Spot Student Mistakes: 1/x < x < 1

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The discussion centers on identifying mistakes in the student's reasoning regarding the inequality 1/x < x < 1. The first error is the incorrect assumption that x is always positive, leading to the false conclusion that 1 < x^2 follows from the initial inequalities. It is highlighted that the implication does not hold for negative values of x, as demonstrated with the example of x = -5. Additionally, the participant emphasizes that if x were negative, the inequalities would not yield valid results, further complicating the student's conclusions. The overall consensus is that the student's assumptions about the positivity of x led to multiple logical errors in their reasoning.
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Homework Statement


Spot the mistakes of a student


Homework Equations


1/x < x < 1
therefore 1<x^2
therefore 1<x
but x<1 therefore there are no solution


The Attempt at a Solution


the questions requires me to spot the mistake made by a student
so first of all
in the 2nd line , it reads 1<x^2 . this statement is wrong since the student assumes that x is always >0

i'm not too sure about about the second one , namely 1<x . Since it's related to the first one . What should i put ?
is there any other mistakes i have not spotted ?
 
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If you are given the statement x2 > 1, it doesn't necessarily imply that x>1. Consider the case where x=-5. x2=25 which is certainly greater than 1, but x>1 is not true. Again, the student has assumed that x is positive.
 
so does it mean there's only one mistake ?
 
If x &gt; 0 you can't have

<br /> 0 &lt; \frac 1 x &lt; x &lt; 1 <br />

because this is equivalent to

<br /> 0 &lt; 1 &lt; x^2 &lt; x<br />
 
If 1/x < x < 1, it does not necessarily follow that 1 < x^2.

For example, for x=-1/2, we have -2 < -1/2 < 1, but 1 < 1/4 is false.

In fact, the implication is false for all -1 < x < 0.
 
If the previous post was directed at mine, you missed one of my points.
I said
If x &gt; 0 you can't have

<br /> 0 &lt; \frac 1 x &lt; x &lt; 1<br />

because (if you multiply through the inequality by x) then you would have

<br /> 0 &lt; 1 &lt; x^2 &lt; x<br />

My initial comment ruled out negative values from consideration. logarithmic, if I misunderstood you post by assuming it was meant at me, I apologize.
 
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