The questions can be amended in this spreadsheet, by changing the numbers in the Answers column.

]]>I have included solutions in this spreadsheet although I would be hesitant to display them in this form, as I would prefer that the results are found and discussed as we go along rather than just revealing them at the end.

]]>A colleague of mine @DrPMaths made this impressive collection of triangles with 3 integer side lengths and an integer height. Again, they are a great way to check that students are identifying the perpendicular height and multiplying that by the base rather than just multiplying the numbers they are presented with. There are literally hundreds of them, this is a snapshot from somewhere near the beginning!

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They start off quite easy but have a nice extension into generalisation (Q4) and then geometric reasoning in Q5. It was good to see my students being able to tackle 1-4 without resorting to drawing anything. However, I think a drawing is definitely warranted in Q5 as it highlights how to find the area.

]]>I want students to make the connection that if the y-value doesn’t change for a bunch of coordinates (in this base the y-value is always 4) then the line that those coordinates all sit on is y=4. I’ve even made a little GIF where the point deliberately slides off the the right for a little while. Gotta love a GIF!

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A number of techniques we explored were about getting exact answers, but this section was as rough as you like! Rob introduced us to his idea of “Zequals“. When teaching rounding, I always enjoy introducing my students to the “approximately equals” sign, ≈. I hadn’t really considered how this symbol, on its own, doesn’t give the complete story. All of these statements are true…

7.3 ≈ 7 7.3 ≈ 10 and even 7.3 ≈ 7.4

but they don’t give an explanation of what you have actually done to the 7.3 and in the last example here, it really would require quite a bit of explanation!

So Rob proposes “Zequals” which has a precise method. It looks like this:

and it involves rounding the numbers you calculate with AND the result to 1 significant figure.

The Numberphile video explains in more detail here. An interesting question to ask might be, what calculation would give the biggest discrepancy between the accurate calculation and the Zequals calculation? And what would the % error be? The blog post explains this beautiful graphical representation of that % error, which turns out to be a fractal.

Now, to be honest, I would be hesitant to “teach” non-standard notation and methods as part of the regular timetable of maths. There is already so much to learn and time is so precious, why would I take a lesson explaining something they are unlikely to ever encounter again in this form? But dismissing it on that basis, misses the point, I feel.

Estimation as a topic features in a fairly minor way at GCSE but is a critically important skill in many jobs and life in general. There was some discussion amongst the attendees this morning that as students progress through KS3 to KS4 and A-level they become more and more reliant on their calculator. With the demise of the C1 paper, there is no longer a requirement for a non-calc paper at A level which is inevitably going to mean that our students will get weaker at this skill rather than stronger. This seems like a real mismatch between our education system and the requirements of employees and our broader society.

An idea which might help is to explicitly teach estimation as a technique to be employed when doing calculations with large numbers or decimals. Typically these types of calculations would involve some sort of “ignoring” the decimal point or the zeros, doing the calculation, and then “putting it back”

3.23 × 3 323 × 3 = 949, so 3.23×3 = 9.49 (counting 2 d.p.) 23.1 × 0.31 231 × 31 = 7161, so 23.1 × 0.31 = 7.161 (counting 3 d.p.) 3200 × 40 32 × 4 = 128, so 3200 × 40 = 128000 (counting 3 zeros)

Maybe instead of, or as well as, “counting the decimal places” when doing these calculations we could do some rounding / estimation. So

3.23 × 3 ≈ 3 × 3 = 9 323 × 3 = 949, so 3.23×3 = 9.49 (same order of magnitude) 23.1 × 0.31 ≈ 20 × 0.3 = 6 231 × 31 = 7161, so 0.23 × 0.031 = 7.161 3200 × 40 ≈ 3000 × 40 = 120000 32 × 4 = 128, so 3200 × 40 = 128000

Now I am not claiming that this is a more efficient or reliable method. It does depend to a certain extent on the examples chosen and “counting the decimal places” is a method that will always work. But I feel that the approximation step helps with number sense: the idea that 20 × 0.3 is a bit less than half of 20 so must be 6 is really valuable for life beyond exams. It provides an opportunity to practise these skills, practice which I believe our students currently have precious little of.

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What other questions could be asked here?

Powerpoint file, with solutions here.

]]>Powerpoint file used to create this here.

]]>Maybe this is an important thing to establish before talking about irrational numbers. With irrational numbers, we are effectively trying to convince students that there is a separate class of numbers on the number line that can’t be expressed as a division of two integers.

There is a proof for why ∏ is irrational but it’s not pretty. I’m taking Peter Mitchell’s word for it on that who presented on this topic at the recent MEI conference. He has a proof here, but in his own words “it’s really, really tedious!” So maybe surds are a better place to look for an example of a proof that irrational numbers exist. Although this is an A level topic, I think with the right class this could be used at KS4.

This is a proof by contradiction, which in itself is a bit strange. But the logic is sound: if I assume something to be true and then work through it to show that there is something inherent within it that is false, then I have proved that thing cannot be true therefore it must be false.

In this case, we are going assume that √2 * is* a rational number, prove that that is

If √2 is a rational number, then we can write it √2 = *a/b* where *a*, *b* are whole numbers, *b* not zero.

We additionally assume that this *a/b* is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for *a/b* to be in simplest terms, both of *a* and *b* cannot be even. One or both must be odd. Otherwise, we could simplify *a/b* further.

Going back to our first statement:

√2 = *a/b*

we can square both sides to get:

2 = *a*^{2}/*b*^{2}

or

*a*^{2} = 2*b*^{2}.

So the square of *a* must be an even number since it is two times something. If *a²* is even then *a* itself must also be even. Any odd number time an odd number creates an odd number (some more of these here).

Okay, if *a* itself is an even number, then *a* is 2 times some other whole number. In symbols, *a* = 2k where k is this other number.

If we substitute *a* = 2k into the original equation 2 = *a*^{2}/b^{2}:

2 = (2k)^{2}/b^{2}

2 = 4k^{2}/b^{2}

2b^{2} = 4k^{2}

b^{2} = 2k^{2}

Again, because b is 2 times something, b **must** be an even number.

We have shown that *a* and *b* are both even numbers, but we started saying that *a/b* was a fraction in its simplest form.

I might want students to explore what happens with √4 in this same proof, i.e. prove why √4 is **not** irrational. From there we could go on to look at √3. It’s a bit harder, but only really requires that all odd numbers can be written in the form 2n+1. Here is a spoiler if you are stuck.

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