$ (\log_2 (3))^2 \approx (1.58496)^2 \approx 2.51211$ A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately $2 \log_2 (3) \approx 2 \cdot 1.58496 \approx 3.16992$
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$2^ {\log_2 (3)} = 3$
Do any of those appear to be equal
(whenever you are wondering whether some general algebraic relationship holds, it's a good idea to first try some simple numerical examples to see if it is even possible. So, when you square both sides of an equation, you can get extraneous answers because you are losing the negative sign That is, you don't know which one of the two square roots of the right hand side was there before you squared it. We can square both side like this
$ x^2= 2$ but i don't understand why that it's okay to square both sides What i learned is that adding, subtracting, multiplying, or dividing both sides by the same thing is okay But how come squaring both. We can't simply square both sides because that's exactly what we're trying to prove
$$0 < a < b \implies a^2 < b^2$$ more somewhat related details
I think it may be a common misconception that simply squaring both sides of an inequality is ok because we can do it indiscriminately with equalities. Using the square root extraction algorithm, the question can be answered Since there are many digits involved, the decimal position allows us to limit ourselves to only a suitable final set of these. (i.e the product of four consecutive numbers, plus one, is always a perfect square.) i understand the algebraic proof, turning it into a difference of squares, but i’m looking for a visual proof.
Say we try to find the closest square number to 26 We already know the closest square number is $25$ However, how do i calculate out 25 Because, if i try to prime factorize it like so
The theorem that $\binom {n} {k} = \frac {n!} {k
Otherwise this would be restricted to $0 <k < n$
