- #1
liz
- 23
- 0
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.
does anyone understand this?
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!
(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.
does anyone understand this?
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!
Last edited: