Question

Proof that
• if a is any real number and m, n are positive integers, then a^m × a^n = a^m+n

• if a is any real number and m , n are positive integers, then (a^m)^n = a^mn = (a^n)^m

Answer 1

Step-by-step explanation:

Proof that

• if a is any real number and m, n are positive integers, then a^m × a^n = a^m+n

• if a is any real number and m , n are positive integers, then (a^m)^n = a^mn = (a^n)^m

Answer 2

Proof that -

• If a is any real number and m, n are positive integers, then a^m × a^n = a^m+n

Answer -

By definition , we have

$\;\;\tt{\;\;{ {a}^{m} } \times { {a}^{n} }}$

>> (a×a×a...to m factors) × (a×a×a... to n factors)

$\;\;\tt{\mapsto\;\;{ {a}^{m} } \times { {a}^{n} }}$

>> a×a×a...to (m+n) factors

$\;\;\tt{ \mapsto\;\;{ {a}^{m} } \times { {a}^{n} } = a ^{m + n} }$

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Proof that -

• if a is any real number and m , n are positive integers, then (a^m)^n = a^mn = (a^n)^m

Answer -

We have ,

$\;\;\tt{\;\;{ ({a}^{m} ) ^{n} }}$

>> $\;\;\tt{\;\;{ {a}^{m} \times {a}^{m} \times {a}^{m} \times ..}}$ to n factors

$\;\;\tt{\mapsto\;\;{ ({a}^{m})^{n} }}$

>> (a×a×a...to m factors) × (a×a×a... to m factors) × (a×a×a...to m factors) ... to n factors

$\;\;\tt{\mapsto\;\;{ ({a}^{m})^{n} }}$

>> a×a×a...to (mn) factors = $\;\;\tt{\;\;{ {a}^{nm} }}$

similarly, we have $\;\;\tt{\;\;{ ({a}^{n})^{m} }={a}^{nm}}$

Hence, $\;\;\tt{\;\;{ ({a}^{m})^{n} }={a}^{mn} = ( {a}^{n}) ^{m} }$