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Sir Col
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Simple Proofs
« on: Jun 18th, 2003, 6:01am » |
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Being a mathematics teacher in a secondary/high school, the greatest challenge is not solving problems but explaining solutions to classes with limited mathematical tools and resources. Having just proved, recently, that sqr(2) is irrational to one of my year 8 classes (12/13 year olds) I set them three problems to work on.
1. Find a square number, which when double, produces another square number. For example, 2*25=50, which is only one more than 49.
2. By using a calculator you can see that sqr(18)/3 gives the same answer as sqr(2). Find five more solutions to sqr(a)/b=sqr(2), where a and b are natural numbers.
3. Solve the equation sqr(a)–b=sqr(2). For example, sqr(51)–6 is close, but no cigar!
Obviously I'm not just asking you to solve these problems – they are relativley trivial – but I'm interested in 'child friendly' approaches that we might use. Bearing in mind that for the particular class I presented them to, the irrationality of sqr(2) [showing that sqr(2)=a/b, where HCF(a,b)=1, reduces to the ratio of two even numbers], was the first proof they'd ever really seen – baptism of fire, eh? Their algebaric skills venture little further than solving linear equation in one variable with brackets on both sides, but they can handle eliminating denominators and manipulating simple equations.
I know that this is not everyone's bag, but I'd really appreciate thoughts/ideas on these types of problem.
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| « Last Edit: Jun 18th, 2003, 6:02am by Sir Col » |
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Icarus
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Re: Simple Proofs
« Reply #1 on: Jun 18th, 2003, 4:09pm » |
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Did you really make it sound like 1 and 3 were solvable? That might have been a bit unfair - since their skills are stretched to handle this, those who are struggling to understand are unlikely to catch on, I think.
It would have been better to state them "Can you find a square number..." and "are the integers (or however you want to describe them) a and b such that...". Then more of them are likely to figure it out.
Are the (within reason) comfortable with distributing yet? (My experience is that most students never fully obtain this skill - I think partly because we are too quick to teach FOIL or other shortcut methods without pushing for simple understanding first.)
If they don't know how to distribute, then 3 is likely to prove too much of a challenge - though it could be dragged out when you get to this skill, to show how it can help.
All of my teaching was to college students, so I am talking without any experience with your target audience.
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Sir Col
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Re: Simple Proofs
« Reply #2 on: Jun 18th, 2003, 5:11pm » |
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I greatly appreciate any thoughts, Icarus.
You're right, it was a little unfair to present #1 and #3 in such as way that a numerical solution appears to exist, but one of the most valuable lessons they can learn is that no solution is a solution. It's interesting to hear them form conjectures and organise themselves to work co-operatively.
Upon reflection (of their own work), I'd expect them to solve #1, as it shouldn't take much effort to produce an equation of the form 2b2=a2 and, at least, recognise this as one of the stages from the proof for the irrationality of root 2.
For #2, some of them had already spotted that the numerator had to be the square root of 2 times a square. A little encouragement should lead them to the simple generalisation. A classic opportunity, as teacher, to reinforce the difference between the particular and a generalisation.
#3 is going to be the most challenging and I certainly don't teach gimmicky approaches to mathematics, so any methods they use will be rooted in methods of which they have a solid grasp. Our next topic is the Pythagorean theorem (part of my motivation for exploring irrationals), so techniques in squaring binomials will be developed.
Whenever I set problems like this, I present them as on-going challenges. My thinking behind #1 was to prepare them for a demonstration of the incommensurability of the diagonal in a square. #2 was to encourage some of the more inquisitive to ask their own questions about the ratios of rational and irrational numbers and resulting quotients. #3 is an opportunity for some of the brighter students to explore basic proof, as it makes use of a varied collection of valuable techniques.
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Icarus
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Re: Simple Proofs
« Reply #3 on: Jun 18th, 2003, 8:01pm » |
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As I said, my only experience was with college students, and I have no doubt you are a far better teacher than I ever was!
Quote:| I certainly don't teach gimmicky approaches to mathematics, so any methods they use will be rooted in methods of which they have a solid grasp. |
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Excellent! 
One of my big headaches was students who learned to use the FOIL method, or the "smiley face" method by rote, and therefore were at a total loss as to what to do when there were THREE terms in one of the mutiplicands. (I tried to teach them right, but they already knew all about the FOIL method, & it was given in the book as well, so they just went back to it instead of figuring out what distributing was about).
Mnemonics and calculators - boon and curse to mathematics: Boon because they make it easier to do the calculations and keep track of what you are doing. Curse because they make it too easy to arrive at the answer with understanding WHY it is the answer.
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wowbagger
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Re: Simple Proofs
« Reply #4 on: Jun 19th, 2003, 4:22am » |
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on Jun 18th, 2003, 8:01pm, Icarus wrote:As I said, my only experience was with college students [...]
One of my big headaches was students who learned to use the FOIL method, or the "smiley face" method by rote, and therefore were at a total loss as to what to do when there were THREE terms in one of the mutiplicands. (I tried to teach them right, but they already knew all about the FOIL method, & it was given in the book as well, so they just went back to it instead of figuring out what distributing was about). |
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I don't know the methods/mnemonics you mentioned, but these college students weren't able to use the distributive law? 
What were they studying? Biology?
Now, I don't claim that math education is really great round here (espacially that of Joe Public, of course) - but these are students, right?
Probably my worst experience so far was a student (college/university level) who didn't remember how to differentiate a polynomial. By the way, she studies something called "Marine Environmental Sciences" (I don't know the official translation), which focuses on Chemistry and Biology.
Quote:| Curse because they make it too easy to arrive at the answer with understanding WHY it is the answer. |
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Without? 
(Sorry for being a bit off-topic.)
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Sir Col
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Re: Simple Proofs
« Reply #5 on: Jun 19th, 2003, 11:34am » |
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FOIL stands for First, Outside, Inside, Last. It's a method used by some teachers to help children remember how to multiply out two binomial factors. For example, (a+b)(c+d):
First=ac
Outside=ad
Inside=bc
Last=bd
So (a+b)(c+d)=ac+ad+bc+bd
Although I said that I don't like cheap methods, I do confess to teaching the 'moon' method to weaker students. If you've never heard of it, try this: Write (a+b)(c+d) on a piece of paper, then join a to c and a to d with arcs above the brackets, then join b to c and b to d below the brackets. It should form a convincing moon shape. I would not use this method with more able students, as I believe it is necessary to understand, and talk in terms of, the law of distributivity. I would quickly move on to trinomial factors with them to ensure that they don't invent their own gimmicky methods.
Out of interest, how do we, as well practiced mathematicians, expand brackets? I often find it interesting to compare how I do it, with what I expect of students.
Consider how you would expand (a+b)(c+d) and (a+b)(c+d)(e+f). For the triple factor, do you work out two brackets first, then multiply by the other, or do you expand in one go? What would you do with four factors? Would it matter if you had trinomial factors: e.g. (a+b)(c+d)(e+f+g)?
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| « Last Edit: Jun 19th, 2003, 11:36am by Sir Col » |
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towr
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Re: Simple Proofs
« Reply #6 on: Jun 19th, 2003, 12:01pm » |
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I could do either, expanding two factor at a time is simplest, but in one go is easy enough as well (it's simply an evaluation tree you're working through depthfirst, though I'd rather program it than do it).
But if possible I'd use math or derive or somesuch tool..
maybe a simple way
(a+b)*(c+d)*(e+f)
adding one column at a time, then sum the rows:
a * c * e
a * c * f
a * d * e
a * d * f
b * c * e
b * c * f
b * d * e
b * d * f
________________ +
(actually I've never really thought about it before..)
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| « Last Edit: Jun 19th, 2003, 12:03pm by towr » |
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NickH
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Re: Simple Proofs
« Reply #7 on: Jun 19th, 2003, 1:23pm » |
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Sir Col, have you ever shown your pupils the proof of the irrationality of sqrt(2) using the Fundamental Theorem of Arithmetic? You'd have to skip the proof of the theorem, as it's non-trivial, and that may negate your purpose, but given those points, the proof is wonderfully intuitive. It also makes crystal clear why all non-perfect squares are irrational, and similarly for higher powers.
Here is the proof, for those unfamiliar with it.
The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be written uniquely as a product of finitely many prime numbers.
Now suppose a/b = sqrt(2), where a/b is a fraction in its lowest terms. Then a2 = 2b2, with both sides having the same prime factorization.
On the L.H.S., each prime occurs an even number of times, while, on the R.H.S., 2 occurs an odd number of times. Contradiction!
The proof can also be presented constructively. If a/b = sqrt(n), then a2 = nb2, and so each prime must occur an even number of times in n; that is, n must be a perfect square.
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| « Last Edit: Jun 19th, 2003, 1:39pm by NickH » |
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Sir Col
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Re: Simple Proofs
« Reply #8 on: Jun 19th, 2003, 2:14pm » |
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I hadn't thought of doing that, NickH, thanks. The particular class have already done a little bit of work writing numbers as a product of prime factors, so it shouldn't be too difficult for them to follow the proof.
So you don't think they'd be up for the fundamental theorem of arithmetic as their second ever proof? 
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Sir Col
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Re: Simple Proofs
« Reply #9 on: Jun 19th, 2003, 2:43pm » |
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One of my favourite proofs uses the integer part function. Like your proof, it can be presented in such a way that it avoids the philosophical objections to proofs by contradiction and work constructively to determine the nature of the sqr(n):
Assume that sqr(n)=a/b, where b is the least integer s.t. b*sqr(n)=a is integer.
Let c=b(sqr(n)–[sqr(n)]). As 0<=sqr(n)–[sqr(n)]<1, 0<=c<b and c=b*sqr(n)–b*[sqr(n)] is integer.
c*sqr(n)=b*n–b*sqr(n)*[sqr(n)] is also integer, but b is the least such integer for which b*sqr(2) is integer. Therefore c*sqr(n) can only be integer if c=0 which means that sqr(n)=[sqr(n)]. Hence n must be a perfect square for sqr(n)=a/b.
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Icarus
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Re: Simple Proofs
« Reply #10 on: Jun 19th, 2003, 5:30pm » |
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Quote:
Although I said that I don't like cheap methods, I do confess to teaching the 'moon' method to weaker students. If you've never heard of it, try this: Write (a+b)(c+d) on a piece of paper, then join a to c and a to d with arcs above the brackets, then join b to c and b to d below the brackets. It should form a convincing moon shape. I would not use this method with more able students, as I believe it is necessary to understand, and talk in terms of, the law of distributivity. I would quickly move on to trinomial factors with them to ensure that they don't invent their own gimmicky methods. |
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I don't see a "moon" in that! It is similar to the "smiley-face" method I mentioned: in it you connect a-c and b-d above, and b-c and a-d below. The equation itself forms the eyes. The arcs above are eyebrows (clearly a unibrow guy), the b-c arc is the nose, and the a-d arc is the mouth.
And yes, for all my grousing, I too have succumbed to teaching the FOIL method to students who were just to slow to catch on!
on Jun 19th, 2003, 4:22am, wowbagger wrote:
I don't know the methods/mnemonics you mentioned, but these college students weren't able to use the distributive law?
What were they studying? Biology?
Now, I don't claim that math education is really great round here (espacially that of Joe Public, of course) - but these are students, right? |
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I was a "Graduate Teaching Assistant" at a public university. Kansas has 5 public universities, and they are required to accept everyone who graduates from a Kansas high school. Since all the best-paying jobs require a college degree, the large majority of high school students go on to college, and because of that rule, they all get in.
One consequence of this, and of the "self-esteem is more important than accomplishment" attitude that was particularly in vogue in those days, was a lot of low-achievers in college math classes. Out of all the courses the university offered, the remedial Mathematics courses had the largest enrollment, and because the professors did not care to deal with them, they were given to the GTAs to teach. During my time, I was extremely fortunate to be allowed to teach one semester of true Calculus (I taught several semesters of Business Calculus). I was the only Teaching Assistant ever to teach such a high level course!
Quote:Probably my worst experience so far was a student (college/university level) who didn't remember how to differentiate a polynomial. By the way, she studies something called "Marine Environmental Sciences" (I don't know the official translation), which focuses on Chemistry and Biology.
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It isn't a worst experience, but one of my favorite "examples" was the student who explained to me that the reason he had such a hard time with mathematics was that he had decided to live his life by logic, and math was just so illogical! 
[Curse you, Gene Roddenberry!  ]
I don't think the explosion from me that followed convinced him of the error of his ways!
on Jun 19th, 2003, 11:34am, Sir Col wrote:| Consider how you would expand (a+b)(c+d) and (a+b)(c+d)(e+f). For the triple factor, do you work out two brackets first, then multiply by the other, or do you expand in one go? What would you do with four factors? Would it matter if you had trinomial factors: e.g. (a+b)(c+d)(e+f+g)? |
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It's really hard for me to say here. It depends on the situation what I would do. For the simple examples shown, I would just start pulling terms out in "lexigraphical" fashion:
(a+b)(c+d)(e+f+g) = ace+acf+acg+ade+adf+adg+bce+bcf+bcg+bde+bdf+bdg
But for some situations, I will multiply them out in sequence. It mostly has to do with how complicated it is to keep track of which terms you've done, and which you haven't.
Quote:| The particular class have already done a little bit of work writing numbers as a product of prime factors, so it shouldn't be too difficult for them to follow the proof. |
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This has raised my curiousity. When I was in elementary school (no - no dinosaurs, but we did have the occasional streaker  ) factorization was large part of my 3rd(?) & 4th grade mathematics, since it is needed in simplifying fractions.
Is it really true that your 11/12 year old students have not done much prime factorization before, or am I misunderstanding this?
It has been my observation that kids are being taught mathematics earlier than I was. (For example, when I was in high school only one high school in town was offering any calculus, and it was a special half-semester "Advanced-Placement" course for college-bound seniors. I learned calculus by borrowing my brother's college textbook and reading through it. Today, they are teaching some calculus to 9th graders.) So it comes as a surprise to discover that they are also learning some of it later.
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Sir Col
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Re: Simple Proofs
« Reply #11 on: Jun 20th, 2003, 12:07pm » |
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on Jun 19th, 2003, 5:30pm, Icarus wrote:
I don't see a "moon" in that! It is similar to the "smiley-face" method I mentioned: in it you connect a-c and b-d above, and b-c and a-d below. The equation itself forms the eyes. The arcs above are eyebrows (clearly a unibrow guy), the b-c arc is the nose, and the a-d arc is the mouth. |
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I don't like the smiley face trick, as it requires too much imagination to see it, but I really must get you to see this moon. It's highly visual for less able students who are unlikely to expand anything beyond (a+b)(c+d). Join a-c and a-d above. Join b-c and b-d below. You may need to exaggerate the height of the arc from a-d and similarly with b-d... See it now?
on Jun 19th, 2003, 5:30pm, Icarus wrote:| Is it really true that your 11/12 year old students have not done much prime factorization before, or am I misunderstanding this? |
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True – prime factorisation is not done until secondary school (aged 11+) and does not advance significantly beyond using the p.f.s to find HCF and LCM of pairs of small numbers until aged 14/15; at which point they begin to learn how to simplify surds.
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NickH
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Re: Simple Proofs
« Reply #12 on: Jun 20th, 2003, 1:32pm » |
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Quote:| One of my favourite proofs uses the integer part function... |
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Nice proof, Sir Col! I've seen that before, but only by contradiction.
Out of curiosity, have you ever asked your pupils to prove that gcd(a,b) * lcm(a,b) = ab? If so, how did they get on?
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Sir Col
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Re: Simple Proofs
« Reply #13 on: Jun 20th, 2003, 6:02pm » |
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I've never proved it with a class formally, but as I usually encourage an investigative approach to their learning, some of them discover the principle. They usually verbalise it from a particular case:
E.g. 12=2x2x3 and 18=2x3x3
HCF(12,18)=2x3
LCM(12,18)=2x2x3 x 3
Their reasoning will be something along the lines: the LCM contains all the prime factors from the first number 'plus' the prime factors from the second number that haven't already appeared. Therefore the ones left (not used in the second number) will be found in both and will be the HCF. So LCM * HCF is the complete set of prime factors from both numbers; that is, their product.
Although the most common way that they discover this is by noticing that a*b/HCF is the LCM. This is not surprising for them, as they know that there is a connection with finding the lowest common denominator in fraction arithmetic. They are aware that, although not the lowest, the product of the denominators can be used as a common denominator, then a little work on it (divide it by the HCF) produces the lowest common denominator, LCM.
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| « Last Edit: Jun 20th, 2003, 6:03pm by Sir Col » |
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Sir Col
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Re: Simple Proofs
« Reply #14 on: Jun 26th, 2003, 2:22pm » |
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An interesting discussion took place with one of my classes yesterday (11/12 year olds). They'd been working with square numbers and were trying to find positive solutions to x2=n. When they tried to solve for n=2, they used a decimal search method to refine their 'answer'. Eventually they began to wonder if you could really find the answer or maybe it was just because their calculator couldn't display any more figures. After a series of observations they made, with regards to solving this problem one way or the other, we talked about the last digit. They realised that the last digit of the number you're squaring, even if it contains a decimal fraction, will determine the last digit of your answer. By considering all possible endings of numbers with decimal fractions: 1-9, the square will end in 1,4,9,6 or 5; so it was clear that they could never get the answer 2 by squaring any decimal.
Question: is this a valid proof that sqr(2) is irrational/cannot be expressed in decima form?
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Icarus
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Re: Simple Proofs
« Reply #15 on: Jun 26th, 2003, 7:28pm » |
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No - it isn't sufficient. It shows that sqr(2) cannot be a terminating decimal. But most rational numbers are not terminating decimals either, so it is not enough to show that sqr(2) is not rational.
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Sir Col
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Re: Simple Proofs
« Reply #16 on: Jun 27th, 2003, 2:21pm » |
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Yes, indeed, a good point, Icarus. I wonder if this method can still be rescued though...
If we're considering a/b=sqr(2). We get a2=2b2. As the last digit of any square will be 0,1,4,9,6 or 5 and its double will be 0,2 or 8 we can see that no solution exists; we can rule out 0 as it would require a to be a multiple of 10 and b to be a multiple of 5.
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Re: Simple Proofs
« Reply #17 on: Jul 5th, 2003, 3:13pm » |
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That works - and can be generalized to show that all numbers whose last digit is 2,3, 7, or 8 have irrational square roots.
Another approach, related to the one Nick gave, is to show that if a and b have no common factors, then neither do an and bn, for any n. Therefore (a/b)n = an/bn is in reduced form when a/b is.
Thus, if a reduced fraction p/q is to be a perfect nth power of some reduced fraction a/b, then p=an and q =bn. I.e. the only way a rational number can be a perfect nth power is for its numerator and denominator to be nth powers of integers.
For the particular case of sqr(2): p/q= 2/1 and n=2. The denominator is the square of 1, but 2 is not the square of an integer. So sqr(2) must be irrational.
The advantage of this approach is that it handles not only sqr(2), but also all nth roots of rational numbers. The disadvantage is that showing an is relatively prime to bn requires some knowledge of primes (though not the full fundamental theorem of arithmetic).
In particular, your students will need to know that if two numbers have a common factor > 1, then they also have a prime common factor. And that if p is a prime factor of an, then p is also a factor of a.
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