I'm very confused with an a level maths q

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  • #1
liz
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the question:
(im presuming n^2 means n squared)

prove the following result: "a traingle with sides that can be written in the form n^2+1, n^2-1, and 2n (where n>1) is right angled.
show, by means of a counter example, that the converse is false.

this q was taken from the back of the book "the curious incedent of the dog in the night-time" and there was a full proof but i dont understand some of it.

it start by exmplaining we need to prove which side is the longest by doing:

n^2+1 - 2n = (n-1)^2

if n>1 then (n-1)^2 >0

therefore n^2+1 > 2n

similarily (n^2+1) - (n^2-1) = 2

therefore n^2+1 > n^2-1

so n^2+1 > n^2-1.

does anyone understand this?????

the rest is worked out using pythagoras but then the converse bit has completly lost me.

if anyone would like to explain, then i would be very grateful. thanks!
 
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Answers and Replies

  • #2
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The converse would be:

If we have a right-angled triangle, then it's sides satisfy [tex] n^2 +1, n^2 -1, 2n [/tex]. So you find a single particular example where this fails, and succeed in showing that the converse is not true.

Hope that helps, good luck.

Ro-me-o
 
  • #3
liz
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thanks thats really helpful. do you understand the n^2+1-2n = (n-1)^2 ....... stuff?

liz
 
  • #4
HallsofIvy
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liz said:
thanks thats really helpful. do you understand the n^2+1-2n = (n-1)^2 ....... stuff?

liz

It's really just arithmetic- well, fancied up a little!

If a- b is a positive number, then a> b. Do you understand that?

To show that n2+ 1 is larger than 2n, look at the difference:
n2+ 1 - 2n.

That is equal to n2- 2n+ 1= (n- 1)2 and a square is never negative. (If n= 1, that can be 0 but if n> 1 it is positive and so n2+ 1 is larger than 2n.


Similarly, to show that n2+ 1 is larger than n2- 1 (which is almost obvious anyway) look at the difference (n2+1)- (n2- 1)= (n2- n2)+ (1- (-1))= 2 which is positive.

(The reason I said "almost obvious" is that it should be clear that n2+1> n2> n2-1.)

Once you know that, to use the Pythagorean theorem, look at
(n2+1)2= n4+ 2n2+ 1

(n2-1)2= n4- 2n2+ 1
(2n)2= 4n2

so (n2-1)2+ (2n)2= n4+ 2n2+ 1. Yep, c2= a2+ b2 so this is a right triangle.

The converse would be "If a, b, c are the sides of a right triangle (with a,b,c integer valued?), c being the hypotenuse, then c= n2+ 1, a= n2-1, b= 2n for some integer n. My first thought was the simple 3,4,5 triangle, but that turns out to work: 3= 22-1, 4= 2(2), 5= 22+ 1. Okay, what about 6, 8, 10?
 
  • #5
honestrosewater
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c = 13 is a counterexample.
 
  • #6
liz
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thank you it looked more difficult than it is. but in my defence i showed it to my boyfriend who is just finishing his maths A level and he couldn't understand it either. a counter example that works is 65, 60 and 25.
 
  • #7
honestrosewater
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liz said:
thank you it looked more difficult than it is. but in my defence i showed it to my boyfriend who is just finishing his maths A level and he couldn't understand it either. a counter example that works is 65, 60 and 25.
I wish I knew how to say this, but I think the set of all 3-tuples (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)), for natural m and n, will give you the lengths of the sides of all right triangles with sides of natural length. And I think n = 1 is the only value for which (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1) = (x2 - 1, 2x, x2 + 1), for some natural x > 2. So every element of the set of all 3-tuples (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)), for natural m and n, with n > 1, should be a counterexample. Just some casual observations - maybe someone else could try to prove it. :wink:
 
  • #8
liz
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i dont know i dont really understand how you got that probably cos i dont know what "3- tuples" or "natural m and n" means. i didnt actually think of the counter examle myself. the question with the answer including the counter example was taken from a book. i worked out that the 65, 60 and 25 triangle is a right angle using pythagoras then used 65= n^2+1 to get n=8. then the other sides would be 8^2 -1 = 63 and 2n = 16. which isnt 60 and 25 therefore proving that not all right angled have sides n^2+1, n^2-1 and 2n.
or so i thought. im probably wrong, im a bit out of my depth with this question.
 
  • #9
honestrosewater
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liz said:
i dont know i dont really understand how you got that probably cos i dont know what "3- tuples" or "natural m and n" means.
Don't worry, it's simple. You already know the natural numbers; They're the members of the infinite set {1, 2, 3, 4, ...}. A tuple is just a finite sequence of objects, like (1, 2, 1). Order matters for sequences, so (1, 2, 1) is not the same as (2, 1, 1). There are x many objects in an x-tuple. So a 3-tuple is a sequence of 3 objects, a 10-tuple is a sequence of 10 objects, etc.
i didnt actually think of the counter examle myself. the question with the answer including the counter example was taken from a book. i worked out that the 65, 60 and 25 triangle is a right angle using pythagoras then used 65= n^2+1 to get n=8. then the other sides would be 8^2 -1 = 63 and 2n = 16. which isnt 60 and 25 therefore proving that not all right angled have sides n^2+1, n^2-1 and 2n.
or so i thought. im probably wrong, im a bit out of my depth with this question.
You're correct- your counterexample is actually my counterexample in a way- they form similar triangles (meaning, among other things, the lengths of their sides are proportional). My example was (5, 12, 13)- multiply each term by 5 to get your example.
I'm trying to prove the first part of my suggestion in another thread. I'll post my result, in case you're interested.
 
  • #10
liz
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yes i am interested, thank you.
 
  • #11
honestrosewater
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I was wrong. Though (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)) always gives you the sides of a right triangle, it doesn't give you the sides of all right triangles. You can read all about the solution I was looking for here and here. (BTW, the 3-tuples I've been talking about are called Pythagorean Triples, and I discovered they have a long and interesting history.)
 

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