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I have this problem:

$$\{x: (x \in\mathbb{Z})\land(x^2<8) \}$$

Originally I thought the solution was $\{-4,-1,0,1,4\}$ but $-2^2$ is $-4$ not $4$ so the answer would be infinite. (my original assumption was that a negative number squared was positive because $-2 \cdot -2=4$, but I've since found out this is not the case).

This is a new subject in my discrete maths exam that our lecturer threw at us today. I've never worked with set notation before. The exam is tomorrow.

Also... just to confirm I'm actually working out the answers correctly...

$$\{x: (x \in\mathbb{N})\land(3x<10) \}$$

is

$$\{0,3,6,9\}?$$

Siong Thye Goh
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blackcat47
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  • You're having a set of $x$'s and not $x^2$'s so in the first set it should be {-2,-1,0,1,2} other then that $-2^2=(-1)\cdot (2^2)$ while $(-2)^2=4$ And similarly for the second one you're listing $3x$ instead of $x$'s so it should be ${0,1,2,3}$. – kingW3 Dec 13 '17 at 19:25
  • In the sense of "$x^2$ when $x=-2$," we mean $(-2)^2=4$, not $-2^2=-(2^2)=-4$. – Thomas Andrews Dec 13 '17 at 19:25
  • $-2^2$ is NOT a square number. There is no possible real value of $x$ so that $x^2 = -2^2 = -(2*2) = -4$. – fleablood Dec 13 '17 at 19:37
  • "but I've since found out this is not the case" NOOOOOOOOOO!!!!! and in a distant galaxy another math teacher dies a little bit inside. It most certainly **!!!!!! IS !!!!!!!** the case. Don't, don't, don't don't! convince yourself otherwise. – fleablood Dec 13 '17 at 19:54

4 Answers4

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You are looking for all integers $x$ such that $x^2<8.$ This is the set consisting of $-2,-1,0,1,$ and $2.$

Reiner Martin
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  • $$\{x: (x \in\mathbb{Z})\land(x^2<8) \}= \{ -2, -1, 0, 1,2\}$$

  • Squaring of negative number does give you positive number.

$$-2^2=-4$$

but we have $$(-2)^2=4$$

If $x=-2$, $x^2$ means $(-2)^2$.

  • Note that $4^2=16 \ge 8$, hence $4 \not \in \{x: (x \in\mathbb{Z})\land(x^2<8) \}$

  • Note that $3x < 10$ is equivalent to $x < \frac{10}3$.

  • Depending on your definition of $\mathbb{N}$,

If you consider $0 \in \mathbb{N}$ $$\{x: (x \in\mathbb{N})\land(3x<10) \}=\{0,1,2,3\}$$ if you consider $0 \not \in \mathbb{N}$ $$\{x: (x \in\mathbb{N})\land(3x<10) \}=\{1,2,3\}$$

Siong Thye Goh
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Your original thought process is correct.

The only thing to look out for is order of operations: You're right that the square of a negative number is a positive number. After all, $(-2)^2 = (-2)\times (-2) = 4$. But if you type $-2^2$ into your calculator, then due to order of operations, it will be interpreted as $-(2^2)=-4$, where the squaring unexpectedly happens before the minus sign is applied.

If set notation is confusing, you can put it into words:

$$\{x : (x \in \mathbb{Z}) \wedge (x^2 < 8)\}$$

means "Find every possible $x$, where $x$ is a positive-or-negative integer and $x^2 < 8$". The answer is $\{-2,-1,0,+1,+2\}$.

Similarly,

$$\{x : (x \in \mathbb{N})\wedge 3x < 10\}$$

means "Find every possible $x$, where $x$ is a whole number, and $3x < 10$." Your process looks exactly right: $\{0,1,2,3\}$.

Just make sure in each case that you're listing the values of $x$ itself, rather than listing $x^2$ or $3x$. To check your answer, like when you say the answer is $\{0,3,6,9\}$, take each of those values and plug them in to the equation $3x < 10$. For example, you can plug in $x=6$ and you find that $3x = 3(6) = 18 \not < 10$. So $x=6$ can't be an answer.

user326210
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  • "Your original solution is correct. " Not quite. The solution was {-2,-1,0,1,2}. The OPs original solution was {-4,-1,0,1,4} which confuses the set being the values of $x$ (correct) with the set being the values of $x^2$ (incorrect) and with $-x^2$ with $(-x)$. – fleablood Dec 13 '17 at 19:45
  • @fleablood Good point. I meant that the approach, with squares of negative numbers being positive, was correct. Thanks for the catch! – user326210 Dec 13 '17 at 19:46
  • I must say I died a little inside when I read "my original assumption was that a negative number squared was positive .. but I've since found out this is not the case" NOOOOOOO!!!!!! That IS the case! Go back ! Go back!. It's so sad to see a student be right, but reasonably confused and thus assume s/he is completely wrong and then go to convoluted steps to convince him/herself that something completely wrong must be right because they no longer trust their basic intelligence. Very very sad. – fleablood Dec 13 '17 at 19:51
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but $-2^2$ is $-4$ and not $4$

And $-2^2$ is not $(-2)^2$.

$-2^2$ is not a square number at all.

If $x = -2$ then $x^2 = (-2)^2 \ne -2^2$.

$(-2)^2 = 4$. There is NO possible number so that $anything^2 = -2^2$. That can NEVER happen.

(Err... that can never happen if we are working with real numbers. If we are working with complex or imaginary numbers than... well, then we are taking a different class....)

(my original assumption was that a negative number squared was positive because −2∗−2=4, but I've since found out this is not the case).

I sincerely hope you haven't! Go back! You were right the first time.

$(-2)^2 = -2*-2 = 4$

$x^2 \ge 0$.

And $(-3)^2 = (-3)*(-3) = 9 > 8$.

So no, it is not infinite as all $x < -3$ are such that $x^2 > 9$.

.....

Originally I thought the solution was {−4,−1,0,1,4}

The set is a set of the $x$s. Not of the $x^2$s.

So since $(-2)^2 = 4 < 8; (-1)^2 = 1 < 8; 0^2 = 0 < 8; 1^2 = 1 < 8; 2^2 = 4 < 8$ whereas $x \ge 3 \implies x^2 \ge 9 > 8$ and $x \le -3 \implies x^2 \ge 9 > 8$ it follows that acceptable values for $x$ are $-2,-1,0,1,2$ and all others are not acceptable.

.....

$\{x: (x \in\mathbb{Z})\land(x^2<8)\}$

means

All $x$ so that 1) $x$ is an integer and 2) $x^2 < 8$

It should be clear that that is $\{-2,-1,0,1,2\}$.

To be formal and pedantic $x^2 < 8 \iff -\sqrt 8 < x < \sqrt 8$ so $-3 < \sqrt 8 < x < \sqrt 8 < 3$ and as $x\in \mathbb Z$ then the set is $\mathbb Z \cap (-\sqrt 8, \sqrt 8)= \{-2,-1,0, 1,2\}$.

fleablood
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