Right, so I’m in the process of putting together a site for our A level students. In it I am going to include some pages to help with their studies. I first thought I should cover some of the basics that every year students arrive from GCSE struggling with. This is the first draft of the fractions help page and I’d love to hear your thoughts on it. Is it clear? Does it need amending? Have I overcolicated things? Have I missed anything? I’d also love suggestions of other topics to cover. In the first instance I’m thinking to of an LCM post to act as a companion to this one. An indices post, a quadratics post and perhaps one around trigonometry. Anyway, here’s the fractions post:

Ok, so you should really be able to handle fractions by now. The fact you’re on the A level course means you’ve got a decent GCSE pass, which in turn implies a grasp of the basics, but we know that fractions can be confusing and every year we encounter A Level students who have trouble working with them.

What are fractions?

I hope you know this, but just incase, a fraction is a rational number. It can be expressed as a/b. The denominator (that’s the bottom number) tells you the number of equal parts you split 1 into to find 1 of them, and the numerator (top number) tells you how many you have. I.e. 1/4 means you split 1 into 4 equal parts and have 1 of it, so 0.25 as a decimal.

And the top has to be smaller right?

NO, NO, NO, NO, NO. Improper fractions are absolutely fine, in fact they are infinitely more useful than mixed numbers and I’ve no idea why you need to spend so much time changing improper fractions to mixed numbers at GCSE. Now you’re doing A Level you should avoid mixed numbers like the plague and always change them to improper fractions. Also, stop changing everything to decimals. Fractions are tge more exact form and decimals can give rise to rounding errors. Fractions are much, much better.

How do I add 1 to a fraction?

To add 1 to a fraction you just need to add the amount in the denominator to the numerator. To add 1 to 3/4 add 4 to the numerator to get 7/4. To add 1 to 3/2 add 2 to the numerator to get 5/2. Simples.

This is because anything over anything is equal to 1. 4/4 = 1 5/5 =1 x/x=1 etc. This is the reason why equivalent fractions work:

Equivalent Fractions

You can multiply or divide the numerator and denominator by the same number to get an equivalent fraction. So 4/8 = 1/2 (numerator and denominator divided by 4) that’s because 4/4 is 1 so you are essentially dividing the fraction by 1.

Adding Fractions

Adding fractions with the same denominator is easy, you just add the numerators. If you gave a fifth, then add another you have two fifths. (1/5 + 1/5 = 2/5). Subtraction is the inverse of addition so it works the same.

The problem comes when we need to add fractions with different denominators. We do this using equivalent fractions to make two fractions with the same denominator. It’s usually best to use the lowest common denominator, this isn’t too important when using numbers, as you can easily simplify later, but when dealing with algebraic fractions it’s always better to find the lowest common denominator as otherwise you may end up with a high order polynomial.

The lowest common denominator is the lowest common multiple of of the denominators. So we would change 1/4 + 1/3 to 3/12 + 4/12, and 1/(x + 1) + 1/(x -1) would become (x – 1)/(x + 1)(x – 1) + (x + 1)/(x + 1)(x – 1)

Multiplying Fractions

This is easy to do, you just multiply the numerators together and the denominators together. So 2/5 x 3/7 = 6/35. The problem is many people forget this, possibly because they never bothered to really understand why. The denominator of a fraction is how many equal parts 1 is broken up into to get the “unit” we are dealing with. So multiplying the units together gives us how many equal parts our answer will deal with. If you break up 1/4 into thirds you get 1/12 (as 3/12 = 1/4). The same works for the numerators:

Ed: I’m going to include a diagram here, and maybe a few others throughout the post.

Dividing Fractions

This is often a source of confusion. And to really understand it you need to really understand what a fraction is. We know the denominator is the amount of equal parts 1 is split into, and the numerator is how many of those parts we have. Which is the same as saying a fraction is what we’d get is we split the numerator into the amount of equal parts specified in the denominator. So 3/4 is 3 divided by 4.

If we imagine dividing 1 by a half, it’s the same as saying how many halves are there in 1. As we get a half by splitting 1 into 2 equal parts, the answer must be 2. Dividing and multiplying are inverse operations, and that is why dividing by 2 is the same as halving something. Multiplying by somethings reciprocal is the same as dividing by it. This works with fractions too, so if we want to divide by 3/4 we can multiply by its reciprocal (4/3). Remember, reciprocal means 1/ and the reciprocal of a fraction is found by switching the numerator and denominator around, or “flipping” it.

Remember, these rules work for all fractions, whether you know the numbers or have them in algebraic form.