# Lesson ideas for 10-11 year old Children (MATHS)

#### NewScientist

Hey,

I'm going into a school tomorrow an dhad a plan to teach the kids (10-11) about algebra in a different weay to they are used to. I.e. give them a deeper understanding of what is going on - to try and liven things up a bit. For example, most children can give a reasonable definition of what x means in a mathematical concept. Of course, I do not want thme to spurt out that it is the dependant variable - but I would like them to be able say, well if your just giving me x then x can be any number in the whole universe!

anyway, I was wondering what ideas you guys had - doesn't have to be algebra based - I have an hour with them and want to have an impact.

-NS

#### mathwonk

Homework Helper
well i liked the picture of the squaring function (x+a)^2 = x^2 + 2ax + a^2.

when you drawe a picture of a square of side length x+a, it contains a square of side a and one of side x and two strips that measure a by x.

then they can see the formula for the binomial expansion. let them guess the one for a cube (x+a)^3.

#### neurocomp2003

if your looking for application...game development may be a good place for you to start...
ie...building the standard physics motion equations OR just a ordinary 3D camera setup.

where x would be any vector position rather than just 1 #...
mmm or you cna relate to the bottom less pit. How deep does the well go type of thing

#### Crosson

Ask the children if they know what 3^2 is. Ask them if they know what 3^3 is. After some discussion, ask them what 3^20 would mean. After you felt that they understand what it means to raise a number to an integer power (x^n), you ask them: "what would 3^(-2) mean? How about 3^(1/2)? or x^0?"

Give them some time to bite in to the mystery, entertain there guesses. Then show them how to figure out what these exponents mean. Show them:

$$(x^a )^b = x^{ab}$$

This is pretty easy to digest just by writing it out by hand.

Of course, from here you could let x = 2, a = 1/2 and b = 2 in the above equation. This shows them that "2^(1/2) is the number which when squared gives you 2". From here it is easy to discuss any fractional powers (discuss examples like 5^(3/2).

After they are comfortable with fractional powers, you can talk about what it would mean to have something like x^(1/100000), "the number which when raised to the millionth power gives you x".

Point out that no matter what x is, x^(1/1000000) has to be a small number (or it would get "too big" being raised to the millionth power). Also point out that x^(1/100000) can't be less then one (fractions get smaller when raised to large powers). The whole point is to convince the children that x^b gets closer to one as b gets smaller, for all x. From here, you can conclude that x^0 = 1 for all x.

Now, armed with x^0 = 1, you should bring up:

$$x^a x^b =x^{a + b}$$

Verify it for examples where x, a and b are small integers. Now let a = 1 and b = -1 to show them:

$$x^1 x^{-1}=x^{1 - 1}=x^0=1$$

Solve this algebraicly to show:

$$x^(-1) = \frac{1}{x}$$

which some of them may have already guessed (and those will be the most delighted). Of course, all of this material works even better (and much faster) if the children have already memorized the meanings of fractional, negative and zero powers.

Notice that aside from the relatively advanced idea of a limit, the point of the exercise is to show how, in math, creativity is guided by logic. In addition, it teaches them important things about exponents in a forgetfullness-resistant way. I have used this lesson many times, but never on 10 year olds. The hardest part to sell is the limit of x^b as b gets small, but this is obviously crucial to understand negative exponents.

Another way to sell x^0 = 1 is by assuming that the real numbers form a vector space with vector addition defined as normal multiplacation, and scalar multiplication being defined as raising a "vector" to a power. If you understand that, and you know how to prove that scalar multiplacation by zero always gives the vector zero, then you should be able to explain to the kids quite easily. Yet another way is by explaining what an empty product is.

#### matt grime

Homework Helper
i like this famous bit of combinatorics:

ask them to draw a circle and mark 6 points on the perimeter. then get them to take 2 different colourd pens and join up each point to every other point using either of the two colours (with straight lines) in anyway they want. when they've done that ask them to put their hand up if they#ve got a triangle in either colour. they sohuld all put the ir hands up, obviously. ask them f the think that;s odd, then ask if they think it'll always have a triangle and then explain to them why it has happened.

there may be other simple demonstrations of the pigeon hole principle you can use as well.

#### LittleWolf

Start with an easy sequence {2,4,6,...} and have the children give you the Nth term. Afterward, you might try the secret number approach. Write X on the board and ask if anyone can guess what is your secret number. Ask if they need more of a hint then depending on the skills of the children write an equation: 2^x=8 or 2+x=8 or whatever. Take all answers and write them on the board and then have them prove and disprove the answers(leave everything on the board). Move on to next secret number Y. Give a hint such as Y+X= 15. Keep inventing secret numbers. The last 10 minutes challenge the children to a game of one pile Nim. Pick one student. Start with 10 sticks and let the child go first, last stick loses, must take at least 1 stick but no more than 2 sticks. Losing positions are {1,4,7,10,...}. What ever you do, win quickly and pick another child. If they realize that going first loses, change the number of sticks to 12. Have fun!

#### mathwonk

Homework Helper
or do another algebra trick, where they input various numbers incuding their age and you end by guessing their age.

#### mathwonk

Homework Helper
or play the famous birthday trick, wherein if there are say 35 or more of them at least two of them are very likely to share a birthday.

as i recall even 25 makes it over 50/50.

#### Borogoves

mathwonk said:
or play the famous birthday trick, wherein if there are say 35 or more of them at least two of them are very likely to share a birthday.

as i recall even 25 makes it over 50/50.

And in what way will that benefit the children ?

There could be a total of 100 children, and yet none of them sharing a birthday.

There could be 10 people, and 5 sharing birthdays.

Can you see now how speculative this is ?

It will only confuse the children.

#### mathwonk

Homework Helper
well my class liked it. is confusion bad? why are you so negative? had a bad day? or do, you just not understand then power of probability? i.e. the scenarios you describe are not too likely.

another thing that worked in my third grade class was to hand out colored polyhedra made of cardboard and ask them to count the faces and edges and vertices. eventually one girl noticed the euler formula V-E+F = 2, and today she is an aerospace engineer.

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#### quetzalcoatl9

the birthday thing is a good one because it teaches them that what may seem improbable isn't necessarily.

another thing would be to show how the game of tic-tac-toe could be done in set-theoretic terms. for example, assign each square:

11 12 13
21 22 23
31 32 33

and then show that "moves" are nothing more than removing an element from the set. then show how a winning algorithm works.

you could even start by having them play each other first, then transcript the games. finally, you could get them to play without a board!

#### Borogoves

mathwonk said:
well my class liked it. is confusion bad? why are you so negative? had a bad day? or do, you just not understand then power of probability? i.e. the scenarios you describe are not too likely.

another thing that worked in my third grade class was to hand out colored polyhedra made of cardboard and ask them to count the faces and edges and vertices. eventually one girl noticed the euler formula V-E+F = 2, and today she is an aerospace engineer.
I have been diagnosed with various mental conditions in the past, but of course these so called psychiatrists are wrong.

It is obviously not good to confuse the kids, but teaching them to manipulate powers is surely a bit much for third grade ?

You see the trouble with you is that you are using rather vague terminology.

Could you even define what you mean by '' the scenarios you describe are not too likely'' ?

#### quetzalcoatl9

Borogoves said:
I have been diagnosed with various mental conditions in the past, but of course these so called psychiatrists are wrong.

It is obviously not good to confuse the kids, but teaching them to manipulate powers is surely a bit much for third grade ?
i personally feel that children are much more capable than we often realize. unfortunately, their exceptionality is often times washed out in a sea of mediocrity. young children have a great capacity for learning abstract concepts, it's just that they are rarely challenged (or interested) to do so.

Borogoves said:
Could you even define what you mean by '' the scenarios you describe are not too likely'' ?
i think that what mathwonk was saying is that your statements like "There could be 10 people, and 5 sharing birthdays" are unlikely in terms of probability. the point of the birthday game is that it is a problem of $$\frac{n!}{r!(n - r)!}$$ so the odds of two people having the same birthday, given a large enough class, are good.

#### mathwonk

Homework Helper
i tell you what borogoves. if you do not approve of my ideas, how about making a positive suggestion for the gentleman, and let him sort out which ones to accept? mine are based on experience by the way, of over 40 years of teaching from second grade to junior high and high school to college and graduate school, and to returning teachers, but so what, you may surely disagree. #### mathwonk

Homework Helper
i do not recommend this for chidren, but when i taught college i brought in two dice, and played craps briefly with the students, amazing them by predicting which points would be made much of the time (6 and 8), and which would not (4 and 10).

#### NewScientist

Hey guys,

Thank you for your suggestions but this is what I did in the lesson.

I started off by just laying some basic ground levels. i.e. I wrote x + 3 = 4 what is x, they all responded 1, so i moved on to x - 3 = 4 etc.

They all said 'This is easy'. I said okay, why don't I give you a problem to do? They all said - is it algebra and I said 'no'.

The problem was thus, (stolen from Nrich!!)

George and Jim go to a sweet shop and want to buy a chocolate bar.
George needs 2 more p, Jim needs 50 more p. (p is pence - a unit of currency in UK)
They put their money together and still do not have enough.
How much is the chocolate bar.

They stared at me as though I was an alien. So, I started going through it and asked what we new. They instantly said that George needed 2p, and Jim needed 50p and that together they did not have enough.

I now said - okay I now start the clock on solving this problem. How shall we do it? Somebody said, try different amounts and get to an answer by changing it. I said fine - so we developed a table with Chocolate bar cost on the left, with the amount of money george must have in the next column ,and the amount of money jim had in the next, and in the final column, the addition of George and Jim.

Then I said - what is an approximate cost of a chocolate bar - somebody shouted out 30p so we used that. Obviously, it made Jim have -20p, I said, is this a problem and they instantly said it was as Jim cannot have -20p. So I said, what makes him have -20p when the chocolate bar costs 30p, they said it was the idea that Jim has 50p less.
We then reasoned that the smallest amount of money the chocolate bar could be was 50p, and so we used 50p , then 51p, and then 52p. For the first two we got numbers that fitted our conditions, but for 52p we didn't. So I asked, are 50p and 51p the only solutions - they said yes, probably. I said, were they sure and they said no.

I then said does anybody now a way to give an answer that must be write and that gives ALL the answers? 'Algebra' somebody whispered, and so I dived straight in!

At this point I was hoping someone would make the leap to an equation but nobody did, so I suggested, looking back a the 'x + 3' stuff, is there anyway we could use algebra here to show the link between the amount of money george has and the cost of a chocolate bar.

One child tentatively said, 'The chocolate bar is equal to george's money plus 2p' Some other members of the class looked a bit worried and said - thats not very 'mathy'. So I wrote it on the board and said what is an easier way to write the chocolate bar, they soon twigged I wanted a symbol so they chose C, I did the same for george's money and then Jim.

c = g + 2
c= j + 50

I said good and what was the only piece of information we hadn't used? THey said we haven't used the adding together bit. So I asked them how would use it?

We got c = g + j

I said 'hmmmm, what would that mean' They soon saw the error but couldn't make the leap to a > sign. So said is C bigger than g + j....yes....and what is a mathematical way for writing bigger than? '>'

c = g +2
c = j + 50
c > g + j

At this point they had no idea where to go - and I was happy with that.

So I said, look at your example, x + 3 = 4, how many bits of that equation are not numbers? I then led them to the conclusion that you can only have 1 variable in an equation to solve it. So I said in c > g + j, which bit shall we leave? C was decided upon due to the fact it was in the other 2 equations.

So I said, we have c = g +2, how do we get g on its own (make the subject) so we can use what is left in the 3rd equation. Blank faces - they had never really rearranged equations. So i talked through that and soon we had

g = c -2
and j = c -50

Now, they were quite sheepish to add these into the inequality as they had never used an inequality for manipulation before but I told them that the rules for a > sign were the same as that for =. So we soon reached

c > 2c - 52

This was manipulated to

52>c

So I said what does this tell us?
-It means that the chocolate bar costs less than 52p.
'Good' I said. 'Now, looking at the other method we used, and used the idea that Jim has to have at least 0p, what is the minimum c can be?'
-50p

so, what are the possible solutions? 50, 51.

The trial and improvement took 10 minutes and the algebra 15. So I said, shall we try again with different numbers?

And so I took another example through on the board in both methods.

Then i set them an example with these conditions - and timed them.
c = g + 4
and c = j + 6

This gives 15 solutions, and I set the most able girl the task of doing it by trial and improvement and the rest by algebra as a later test of how fast algebra is.

THe slowest kid finished in 5 minutes, the rest in about 3/4. The trial and improvement took about about 7, there were no calculators. I pointed out that this was only the third time they were dong things like this and they were still slow as compared to how fast you can become, but that they should realise how fast algebra is - they all agreed.

I was just wrapping up the lesson (it lasted about 1 hour) and a girl at the back who had been quite quiet piped up:
'Ben, algebra is more than quick'
'What do you mean?' I replied
'Its always right, gives you all the answers AND is quick' she said - and this reinforced for me the message with the class and the girl was only of average ability and so the class left buzzing for doing algebra and seeing a huge decrease in the time it took to solve a problem that looked huge before they started. Later on in the day they actually asked for more algebra!!

-I thoroughly enjoyed it and think the kids learnt a lot.

-NS

#### mathwonk

Homework Helper
wow! that sounds great! congratulations.

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