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Course: 8η τάξη > Ενότητα 1

Μάθημα 6: Αρνητικοί εκθέτες
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8η τάξη
Numbers and operations
Αρνητικοί εκθέτες
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Negative exponents review

Google Τάξη
Review the basics of negative exponents and try some practice problems. 

Definition for negative exponents

We define a negative power as the multiplicative inverse of the base raised to the positive opposite of the power:
x, start superscript, minus, n, end superscript, equals, start fraction, 1, divided by, x, start superscript, n, end superscript, end fraction
Want to learn more about this definition? Check out this video.

Παραδείγματα:

  • 3, start superscript, minus, 5, end superscript, equals, start fraction, 1, divided by, 3, start superscript, 5, end superscript, end fraction
  • start fraction, 1, divided by, 2, start superscript, 8, end superscript, end fraction, equals, 2, start superscript, minus, 8, end superscript
  • y, start superscript, minus, 2, end superscript, equals, start fraction, 1, divided by, y, squared, end fraction
  • left parenthesis, start fraction, 8, divided by, 6, end fraction, right parenthesis, start superscript, minus, 3, end superscript, equals, left parenthesis, start fraction, 6, divided by, 8, end fraction, right parenthesis, cubed

Εξάσκηση

Πρόβλημα 1
  • Τρέχων
Select the equivalent expression.
4, start superscript, minus, 3, end superscript, equals, question mark
Επιλέξτε 1 απάντηση:

Want to try more problems like these? Check out this exercise.

Some intuition

So why do we define negative exponents this way? Here are a couple of justifications:

Justification #1: Patterns

n2, start superscript, n, end superscript
32, cubed, equals, 8
22, squared, equals, 4
12, start superscript, 1, end superscript, equals, 2
02, start superscript, 0, end superscript, equals, 1
minus, 12, start superscript, minus, 1, end superscript, equals, start fraction, 1, divided by, 2, end fraction
minus, 22, start superscript, minus, 2, end superscript, equals, start fraction, 1, divided by, 4, end fraction
Notice how 2, start superscript, n, end superscript is divided by 2 each time we reduce n. This pattern continues even when n is zero or negative.

Justification #2: Exponent properties

Recall that start fraction, x, start superscript, n, end superscript, divided by, x, start superscript, m, end superscript, end fraction, equals, x, start superscript, n, minus, m, end superscript. So...
2223=22−3=2−1\begin{aligned} \dfrac{2^2}{2^3}&=2^{2-3} \\\\ &=2^{-1} \end{aligned}
We also know that
2223=2⋅22⋅2⋅2=12\begin{aligned} \dfrac{2^2}{2^3}&=\dfrac{\cancel 2\cdot\cancel 2}{\cancel 2\cdot\cancel 2\cdot 2} \\\\ &=\dfrac12 \end{aligned}
And so we get 2, start superscript, minus, 1, end superscript, equals, start fraction, 1, divided by, 2, end fraction.
Also, recall that x, start superscript, n, end superscript, dot, x, start superscript, m, end superscript, equals, x, start superscript, n, plus, m, end superscript. So...
22⋅2−2=22+(−2)=20=1\begin{aligned} 2^2\cdot 2^{-2}&=2^{2+(-2)} \\\\ &=2^0 \\\\ &=1 \end{aligned}
And indeed, according to the definition...
22⋅2−2=22⋅122=2222=1\begin{aligned} 2^2\cdot 2^{-2}&=2^2\cdot\dfrac{1}{2^2} \\\\ &=\dfrac{2^2}{2^2} \\\\ &=1 \end{aligned}

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