- #1
geoman
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1. show that for each positive integer n and each real number x, f(nx)=nf(x).
f(x) is an additive function (f(x+y)=f(x)+f(x))
Well I'm thinking that I can just use mathematical induction to show that:
1) f((1)x)=1f(x)
2)f(nx)=nf(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+...+f[tex]_{}n[/tex](x) (assume this is true)
3)f((n+1)x)=(n+1)f(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+...+f[tex]_{}n+1[/tex](x)
from here it would just be algebra. is this right?
note: the numbered functions above are supposed to be subscripted but i messed it up somehow. They are only supposed to number the functions as I add the function n times and then n+1 times
Homework Equations
f(x) is an additive function (f(x+y)=f(x)+f(x))
The Attempt at a Solution
Well I'm thinking that I can just use mathematical induction to show that:
1) f((1)x)=1f(x)
2)f(nx)=nf(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+...+f[tex]_{}n[/tex](x) (assume this is true)
3)f((n+1)x)=(n+1)f(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+...+f[tex]_{}n+1[/tex](x)
from here it would just be algebra. is this right?
note: the numbered functions above are supposed to be subscripted but i messed it up somehow. They are only supposed to number the functions as I add the function n times and then n+1 times
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