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# wierd proofs

$To\quad Prove\quad :\quad Any\quad power\quad of\quad 1\quad is\quad 1\\ we\quad all\quad are\quad aware\quad of\quad the\quad fact\quad that\\ { ({ a }^{ m }) }^{ n }\quad =\quad { ({ a }^{ n }) }^{ m }\\ Let\quad us\quad put\quad m\quad =\quad 0,\quad then\quad \\ { ({ a }^{ 0 }) }^{ n }\quad =\quad ({ something) }^{ 0 }\\ { 1 }^{ n }\quad =\quad 1\\ Hence\quad proved.$ Proof of the assumption I have used in the proof. ${ a }^{ 0 }\quad =\quad { a }^{ b-b }\\ { a }^{ b-b }\quad =\quad \frac { { a }^{ b } }{ { a }^{ b } } \quad =\quad \boxed { 1 }$ Credit to @ArchitBoobna

Note by Rajdeep Dhingra
2 years, 4 months ago

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How will you prove that 1^n=1 without proving a^0=1 · 2 years, 4 months ago

${ a }^{ 0 }\quad =\quad { a }^{ b-b }\\ { a }^{ b-b }\quad =\quad \frac { { a }^{ b } }{ { a }^{ b } } \quad =\quad \boxed { 1 }$ · 2 years, 4 months ago

Your explanation for a^0 is correct,but you have to prove this in your note. · 2 years, 4 months ago

Rajdeep, this proof is created by me, so please mention it. · 2 years, 4 months ago

Of course, we're making the assumption that $$0^{0}$$ is 1, correct? · 2 years, 4 months ago

No we are not Please tell in which step am i using it · 2 years, 4 months ago

No we are not... we can consider a is not equal to 0... instead of "a" we could have used any constant · 2 years, 4 months ago

Alright, just making sure... · 2 years, 4 months ago