- #1
Ulagatin
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Hi everyone,
Any assistance with this following problem would be greatly appreciated. I'm in Year 11 and working through Apostol volume 1.
sin n*pi = 0, where n is an integer
sin n*pi =/= 0, where n is not an integer.
Prove these statements...
Hmm...
My idea is that perhaps I need to use set theoretic ideas or mathematical induction, initially to show that the first case holds for all n.
sin 0*pi = 0
sin 1*pi = 0
sin (1+1)pi = 0 = sin (n+1)*pi
sin n*pi = 0
So, sin (n+1)*pi = 0 (where n is an integer)
Now, to use set theoretic notation:
Let sin n*pi = A.
A = {n is an element of R|sin n*pi = 0, n is an integer}
B = {n is an element of R|sin n*pi =/= 0, n is not an integer}
.: A = {0}
.: B =/= {0}
Now, this is circular - I haven't proved anything at all. (sorry for not using latex, unsure about how to write set notation).
Any assistance would be grand.
Cheers,
Davin
Any assistance with this following problem would be greatly appreciated. I'm in Year 11 and working through Apostol volume 1.
Homework Statement
sin n*pi = 0, where n is an integer
sin n*pi =/= 0, where n is not an integer.
Prove these statements...
Homework Equations
Hmm...
The Attempt at a Solution
My idea is that perhaps I need to use set theoretic ideas or mathematical induction, initially to show that the first case holds for all n.
sin 0*pi = 0
sin 1*pi = 0
sin (1+1)pi = 0 = sin (n+1)*pi
sin n*pi = 0
So, sin (n+1)*pi = 0 (where n is an integer)
Now, to use set theoretic notation:
Let sin n*pi = A.
A = {n is an element of R|sin n*pi = 0, n is an integer}
B = {n is an element of R|sin n*pi =/= 0, n is not an integer}
.: A = {0}
.: B =/= {0}
Now, this is circular - I haven't proved anything at all. (sorry for not using latex, unsure about how to write set notation).
Any assistance would be grand.
Cheers,
Davin