题目设a0,α,βa0,\alpha,\betaa0,α,β为任意两个实数则有aα⋅aβaαβa^\alpha\cdot a^\betaa^{\alpha\beta}aα⋅aβaαβ证明不妨设a1a1a1根据实指数乘幂的定义知aαsup{ar∣r≤α,r为有理数}aβsup{ar∣r≤β,r为有理数} a^\alphasup\{a^r|r\leq \alpha,r为有理数\}\\ a^\betasup\{a^r|r\leq \beta,r为有理数\}\\aαsup{ar∣r≤α,r为有理数}aβsup{ar∣r≤β,r为有理数}根据上确界的定义可知对于任意给定的ε0\varepsilon0ε0存在有理数rα,sβr\alpha,s\betarα,sβ使得aα−εaraβ−εas a^\alpha-\varepsilona^r\\ a^\beta-\varepsilona^saα−εaraβ−εas由于rsαβrs\alpha\betarsαβ所以ar⋅asarsaαβ⇒(aα−ε)(aβ−ε)aαβa^r\cdot a^sa^{rs}a^{\alpha\beta}\Rightarrow(a^\alpha-\varepsilon)(a^\beta-\varepsilon)a^{\alpha\beta}ar⋅asarsaαβ⇒(aα−ε)(aβ−ε)aαβ由ε\varepsilonε的任意性可推出aα⋅aβ≤aαβ(1) a^\alpha\cdot a^\beta\leq a^{\alpha\beta}\tag1aα⋅aβ≤aαβ(1)又根据实指数乘幂的定义知aαβsup{ar∣r≤αβ,r为有理数} a^{\alpha\beta}sup\{a^r|r\leq \alpha\beta,r为有理数\}aαβsup{ar∣r≤αβ,r为有理数}根据上确界的定义可知对于上述任意给定的ε0\varepsilon0ε0存在有理数prsαβprs\alpha\betaprsαβ且有rα,sβr\alpha,s\betarα,sβ使得aαβ−εaparsar⋅as≤aα⋅aβ a^{\alpha\beta}-\varepsilona^pa^{rs}a^r\cdot a^s\leq a^\alpha\cdot a^\betaaαβ−εaparsar⋅as≤aα⋅aβ由ε\varepsilonε的任意性可推出aαβ≤aα⋅aβ(2) a^{\alpha\beta}\leq a^\alpha\cdot a^\beta\tag2aαβ≤aα⋅aβ(2)由1、2可知aα⋅aβaαβ a^\alpha\cdot a^\betaa^{\alpha\beta}aα⋅aβaαβ