Base ten decimal number 0.142 857 142 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number 0.142 857 142(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 0. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

0(10) =


0(2)

3. Convert to binary (base 2) the fractional part: 0.142 857 142. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

  • #) multiplying = integer + fractional part;
  • 1) 0.142 857 142 × 2 = 0 + 0.285 714 284;
  • 2) 0.285 714 284 × 2 = 0 + 0.571 428 568;
  • 3) 0.571 428 568 × 2 = 1 + 0.142 857 136;
  • 4) 0.142 857 136 × 2 = 0 + 0.285 714 272;
  • 5) 0.285 714 272 × 2 = 0 + 0.571 428 544;
  • 6) 0.571 428 544 × 2 = 1 + 0.142 857 088;
  • 7) 0.142 857 088 × 2 = 0 + 0.285 714 176;
  • 8) 0.285 714 176 × 2 = 0 + 0.571 428 352;
  • 9) 0.571 428 352 × 2 = 1 + 0.142 856 704;
  • 10) 0.142 856 704 × 2 = 0 + 0.285 713 408;
  • 11) 0.285 713 408 × 2 = 0 + 0.571 426 816;
  • 12) 0.571 426 816 × 2 = 1 + 0.142 853 632;
  • 13) 0.142 853 632 × 2 = 0 + 0.285 707 264;
  • 14) 0.285 707 264 × 2 = 0 + 0.571 414 528;
  • 15) 0.571 414 528 × 2 = 1 + 0.142 829 056;
  • 16) 0.142 829 056 × 2 = 0 + 0.285 658 112;
  • 17) 0.285 658 112 × 2 = 0 + 0.571 316 224;
  • 18) 0.571 316 224 × 2 = 1 + 0.142 632 448;
  • 19) 0.142 632 448 × 2 = 0 + 0.285 264 896;
  • 20) 0.285 264 896 × 2 = 0 + 0.570 529 792;
  • 21) 0.570 529 792 × 2 = 1 + 0.141 059 584;
  • 22) 0.141 059 584 × 2 = 0 + 0.282 119 168;
  • 23) 0.282 119 168 × 2 = 0 + 0.564 238 336;
  • 24) 0.564 238 336 × 2 = 1 + 0.128 476 672;
  • 25) 0.128 476 672 × 2 = 0 + 0.256 953 344;
  • 26) 0.256 953 344 × 2 = 0 + 0.513 906 688;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)

4. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.142 857 142(10) =


0.0010 0100 1001 0010 0100 1001 00(2)

Positive number before normalization:

0.142 857 142(10) =


0.0010 0100 1001 0010 0100 1001 00(2)

5. Normalize the binary representation of the number, shifting the decimal mark 3 positions to the right so that only one non zero digit remains to the left of it:

0.142 857 142(10) =


0.0010 0100 1001 0010 0100 1001 00(2) =


0.0010 0100 1001 0010 0100 1001 00(2) × 20 =


1.0010 0100 1001 0010 0100 100(2) × 2-3

Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): -3


Mantissa (not normalized): 1.0010 0100 1001 0010 0100 100

6. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-3 + 2(8-1) - 1 =


(-3 + 127)(10) =


124(10)


  • division = quotient + remainder;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


124(10) =


0111 1100(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, only if necessary (not the case here):

Mantissa (normalized) =


1. 001 0010 0100 1001 0010 0100 =


001 0010 0100 1001 0010 0100

Conclusion:

The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
0111 1100


Mantissa (23 bits) =
001 0010 0100 1001 0010 0100

Number 0.142 857 142, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


0 - 0111 1100 - 001 0010 0100 1001 0010 0100

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 0

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

Convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits), mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111