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  2. Power of two - Wikipedia

    en.wikipedia.org/wiki/Power_of_two

    [1] [2] The first ten powers of 2 for non-negative values of n are: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, ... (sequence A000079 in the OEIS) By comparison, powers of two with negative exponents are fractions: for positive integer n, 2 −n is one half multiplied by itself n times. Thus the first few negative powers of 2 are ⁠ 1 / 2 ...

  3. Exponentiation - Wikipedia

    en.wikipedia.org/wiki/Exponentiation

    Graphs of y = b x for various bases b: base 10, base e, base 2, base ⁠ 1 / 2 ⁠. Each curve passes through the point (0, 1) because any nonzero number raised to the power of 0 is 1. At x = 1, the value of y equals the base because any number raised to the power of 1 is the number itself.

  4. Finger binary - Wikipedia

    en.wikipedia.org/wiki/Finger_binary

    The values of each raised finger are added together to arrive at a total number. In the one-handed version, all fingers raised is thus 31 (16 + 8 + 4 + 2 + 1), and all fingers lowered (a fist) is 0. In the two-handed system, all fingers raised is 1,023 (512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1) and two fists (no fingers raised) represents 0.

  5. Modular exponentiation - Wikipedia

    en.wikipedia.org/wiki/Modular_exponentiation

    The exponent is 1101 in binary. There are four binary digits, so the loop executes four times, with values a 0 = 1, a 1 = 0, a 2 = 1, and a 3 = 1. First, initialize the result to 1 and preserve the value of b in the variable x: (=).

  6. Scientific notation - Wikipedia

    en.wikipedia.org/wiki/Scientific_notation

    The number 1 230 400 is usually read to have five significant figures: 1, 2, 3, 0, and 4, the final two zeroes serving only as placeholders and adding no precision. The same number, however, would be used if the last two digits were also measured precisely and found to equal 0 – seven significant figures.

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  8. Binary logarithm - Wikipedia

    en.wikipedia.org/wiki/Binary_logarithm

    Start with a real number y in the half-open interval [1, 2). If y = 1, then the algorithm is done, and the fractional part is zero. Otherwise, square y repeatedly until the result z lies in the interval [2, 4). Let m be the number of squarings needed. That is, z = y 2 m with m chosen such that z is in [2, 4).

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