-0.000 000 000 864 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 864(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 864(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 864| = 0.000 000 000 864


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when 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.

Take 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: 0.000 000 000 864.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 864 × 2 = 0 + 0.000 000 001 728;
  • 2) 0.000 000 001 728 × 2 = 0 + 0.000 000 003 456;
  • 3) 0.000 000 003 456 × 2 = 0 + 0.000 000 006 912;
  • 4) 0.000 000 006 912 × 2 = 0 + 0.000 000 013 824;
  • 5) 0.000 000 013 824 × 2 = 0 + 0.000 000 027 648;
  • 6) 0.000 000 027 648 × 2 = 0 + 0.000 000 055 296;
  • 7) 0.000 000 055 296 × 2 = 0 + 0.000 000 110 592;
  • 8) 0.000 000 110 592 × 2 = 0 + 0.000 000 221 184;
  • 9) 0.000 000 221 184 × 2 = 0 + 0.000 000 442 368;
  • 10) 0.000 000 442 368 × 2 = 0 + 0.000 000 884 736;
  • 11) 0.000 000 884 736 × 2 = 0 + 0.000 001 769 472;
  • 12) 0.000 001 769 472 × 2 = 0 + 0.000 003 538 944;
  • 13) 0.000 003 538 944 × 2 = 0 + 0.000 007 077 888;
  • 14) 0.000 007 077 888 × 2 = 0 + 0.000 014 155 776;
  • 15) 0.000 014 155 776 × 2 = 0 + 0.000 028 311 552;
  • 16) 0.000 028 311 552 × 2 = 0 + 0.000 056 623 104;
  • 17) 0.000 056 623 104 × 2 = 0 + 0.000 113 246 208;
  • 18) 0.000 113 246 208 × 2 = 0 + 0.000 226 492 416;
  • 19) 0.000 226 492 416 × 2 = 0 + 0.000 452 984 832;
  • 20) 0.000 452 984 832 × 2 = 0 + 0.000 905 969 664;
  • 21) 0.000 905 969 664 × 2 = 0 + 0.001 811 939 328;
  • 22) 0.001 811 939 328 × 2 = 0 + 0.003 623 878 656;
  • 23) 0.003 623 878 656 × 2 = 0 + 0.007 247 757 312;
  • 24) 0.007 247 757 312 × 2 = 0 + 0.014 495 514 624;
  • 25) 0.014 495 514 624 × 2 = 0 + 0.028 991 029 248;
  • 26) 0.028 991 029 248 × 2 = 0 + 0.057 982 058 496;
  • 27) 0.057 982 058 496 × 2 = 0 + 0.115 964 116 992;
  • 28) 0.115 964 116 992 × 2 = 0 + 0.231 928 233 984;
  • 29) 0.231 928 233 984 × 2 = 0 + 0.463 856 467 968;
  • 30) 0.463 856 467 968 × 2 = 0 + 0.927 712 935 936;
  • 31) 0.927 712 935 936 × 2 = 1 + 0.855 425 871 872;
  • 32) 0.855 425 871 872 × 2 = 1 + 0.710 851 743 744;
  • 33) 0.710 851 743 744 × 2 = 1 + 0.421 703 487 488;
  • 34) 0.421 703 487 488 × 2 = 0 + 0.843 406 974 976;
  • 35) 0.843 406 974 976 × 2 = 1 + 0.686 813 949 952;
  • 36) 0.686 813 949 952 × 2 = 1 + 0.373 627 899 904;
  • 37) 0.373 627 899 904 × 2 = 0 + 0.747 255 799 808;
  • 38) 0.747 255 799 808 × 2 = 1 + 0.494 511 599 616;
  • 39) 0.494 511 599 616 × 2 = 0 + 0.989 023 199 232;
  • 40) 0.989 023 199 232 × 2 = 1 + 0.978 046 398 464;
  • 41) 0.978 046 398 464 × 2 = 1 + 0.956 092 796 928;
  • 42) 0.956 092 796 928 × 2 = 1 + 0.912 185 593 856;
  • 43) 0.912 185 593 856 × 2 = 1 + 0.824 371 187 712;
  • 44) 0.824 371 187 712 × 2 = 1 + 0.648 742 375 424;
  • 45) 0.648 742 375 424 × 2 = 1 + 0.297 484 750 848;
  • 46) 0.297 484 750 848 × 2 = 0 + 0.594 969 501 696;
  • 47) 0.594 969 501 696 × 2 = 1 + 0.189 939 003 392;
  • 48) 0.189 939 003 392 × 2 = 0 + 0.379 878 006 784;
  • 49) 0.379 878 006 784 × 2 = 0 + 0.759 756 013 568;
  • 50) 0.759 756 013 568 × 2 = 1 + 0.519 512 027 136;
  • 51) 0.519 512 027 136 × 2 = 1 + 0.039 024 054 272;
  • 52) 0.039 024 054 272 × 2 = 0 + 0.078 048 108 544;
  • 53) 0.078 048 108 544 × 2 = 0 + 0.156 096 217 088;
  • 54) 0.156 096 217 088 × 2 = 0 + 0.312 192 434 176;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 864(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0101 1111 1010 0110 00(2)

6. Positive number before normalization:

0.000 000 000 864(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0101 1111 1010 0110 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 864(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0101 1111 1010 0110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0101 1111 1010 0110 00(2) × 20 =

1.1101 1010 1111 1101 0011 000(2) × 2-31


8. 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): -31

Mantissa (not normalized):
1.1101 1010 1111 1101 0011 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1101 0111 1110 1001 1000 =

110 1101 0111 1110 1001 1000


13. 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 0000

Mantissa (23 bits) =
110 1101 0111 1110 1001 1000


Decimal number -0.000 000 000 864 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 1101 0111 1110 1001 1000

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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