I am trying to find a counter example to show that the statement below is false, but I am having difficulty in trying to find a reasonable argument. Here is the statement:
$n^2-12n + 35 \geq 0$ for every positive integer $n$.
This is all I have thus far upon trying to show that the statement is false -
Let $n = 6$.
Then, we have the following:
$(6)^2 - 12(6) + 35 = -1$.
Would anyone be able to point me in the right direction or possibly give me some hints to further disprove this statement?
On
The equation $\bf f(n)= n^2-12n + 35$ is a quadratic equation in $\bf n$, Discriminant of which is $\bf D=\sqrt{12^2-4.35}=2$ which is positive and thus there must exist two distinct roots of it which are $\bf 5$ and $\bf 7$ [remember $\alpha,\beta=\{12\pm\sqrt{4}\}/2$ from quadratic formula];Since it is a parabola facing upwards it must have negative values in between $\bf 5$ and $\bf 7$ which includes your $\bf 6$.
To prove that the statement:
$$\forall n\in\Bbb N,\quad n^2-12n+35\ge0$$ isn't true it suffices to show that its negation: $$\exists n_0\in\Bbb N,\quad n_0^2-12n_0+35<0$$ is true and your choice $n_0=6$ does the job.