Question
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Ett av alternativen är inte sant. Vilket? \( \begin{array}{l}0^{a}=0 \\ a^{m} \cdot a^{-m}=0 \\ \left(a^{m}\right)^{n}=a^{m n} \\ a^{0}=1\end{array} \)

Ask by Hanson Flynn. in Sweden
Jan 23,2025

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Answer

Det andra påståendet \( a^{m} \cdot a^{-m}=0 \) är inte sant.

Solution

Låt oss analysera varje påstående ett i taget för att identifiera vilket som inte är sant. 1. **\( 0^{a}=0 \)**: Detta påstående är sant för alla positiva värden av \( a \). Men om \( a = 0 \) så är \( 0^{0} \) odefinierat i många sammanhang, och om \( a < 0 \) så är \( 0^{a} \) odefinierat. Så detta påstående är inte alltid sant. 2. **\( a^{m} \cdot a^{-m}=0 \)**: Detta påstående är falskt. Enligt exponentreglerna är \( a^{m} \cdot a^{-m} = a^{m - m} = a^{0} = 1 \) (för \( a \neq 0 \)). Så detta påstående är inte sant. 3. **\( \left(a^{m}\right)^{n}=a^{m n} \)**: Detta påstående är sant enligt exponentreglerna. 4. **\( a^{0}=1 \)**: Detta påstående är sant för alla \( a \neq 0 \). För \( a = 0 \) är det odefinierat, men i allmänhet gäller detta för alla andra värden. Sammanfattningsvis är det andra påståendet \( a^{m} \cdot a^{-m}=0 \) det som inte är sant.

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Det falska alternativet i listan är \( a^{m} \cdot a^{-m}=0 \). Detta påstående är felaktigt eftersom produkten av ett tal och dess invers faktiskt ger 1 (om \( a \neq 0 \)), alltså \( a^{m} \cdot a^{-m} = a^{m-m} = a^{0} = 1 \). För att förtydliga, så är det värt att notera att uttrycket \( 0^a = 0 \) gäller när \( a \) är ett positivt tal, men ger mer komplicerade resultat för andra värden av \( a \). Det är viktigt att vara medveten om dessa fall när man arbetar med exponenter och noll.

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