For this question:
$$\forall n \in Z, \exists y \in R -{0}$$ such that $$y^n \leq y$$
Why is the following explanation wrong?
$$y^n\leq y=nln(y)\leq ln(y)= n \leq 1$$ Hence the statement is false
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Also, I negated the statement and got: $\exists n \in Z s.t \forall y \in R - {0}, y^n \geq y$ If I pick $n=2$ Then, the negated statement would be true. Hence the original statement is false, right? – Mengen Liu Oct 02 '22 at 00:42
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IF $n > 1$, then choose $y < 1$. If $n \leq 0$, choose $y > 1$. – Nick Matteo Oct 02 '22 at 00:45
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Hint: Doesn't this inequality hold for all positive integers when $y=1/2$? Can you find an example when $b$ is negative? – Ethan Bolker Oct 02 '22 at 00:45
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My bad, I didn't realize that, What about the explanation provided on the question? – Mengen Liu Oct 02 '22 at 00:47
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I read the $\mathbb R-0$ as $\mathbb R^{-}$. Sorry for this answer. – lone student Oct 02 '22 at 01:09
1 Answers
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The line of reasoning you presented is incorrect because $\ln y$ is less than $0$ for $y \in (0,1)$. [Recall $am < an$; $a$ negative; $\implies$ $m > n$.]
You should apply this to your OP to note the following:
If $n$ positive and $\not =1$ then $y^n < y$ for all $y \in (0,1)$. [Use $n >1$ and multiply both sides of $n >1$ by $\ln y$ and then note $\ln y < 0$ to conclude $n \ln y < \ln y$.]
If $n$ is nonpositive then $y^n < y$ for all $y > 1$. [Use $\ln y$ positive and $n$ nonpositive to conclude $n \ln y < \ln y$ i.e., a nonpositive number namely here $n \ln y$ is less than a positive one namely here $\ln y$.
What about $n=1$?
Or starting from the top, you could simply note directly that $(1/2)^n < 1/2$ for each integer $n \ge 2$ and $2^n < 2$ for each integer $n \le 0$, and $1^n \le 1$ [actually equals of course] for the remaining $n=1$.
Mike
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1@EthanBolker I am only trying to fix the OP's approach. Of course logs are certainly not needed. – Mike Oct 02 '22 at 01:37