Equality of Algebraic Expressions Involving Exponents: When Does ( a cdot b^n a^n cdot b^n ) Hold?
To determine whether the expression ( a cdot b^n ) can equal ( a^n cdot b^n ) for all ( n ), we must consider the specific conditions under which this equality holds.
Conditions for Equality
The expression ( a cdot b^n ) is equal to ( a^n cdot b^n ) only under specific conditions. Let's analyze this in detail.
Binomial Theorem
According to the Binomial Theorem,
[ a cdot b^n sum_{k0}^n binom{n}{k} a^{n-k} b^k ]This expansion includes terms where both ( a ) and ( b ) are raised to various powers, multiplied by binomial coefficients.
Equality Condition
For ( a cdot b^n ) to equal ( a^n cdot b^n ), all intermediate terms in the expansion must be zero. This happens when:
Either ( a 0 ) or ( b 0 ) Or when ( n 1 )Specific Cases
Let's explore some specific cases to understand this better:
If ( n 1 ), then ( a cdot b^1 a cdot b ) which equals ( a^1 cdot b^1 ). If ( a 0 ), then ( 0 cdot b^n b^n ) equals ( 0^n cdot b^n ). If ( b 0 ), then ( a cdot 0^n a^n ) equals ( a^n cdot 0^n ).Counterexample
For ( n 2 ),
[ a cdot b^2 a^2 2ab b^2 ]This is not equal to ( a^2 cdot b^2 ) unless ( ab 0 ), i.e., one of ( a ) or ( b ) is zero.
Algebraic Settings
Depending on the algebraic setting, the equality ( a cdot b^n a^n cdot b^n ) may or may not hold:
Over ( mathbb{Z} ) or ( mathbb{R} )
Over the integers ( mathbb{Z} ) or the real numbers ( mathbb{R} ), the statement is generally false. Plenty of counterexamples can be found by substituting values of ( a ) and ( b ).
In fact, by the Binomial Theorem:
[ a cdot b^n sum_{k0}^n binom{n}{k} a^{n-k} b^k ]And for all ( n in mathbb{Z} ),
[ binom{n}{k} frac{n!}{k!(n-k)!} in mathbb{Z} ]Over ( mathbb{Z}_n ) or ( mathbb{R}_n )
When working in the modular arithmetic context ( mathbb{Z}_n ) or ( mathbb{R}_n ), the situation changes:
Applying modular division by ( n ):
[ a cdot b^n sum_{k0}^n binom{n}{k} a^{n-k} b^k ][ a cdot b^n a^n displaystylesum_{k1}^{n-1} binom{n}{k} a^{n-k} b^k b^n ]
[ implies a cdot b^n equiv a^n displaystylesum_{k1}^{n-1} binom{n}{k} a^{n-k} b^k b^n mod n ]
[ implies a cdot b^n equiv a^n 0 cdot b^n mod n ]
[ therefore a cdot b^n equiv a^n cdot b^n mod n ]
This means the equality ( a cdot b^n a^n cdot b^n ) holds for all ( a, b in mathbb{Z}_n ).
This completes the analysis. (blacksquare)