I'm veryconfused with an a level maths q

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In summary, a triangle with sides that can be written in the form n^2+1, n^2-1, and 2n (where n>1) is right angled. This is proven by showing that the side n^2+1 is greater than or equal to the side 2n. However, the converse is false as shown by a counter example. The rest of the proof involves using Pythagoras' theorem, but the details are not fully understood.
  • #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 don't 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.

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

if anyone would like to explain, then i would be very grateful. thanks!
 
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  • #2
He is simply showing which side is longer,

[tex](n^2+1) - (2n) = (n-1)^2[/tex]

The left hand side is the difference between 2 of the sides.

However, [itex](n-1)^2 \geq 0[/itex]

So you can replace [itex](n-1)^2 [/itex] with: [itex]\geq 0[/itex] and you get:

[tex](n^2+1) - (2n) \geq 0[/tex]

Adding 2n to both sides:

[tex]n^2+1 \geq 2n [/tex]

So you know that the n2 + 1 side is greater or equal to the 2n side.
 
  • #3
Please,do not double post.It's not really fair.

Daniel.
 

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