Base ten decimal number -0.000 011 011 1 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number -0.000 011 011 1(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. We start with the positive version of the number:

|-0.000 011 011 1| = 0.000 011 011 1

2. 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;

3. 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)

4. Convert to binary (base 2) the fractional part: 0.000 011 011 1. 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) 0.000 011 011 1 × 2 = 0 + 0.000 022 022 2;
  • 2) 0.000 022 022 2 × 2 = 0 + 0.000 044 044 4;
  • 3) 0.000 044 044 4 × 2 = 0 + 0.000 088 088 8;
  • 4) 0.000 088 088 8 × 2 = 0 + 0.000 176 177 6;
  • 5) 0.000 176 177 6 × 2 = 0 + 0.000 352 355 2;
  • 6) 0.000 352 355 2 × 2 = 0 + 0.000 704 710 4;
  • 7) 0.000 704 710 4 × 2 = 0 + 0.001 409 420 8;
  • 8) 0.001 409 420 8 × 2 = 0 + 0.002 818 841 6;
  • 9) 0.002 818 841 6 × 2 = 0 + 0.005 637 683 2;
  • 10) 0.005 637 683 2 × 2 = 0 + 0.011 275 366 4;
  • 11) 0.011 275 366 4 × 2 = 0 + 0.022 550 732 8;
  • 12) 0.022 550 732 8 × 2 = 0 + 0.045 101 465 6;
  • 13) 0.045 101 465 6 × 2 = 0 + 0.090 202 931 2;
  • 14) 0.090 202 931 2 × 2 = 0 + 0.180 405 862 4;
  • 15) 0.180 405 862 4 × 2 = 0 + 0.360 811 724 8;
  • 16) 0.360 811 724 8 × 2 = 0 + 0.721 623 449 6;
  • 17) 0.721 623 449 6 × 2 = 1 + 0.443 246 899 2;
  • 18) 0.443 246 899 2 × 2 = 0 + 0.886 493 798 4;
  • 19) 0.886 493 798 4 × 2 = 1 + 0.772 987 596 8;
  • 20) 0.772 987 596 8 × 2 = 1 + 0.545 975 193 6;
  • 21) 0.545 975 193 6 × 2 = 1 + 0.091 950 387 2;
  • 22) 0.091 950 387 2 × 2 = 0 + 0.183 900 774 4;
  • 23) 0.183 900 774 4 × 2 = 0 + 0.367 801 548 8;
  • 24) 0.367 801 548 8 × 2 = 0 + 0.735 603 097 6;
  • 25) 0.735 603 097 6 × 2 = 1 + 0.471 206 195 2;
  • 26) 0.471 206 195 2 × 2 = 0 + 0.942 412 390 4;
  • 27) 0.942 412 390 4 × 2 = 1 + 0.884 824 780 8;
  • 28) 0.884 824 780 8 × 2 = 1 + 0.769 649 561 6;
  • 29) 0.769 649 561 6 × 2 = 1 + 0.539 299 123 2;
  • 30) 0.539 299 123 2 × 2 = 1 + 0.078 598 246 4;
  • 31) 0.078 598 246 4 × 2 = 0 + 0.157 196 492 8;
  • 32) 0.157 196 492 8 × 2 = 0 + 0.314 392 985 6;
  • 33) 0.314 392 985 6 × 2 = 0 + 0.628 785 971 2;
  • 34) 0.628 785 971 2 × 2 = 1 + 0.257 571 942 4;
  • 35) 0.257 571 942 4 × 2 = 0 + 0.515 143 884 8;
  • 36) 0.515 143 884 8 × 2 = 1 + 0.030 287 769 6;
  • 37) 0.030 287 769 6 × 2 = 0 + 0.060 575 539 2;
  • 38) 0.060 575 539 2 × 2 = 0 + 0.121 151 078 4;
  • 39) 0.121 151 078 4 × 2 = 0 + 0.242 302 156 8;
  • 40) 0.242 302 156 8 × 2 = 0 + 0.484 604 313 6;

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...)

5. 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:

0.000 011 011 1(10) =


0.0000 0000 0000 0000 1011 1000 1011 1100 0101 0000(2)

Positive number before normalization:

0.000 011 011 1(10) =


0.0000 0000 0000 0000 1011 1000 1011 1100 0101 0000(2)

6. Normalize the binary representation of the number, shifting the decimal mark 17 positions to the right so that only one non zero digit remains to the left of it:

0.000 011 011 1(10) =


0.0000 0000 0000 0000 1011 1000 1011 1100 0101 0000(2) =


0.0000 0000 0000 0000 1011 1000 1011 1100 0101 0000(2) × 20 =


1.0111 0001 0111 1000 1010 000(2) × 2-17

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: 1 (a negative number)


Exponent (unadjusted): -17


Mantissa (not normalized): 1.0111 0001 0111 1000 1010 000

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:

Exponent (adjusted) =


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


-17 + 2(8-1) - 1 =


(-17 + 127)(10) =


110(10)


  • division = quotient + remainder;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


110(10) =


0110 1110(2)

8. 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. 011 1000 1011 1100 0101 0000 =


011 1000 1011 1100 0101 0000

Conclusion:

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

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


Exponent (8 bits) =
0110 1110


Mantissa (23 bits) =
011 1000 1011 1100 0101 0000

Number -0.000 011 011 1, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


1 - 0110 1110 - 011 1000 1011 1100 0101 0000

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 0

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 0

      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