# 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
6 years, 5 months ago

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## Comments

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How will you prove that 1^n=1 without proving a^0=1

- 6 years, 5 months ago

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${ a }^{ 0 }\quad =\quad { a }^{ b-b }\\ { a }^{ b-b }\quad =\quad \frac { { a }^{ b } }{ { a }^{ b } } \quad =\quad \boxed { 1 }$

- 6 years, 5 months ago

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Your explanation for a^0 is correct,but you have to prove this in your note.

- 6 years, 5 months ago

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Rajdeep, this proof is created by me, so please mention it.

- 6 years, 5 months ago

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Of course, we're making the assumption that $0^{0}$ is 1, correct?

- 6 years, 5 months ago

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No we are not... we can consider a is not equal to 0... instead of "a" we could have used any constant

- 6 years, 5 months ago

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Alright, just making sure...

- 6 years, 5 months ago

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@Archit Boobna Whats ur score in NTSE Delhi held on 18 Dec. Btw what is score highest in FSD?

- 4 years, 5 months ago

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No we are not Please tell in which step am i using it

- 6 years, 5 months ago

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