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Okay I'm embarrassed to even ask this clearly I am not going to win a Nobel prize in my life.

Question

What values of $x$ satisfy:

$$\frac{x^2}{x} \le 0 $$

My attempt

I'm very tempted to simplify the LHS and say the answer is $x \le 0$

But I have this nagging concern that the answer is actually $x<0$

If I start from the other direction, then I have

$$x \le 0$$

$$x*1 \le 0*1$$

$$x\frac{x}{x} \le 0 \frac{x}{x}$$

And the last line only holds if $x$ is not zero and therefore it changes to

$$x\frac{x}{x} < 0 \frac{x}{x}$$

$$\frac{x^2}{x} < 0 $$

And now that there is no threat of dividing by zero I can reduce it to $x<0$

Is this right?

I feel like in general when doing math I multiply top and bottom by $x$ all the time and I never really think about explicitly calling out that $x$ can't be zero.

Thanks for your help/patience.

HJ_beginner
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    People can certainly improve throughout their life. But I have a suspicion you won't get a Nobel prize in math. – mathworker21 Mar 23 '18 at 00:54
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    @mathworker21 I don't think there is a Nobel prize in math. – valer Mar 23 '18 at 00:55
  • Thanks for all the help. In general when you have $5x = 15$ can you do $5x \frac{x}{x} = 15 \frac{x}{x} $ so $x=3$ and $x \ne 0$ therefore $x=3$. I know that is kind of ridiculous but is it the right thinking? – HJ_beginner Mar 23 '18 at 00:56
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    It is confusing because the inequality, $$\frac{x^2}{x}\leqslant 0$$ is incorrect since division by $0$ is impossible, and $\dfrac{x^2}{x} = x^{2-1} = x$. – Mr Pie Mar 23 '18 at 01:05
  • @user477343 Interesting... so the values of $x$ that "satisfy" the inequality are $x<0$ but you are right you will never "fully satisfy" the original inequality since you can't get $0$ to do the full "$\le$"... is that what you're talking about? – HJ_beginner Mar 23 '18 at 01:12
  • @valer https://imgur.com/gallery/utzTCyo – Umberto P. Mar 23 '18 at 01:17
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    @HJ_beginner yes. This proves that if $f(x) = \dfrac{x^2}{x}$, it does not necessarily mean that $f(x) = x$. In this case, we just have an inequality. – Mr Pie Mar 23 '18 at 01:26
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    @valer that's the point – mathworker21 Mar 23 '18 at 02:06
  • @user477343 Thanks that is a good way of putting it. – HJ_beginner Mar 23 '18 at 02:11
  • @valer @ mathworker I stand corrected I will not win a Fields medal. I also won't win a Nobel either. Honestly I don't think I'm going to win anything in life except death and freedom from the 3 lbs of fat I'm trapped in. – HJ_beginner Mar 23 '18 at 02:12

3 Answers3

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If $x=0$, $\frac{x^2}{x}$ is undefined.

If $x>0$, $\frac{x^2}{x}=x >0$.

If $x<0$, $\frac{x^2}{x} = x<0$.

Hence the answer is $x<0$.

Siong Thye Goh
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  • Thanks Siong as always you are the man. In general when you have $5x = 15$ can you do $5x \frac{x}{x} = 15 \frac{x}{x} $ so $x=3$ and $x \ne 0$ therefore $x=3$. I know that is kind of ridiculous but is it the right thinking? – HJ_beginner Mar 23 '18 at 00:57
  • If $5x=15$, I don't see how multiply and divide by $x$ helps,I would just divide both sides by $5$ and conclude that $x=\frac{15}3=5$. – Siong Thye Goh Mar 23 '18 at 01:00
  • Yeah I agree that multiplying by $\frac{x}{x}$ is ridiculous... I'm just trying to solidify the rules in my head. – HJ_beginner Mar 23 '18 at 01:01
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    You can do that after you check that $x \ne 0$. but yup, doesn't seem to simplify it though it is alright to do so. – Siong Thye Goh Mar 23 '18 at 01:02
  • Yes I'm thinking since $x \ne 0$ and $x = 3$ these two statements reduce to $x=3$ – HJ_beginner Mar 23 '18 at 01:04
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The top is always non-negative, so you need the bottom to be purely negative to make the whole thing less than $0$. Hence $x <0$.

Note that having it equal to $0$ is out of the question because this could only occur when the top is $0$ but the bottom isn't, which is impossible because the top is $0$ only when the bottom is.

Randall
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    Ahhh I was so caught up in solving the problem by manipulating things that I forgot about just looking at the problem. Thanks that helps a lot! – HJ_beginner Mar 23 '18 at 00:56
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    We've all been there. – Randall Mar 23 '18 at 00:57
  • In general when you have $5x = 15$ can you do $5x \frac{x}{x} = 15 \frac{x}{x} $ so $x=3$ and $x \ne 0$ therefore $x=3$. I know that is kind of ridiculous but is it the right thinking as I'm trying to solidify the rules of when you can multiply top and bottom by $x$. – HJ_beginner Mar 23 '18 at 01:02
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For

$$\frac{x^2}{x} \le 0 $$

we need $x\neq0$ then numerator is always positive and the sign depends by the denominator then

$$\frac{x^2}{x} < 0 \iff x< 0$$

and never $\frac{x^2}{x} =0 $.

user
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  • Appreciate it. I think for novices like me it's just so tempting to simply the LHS to $x$ and I couldn't articulate in my head why that is a no no. – HJ_beginner Mar 23 '18 at 01:03
  • I can add a solution by simplification explaining when and how it can be done, if useful. – user Mar 23 '18 at 01:05
  • Can you help me with another thought experiment. If you have $5x = 15$ can you do $5x \frac{x}{x} = 15 \frac{x}{x} $ so $x=3$ and $x \ne 0$ therefore these two statements about $x$ become one, namely $x=3$. I know that is a strange thing to do but I'm trying to solidify the rules of multiplying top and bottom by $x$. – HJ_beginner Mar 23 '18 at 01:09
  • I don’t understand your step, of course the thing to fo here is $5x=15\iff \frac55x=\frac{15}5$. – user Mar 23 '18 at 01:14
  • Yes I agree it's a bizarre thing to do and nobody would do it. But I believe it is technically legal... and I was just trying to square in my head how introducing $\frac{x}{x}$ into an equation works for a simple example. – HJ_beginner Mar 23 '18 at 01:16
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    For you inequality you csn indeed condider $$\frac{x^2}{x} =\frac{x}{x} x\le 0 $$ and for $x\neq 0$ we have $\frac{x}{x}=1$ and then $\frac{x}{x} x\le 0\implies x<0$. – user Mar 23 '18 at 01:27
  • Thanks for your help gimusi that helps. You are doing the Lord's work. – HJ_beginner Mar 23 '18 at 02:10