24 мар. 2020 г. · Is this proof right? Let sqrt(2)^sqrt(2) be a rational number q/p, where p, q are all integers, p, q are nonzero, and the gcd of p, q is 1.

14 нояб. 2014 г. · Think of it as a limit. Take a sequence of the form an/bn that approaches sqrt(2). Then lim (n->inf) 2an/bn = 2 ^ (sqrt(2)).

11 апр. 2021 г. · I keep getting stuck when using 45 degrees because, to my knowledge, the sides of the triangle are 1,1, and the square root of 2 for the ...

14 июл. 2021 г. · sqrt(2)/2 is preferred to sqrt(1/2) even though they are equal. Perhaps this is because when calculating on pen and paper, it's easier (although not impossible ...

22 июл. 2024 г. · It's not true. If you start with 2 = sqrt(2)*sqrt(2) and multiply both sides by sqrt(2) (which is what I suspect you are doing), ...

14 мар. 2014 г. · The sequence x_{n+1} = sqrt(2 + x_n), x_0 = sqrt(2) is monotonically increasing, and clearly x_0 > 0. So the limit x must be greater than zero ...

27 авг. 2024 г. · Simplifying the innermost square root would simplify the problem greatly. sqrt[6 + 2(sqrt3 + sqrt2 + sqrt6)] = sqrt[6 + 2sqrt6 + 2(sqrt3 + sqrt2)] {expression ...

3 мая 2024 г. · Consider i * sqrt 2. (i * sqrt 2)2 = i2 * (sqrt 2)2 = (-1) * 2 = -2. So we should have i*sqrt 2 = sqrt (-2), and sqrt(-2)2 = -2.

2 окт. 2018 г. · Note that in the very first step we are using the fact that 1/√2 = 1/√2 × 1 (anything × 1 = itself) and that √2/√2 = 1 (anything ÷ itself = 1).

18 окт. 2017 г. · The square root of 1/2 (square root of 2 divided by 2) has some sort of significance in the engineering field, and even has a special name.