How can I prove that the statement:
'there is no integer, $n$, such that $4n^2 + 1 < 4n$'
is true, by contradiction?
I greatly appreciate any help with this. Thanks.
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$f(x)=4x^2-4x+1$ is increasing on $\left[1,\infty\right)$ and $f(1)=1>0$ – Tucker Aug 09 '15 at 01:59
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@Tucker Likewise for $(-\infty,0]$ – ignoramus Aug 09 '15 at 02:03
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@ignoramus it is decreasing on that interval. Take a derivative and check the sign for $x\leq 0$ – Tucker Aug 09 '15 at 02:06
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@Tucker I meant that the result still holds for the negative integers. – ignoramus Aug 09 '15 at 02:09
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It is true that $4n<1+4n^2$ for $n<0$ – Tucker Aug 09 '15 at 02:11
2 Answers
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Assume that there is an $n$ for which $4n^2+1<4n$. Then
$\begin{align} 4n^2-4n+1<0 &\implies\\ (2n-1)^2<0 \end{align}$
which is a contradiction because $(2n-1)$ is an integer and squares of integers can never be negative.
Marconius
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See, I'm such a noob. I had (2n-1)^2 but didn't see that step. Thank you. – MathsNoob2015 Aug 09 '15 at 02:33
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Assume for the sake of contradiction that
$$4n^2+1<4n$$
Well it is certainly the case that
$$4n^2<4n^2+1$$
So we have
$$4n^2<4n^2+1<4n$$
$$4n^2<4n$$ $$4n^2-4n<0$$
$$4n(n-1)<0$$
So $n$ and $n-1$ must be of opposite sign
$n>n-1$ So we take $n-1<0$ and $n>0$ from which we get $0<n<1$ which contradicts our original assumptions that $n\in\mathbb{Z}$
Tucker
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The question states $n \in \mathbb{Z}$ not $n\in\mathbb{N}$. So there is a step where you say if $n = 0$ then $0^2<0$ which is a contradiction, and if $n < 0$ then a square is less than a negative which is also a contradiction. – ignoramus Aug 09 '15 at 02:13
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Thanks . I overlooked that detail but believe I have been been able to salvage the proof. – Tucker Aug 09 '15 at 02:17