Rekenen met machten

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Deze pagina geeft een overzicht van de rekenregels voor machten. De focus ligt op rekenvaardigheid. Voor meer nauwkeurige en theoretische bespreking verwijzen we door naar een handboek.

(Machtsverheffing)

Voor een natuurlijk getal $n$ en een reëel getal $a$ is de $n$-de macht van $a$ het getal

$\relax \fcolorbox {black}{white}{$ a^n \perdef \underbrace {a\cdot a\cdot \ldots \cdot a}_{n\text { factoren}} $}$

$ 2^3=2\times 2\times 2 $

Men noemt $a$ het grondtal, $n$ de exponent en $a^n$ een macht.

Negatieve exponenten geven per definitie het omgekeerde van de macht met de positieve exponent:

$\relax \fcolorbox {black}{white}{$a^{-n}\perdef \frac {1}{a^n}$}$

$2^{-3}=\frac {1}{2^3} = \frac {1}{2\times 2\times 2}$

Als de exponent 0 is (en het grondtal is niet 0), geldt per definitie dat $\relax \fcolorbox {black}{white}{$a^0 = 1$}$, als $a\neq 0$.
De uitdrukkingen $0^0$ en $0^{-n}$ zijn onbepaald en hebben geen betekenis.

Welke formules zijn waar?

Juist Fout $\left (-2\right )^2 = -4$

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

Juist Fout $-2^2 = -4$

$-2^2 = - 2\cdot 2 = - 4 $

Juist Fout $2^{-2} = -4$

$2^{-2} = \frac {1}{2^2} = \frac {1}{4} = 0,25$

Juist Fout $\left (\frac {1}{2}\right )^2 = \frac {1}{4}$

Een kwadraat kan dus kleiner zijn dan het grondtal.

Juist Fout $\left (\frac {1}{2}\right )^{-2} = 4$

$ \left (\frac {1}{2}\right )^{-2} = \left (\frac {2}{1}\right )^{2} = 4$

Juist Fout $\left (\frac {-1}{2}\right )^{-2} = -4$

$\left (\frac {-1}{2}\right )^{-2} = (-2)^2 = 4$

(Machtswortels)

Een reëel getal $b$ is een vierkantswortel van een reëel getal $a$ als

$\relax \fcolorbox {black}{white}{$a = b^2$}$

$2 \text { en } -2 \text { zijn wortels van } 4 \text { want } 2^2 = 4 \text { en } (-2)^2 = 4$

Voor positieve $a$ noemen we het unieke positieve getal $b$ zodat $b^2=a$ de vierkantswortel, en die noteren we als $\sqrt {a}$ of ook als ’$a$ tot de macht een half’:

$\relax \fcolorbox {black}{white}{$ \sqrt {a} \perdef a^\frac 12 = b \iff b^2 = a \text { en } b\geq 0 $}$

$\sqrt {25} \perdef 25^{\frac {1}{2}} = 5$

$\sqrt {a} = a^\frac {1}{2}$ is dus het unieke positieve getal met kwadraat gelijk aan $a$.

Meer in het algemeen schrijven we ook $n$-de machtswortels als ’tot de macht $\tfrac 1n$’:

$\relax \fcolorbox {black}{white}{$ \sqrt [n]{a} \perdef a^\frac 1n = b \iff b^n = a$}$

$\sqrt [3]{8} \perdef 8^\frac {1}{3} = 2 \Twant 2^3=8$

Welke formules zijn waar?

Juist Fout $\sqrt {4} = 2$

$2 ^2 = 4$

Juist Fout $-\sqrt {4} = -2$

$2 ^2 = 4$

Juist Fout $\sqrt {-4} = -2$

$(-2)^2 = 4 \neq -4. \sqrt {-4}$ bestaat niet (in $\R $).

Juist Fout $\sqrt [3]{-8} = -2$

$(-2)^3 = (2)(-2)(-2) = -8$

Juist Fout $\sqrt [3]{9} = 3$

$3\cdot 3\cdot 3 = 9\cdot 3 = 27 \neq 9$

Juist Fout $\sqrt [3]{27} = 3$

$3\cdot 3\cdot 3 = 9\cdot 3 = 27$

Uit deze definities volgen enkele belangrijke rekenregels:

Rekenregels machten

$\relax \fcolorbox {black}{white}{$(xy)^n = x^n y^n \quad \text {(macht van product)} $}$

$(3y)^2 =3y\cdot 3y =3\cdot 3 \cdot y\cdot y = 3^2 y^2=9y^2 $

$\relax \fcolorbox {black}{white}{$x^{m}x^{n} = x^{m+n} \quad \text {(product met gelijk grondtal)} $}$

$x^{3}x^{2} = (x\cdot x\cdot x)\cdot (x\cdot x) = x^{3+2} = x^{5} $

$\relax \fcolorbox {black}{white}{$\left (x^{m}\right )^{n} =x^{m\cdot n}\quad \text {(macht van macht)} $}$

$\left (x^{3}\right )^{2}=x^3 \cdot x^3 = (x \cdot x\cdot x)\cdot (x\cdot x\cdot x)=x^{3\cdot 2}=x^6$

$\relax \fcolorbox {black}{white}{$\left (\frac {x}{y}\right )^{n} = \frac {x^{n}}{y^{n}} \quad \text {(macht van breuk)}$}$

$\left (\frac {x}{2}\right )^{3} = \left (\frac {x}{2}\right ) \cdot \left (\frac {x}{2}\right ) \cdot \left (\frac {x}{2}\right ) = \frac {x\cdot x \cdot x}{2 \cdot 2 \cdot 2} = \frac {x^{3}}{2^{3}}=\frac {x^3}{8} $

Bij het rekenen met breuken zijn er enkele foutieve operaties die voor leerlingen soms erg aanlokkelijk zijn. Doe jezelf (en je leerkracht...) een plezier en overtuig jezelf dat de volgende uitdrukkingen verkeerd zijn.

In het algemeen geldt niet:

\[ \sqrt {a+b} \xcancel {=} \sqrt {a} + \sqrt {b}\]

\[ x^n + x^m \xcancel {=} x^{n+m} \]