Question

Given a single precision floating point number 8.0, what is the smallest precision floating point number that is bigger than 8.0?

Answer 1

Answer:
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8.00000095367431640625

Explanation:
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First, let's 8.0 to single-precision format binary
Converting 8.0 to binary
   Convert decimal part first, then the fractional part
   > First convert 8 to binary
   Divide 8 successively by 2 until the quotient is 0
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 1000
   so, 8.0 in binary is 1000.0
8.0 in simple binary => 1000.0
so, 8.0 in normal binary is 1000.0 => 1. * 2^3

single precision:
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sign bit is 0(+ve)
exp bits are (127+3=130) => 10000010
   Divide 130 successively by 2 until the quotient is 0
      > 130/2 = 65, remainder is 0
      > 65/2 = 32, remainder is 1
      > 32/2 = 16, remainder is 0
      > 16/2 = 8, remainder is 0
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 10000010
   So, 130 of decimal is 10000010 in binary
frac bits are 00000000000000000000000

so, 8.0 in single-precision format is 0 10000010 00000000000000000000000

now smallest number bigger than 8.0 in single-precision binary format is 0 10000010 00000000000000000000001

let's convert this back to decimal
0 10000010 00000000000000000000001
sign bit is 0(+ve)
exp bits are 10000010
   => 10000010
   => 1x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+1x2^1+0x2^0
   => 1x128+0x64+0x32+0x16+0x8+0x4+1x2+0x1
   => 128+0+0+0+0+0+2+0
   => 130
in decimal it is 130
so, exponent/bias is 130-127 = 3
frac bits are 00000000000000000000001

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.00000000000000000000001 * 2^3
1.00000000000000000000001 in decimal is 1.0000001192092896
   => 1.00000000000000000000001
   => 1x2^0+0x2^-1+0x2^-2+0x2^-3+0x2^-4+0x2^-5+0x2^-6+0x2^-7+0x2^-8+0x2^-9+0x2^-10+0x2^-11+0x2^-12+0x2^-13+0x2^-14+0x2^-15+0x2^-16+0x2^-17+0x2^-18+0x2^-19+0x2^-20+0x2^-21+0x2^-22+1x2^-23
   => 1x1+0x0.5+0x0.25+0x0.125+0x0.0625+0x0.03125+0x0.015625+0x0.0078125+0x0.00390625+0x0.001953125+0x0.0009765625+0x0.00048828125+0x0.000244140625+0x0.0001220703125+0x6.103515625e-05+0x3.0517578125e-05+0x1.52587890625e-05+0x7.62939453125e-06+0x3.814697265625e-06+0x1.9073486328125e-06+0x9.5367431640625e-07+0x4.76837158203125e-07+0x2.384185791015625e-07+1x0.00000011920928955078125
   => 1+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.00000011920928955078125
   => 1.00000011920928955078125
so, 1.00000011920928955078125 * 2^3 in decimal is 8.000000953674316
so, 01000001000000000000000000000001 in IEEE-754 single precision format is 8.00000095367431640625
Answer: 8.00000095367431640625