Question 1200073
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{{{root(n,x) = 1-x}}}


{{{root(n,x)-1+x = 0}}}


{{{f(x) = 0}}}


The left hand side is the function {{{f(x) = root(n,x)-1+x}}}


Plug in x = 0
{{{f(x) = root(n,x)-1+x}}}


{{{f(0) = root(n,0)-1+0}}}


{{{f(0) = 0-1+0}}}


{{{f(0) = -1}}}
The result is negative.


Plug in x = 1
{{{f(x) = root(n,x)-1+x}}}


{{{f(1) = root(n,1)-1+1}}}


{{{f(1) = 1-1+1}}}


{{{f(1) = 1}}}
The result is positive.


f(x) changes from negative to positive when going through this interval 0 < x < 1.
Somewhere in the middle we must have f(x) = 0 occur at least once, which leads to f(x) having at least one root in this interval.
This is because f(x) is a continuous curve.


Identities used:
{{{root(n,0) = 0}}}
{{{root(n,1) = 1}}}
Those two identities are based on
{{{0^n = 0}}} where n > 0
{{{1^n = 1}}}
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