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Why is $1$ not considered a prime number? Or, why is the definition of prime numbers given for integers greater than $1$?

Substitute a by 1 and b, c by x, we find both definition of 1 x and x − 1 reduce to the unique y in Y (if exists) such that yx = 1 = xy. As a result, 1 x = x − 1 whenever they make sense.

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How can I prove from first principles that $0!$ is equal to $1$?

Is there a formal proof for $(-1) \\times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?

Hi Andre, sorry for commenting on an old post. As we know, the series 1 − t +t2 −t3 + ⋯ 1 − t + t 2 − t 3 + ⋯ $1-t+{t}^{2}-{t}^{3}+\cdots$ doesn't converge uniformly in |t| <1 | t | <1 $|t|<1$. Then why we can …

what I'm not able to understand is how does the inverse of kA leads to k coming in the denominator

$1$ square is $1$, so is raised $1$ to $123434234$. My maths teacher claims that $1$ raised to infinity is not $1$, but not defined. Is there any reason for this? I know that any number raised to

May 11, 2015 · Why is 1/i 1 / i $1/i$ equal to −i i $-i$? Ask Question Asked 11 years, 2 months ago Modified 1 year, 6 months ago Viewed 117k times 67