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I'm wondering what mathematical rule is applied when negating a polynomial expression. For example, in high school it is taught that $-(-6x^2 + 15x - 5) = 6x^2 -15x + 5$, but what rule(s) is applied here? Some say it's the distributive property (distributing the negation), but I'm not sure about that because negation is an operation, not a number.

It's easier for me to see that $-(-6) = 6$ (for example) because of the rule that the negation of a negative number is the corresponding positive number.

yroc
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  • you can use -X = (-1)*X to make that analogy with distribution precise. –  Mar 04 '13 at 16:08
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    @user58512, so you're saying that the distributive rule is used in combination with the rule -x = -1x? There is an official rule that -x = -1x? – yroc Mar 04 '13 at 16:17
  • if $-x$ is the unique number such that $x + -x = 0$ then $$x + (-1)\cdot x = 1\cdot x + (-1)\cdot x = (1 + (-1))\cdot x = 0\cdot x = 0$$ proves that $(-1)\cdot x = -x$. –  Mar 04 '13 at 16:22

6 Answers6

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Negation is an operation, yes. Specifically, it is multiplication by $-1$. That's where distributivity comes into play.

To see why this is true, note that $$x+-x=0=0\cdot x=1\cdot x+(-1)\cdot x=x+(-1)\cdot x,$$ so $$-x=(-1)\cdot x.$$

Cameron Buie
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The distributive property is used to distribute the factor of $-1$ over all the terms: $$ \begin{align} -1\cdot(-6x^2+15x-5) &=-1(-6x^2)+-1(15x)+-1(-5)\\ &=6x^2-15x+5 \end{align} $$

robjohn
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It's the distributive property of multiplication over addition: $$-(-6x^2 + 15x - 5) = -1\cdot(6x^2 + 15x - 5) = -1\cdot 6x^2 + -1\cdot 15x - (-1)\cdot 5 = -6x^2 -15x + 5$$

amWhy
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$(6x^2-15x+5)+(-6x^2+15x-5)=0$, now add the negative of $-6x^2+15x-5$ to each side.

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That $\rm\ {-}(-6x^2\! + 15x - 5)\, =\, 6x^2\! -15x + 5\ $ does not require the distributive law. Rather, it is true more generally in any abelian (commutative) group that $\rm\:-(a+b) = -a + -b,\:$ because $$\rm\:-a + -b + a + b\, =\, -a + a + -b + b\, =\, 0 + 0\, =\, 0$$

In the special case of a ring, one has that $\rm\: -x = (-1)x,\:$ so one can instead use distributivity $$\rm -(a+b)\, =\, (-1)(a+b)\, =\, (-1)a + (-1)b\, =\, -a + -b$$

which then yields the Law of Signs $\rm\,\ (-a)(-b) = ab\,\ $ and related properties.

But, conceptually, your inference is purely a property of groups, true also in nonabelian groups

$$\rm (abc)^{-1} =\, c^{-1}b^{-1}a^{-1}$$

Community
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Math Gems
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  • so you're saying that $(-a+-b)+(a+b)=0$ (due to commuitative property) shows that $(-a+-b)$ is the negation of $(a+b)$. Thus, $(-a+-b)$ = $-(a+b)$. That makes sense to me -- I just want to be sure I'm not oversimplifying what you're saying... – yroc Mar 05 '13 at 20:02
  • @yroc Yes, that's right. The proof requires only the additive structure of the ring. – Math Gems Mar 05 '13 at 20:11
  • That's really neat. Thanks! – yroc Mar 05 '13 at 21:08
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Suppose we know the rule that $(-a) = (-1) a$ and we know the distributive property for multiplication over addition. Then,

$$-(a+b) = (-1) (a+b) = (-1)a + (-1) b = (-a) + (-b)$$

so we've discovered a new rule: negation distributes over addition. In some form it distributes over subtraction too:

$$-(a-b) = (-1) (a-b) = (-1) - (-1) b = (-a) - (-b) = (-a) + b$$

or across many terms

$$-(a+b-c+d-e) = (-1) (a+b-c+d-e) = \cdots = -a - b + c -d + e $$

Distribution of negation fits into the same spatial reasoning that distribution of multiplication does; visually, one "sees" the same thing happening in both cases. For the vast majority of purposes, it is not useful to distinguish between the two.