0.000 000 110 009 23 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 009 23(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
0.000 000 110 009 23(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 009 23.

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) 0.000 000 110 009 23 × 2 = 0 + 0.000 000 220 018 46;
  • 2) 0.000 000 220 018 46 × 2 = 0 + 0.000 000 440 036 92;
  • 3) 0.000 000 440 036 92 × 2 = 0 + 0.000 000 880 073 84;
  • 4) 0.000 000 880 073 84 × 2 = 0 + 0.000 001 760 147 68;
  • 5) 0.000 001 760 147 68 × 2 = 0 + 0.000 003 520 295 36;
  • 6) 0.000 003 520 295 36 × 2 = 0 + 0.000 007 040 590 72;
  • 7) 0.000 007 040 590 72 × 2 = 0 + 0.000 014 081 181 44;
  • 8) 0.000 014 081 181 44 × 2 = 0 + 0.000 028 162 362 88;
  • 9) 0.000 028 162 362 88 × 2 = 0 + 0.000 056 324 725 76;
  • 10) 0.000 056 324 725 76 × 2 = 0 + 0.000 112 649 451 52;
  • 11) 0.000 112 649 451 52 × 2 = 0 + 0.000 225 298 903 04;
  • 12) 0.000 225 298 903 04 × 2 = 0 + 0.000 450 597 806 08;
  • 13) 0.000 450 597 806 08 × 2 = 0 + 0.000 901 195 612 16;
  • 14) 0.000 901 195 612 16 × 2 = 0 + 0.001 802 391 224 32;
  • 15) 0.001 802 391 224 32 × 2 = 0 + 0.003 604 782 448 64;
  • 16) 0.003 604 782 448 64 × 2 = 0 + 0.007 209 564 897 28;
  • 17) 0.007 209 564 897 28 × 2 = 0 + 0.014 419 129 794 56;
  • 18) 0.014 419 129 794 56 × 2 = 0 + 0.028 838 259 589 12;
  • 19) 0.028 838 259 589 12 × 2 = 0 + 0.057 676 519 178 24;
  • 20) 0.057 676 519 178 24 × 2 = 0 + 0.115 353 038 356 48;
  • 21) 0.115 353 038 356 48 × 2 = 0 + 0.230 706 076 712 96;
  • 22) 0.230 706 076 712 96 × 2 = 0 + 0.461 412 153 425 92;
  • 23) 0.461 412 153 425 92 × 2 = 0 + 0.922 824 306 851 84;
  • 24) 0.922 824 306 851 84 × 2 = 1 + 0.845 648 613 703 68;
  • 25) 0.845 648 613 703 68 × 2 = 1 + 0.691 297 227 407 36;
  • 26) 0.691 297 227 407 36 × 2 = 1 + 0.382 594 454 814 72;
  • 27) 0.382 594 454 814 72 × 2 = 0 + 0.765 188 909 629 44;
  • 28) 0.765 188 909 629 44 × 2 = 1 + 0.530 377 819 258 88;
  • 29) 0.530 377 819 258 88 × 2 = 1 + 0.060 755 638 517 76;
  • 30) 0.060 755 638 517 76 × 2 = 0 + 0.121 511 277 035 52;
  • 31) 0.121 511 277 035 52 × 2 = 0 + 0.243 022 554 071 04;
  • 32) 0.243 022 554 071 04 × 2 = 0 + 0.486 045 108 142 08;
  • 33) 0.486 045 108 142 08 × 2 = 0 + 0.972 090 216 284 16;
  • 34) 0.972 090 216 284 16 × 2 = 1 + 0.944 180 432 568 32;
  • 35) 0.944 180 432 568 32 × 2 = 1 + 0.888 360 865 136 64;
  • 36) 0.888 360 865 136 64 × 2 = 1 + 0.776 721 730 273 28;
  • 37) 0.776 721 730 273 28 × 2 = 1 + 0.553 443 460 546 56;
  • 38) 0.553 443 460 546 56 × 2 = 1 + 0.106 886 921 093 12;
  • 39) 0.106 886 921 093 12 × 2 = 0 + 0.213 773 842 186 24;
  • 40) 0.213 773 842 186 24 × 2 = 0 + 0.427 547 684 372 48;
  • 41) 0.427 547 684 372 48 × 2 = 0 + 0.855 095 368 744 96;
  • 42) 0.855 095 368 744 96 × 2 = 1 + 0.710 190 737 489 92;
  • 43) 0.710 190 737 489 92 × 2 = 1 + 0.420 381 474 979 84;
  • 44) 0.420 381 474 979 84 × 2 = 0 + 0.840 762 949 959 68;
  • 45) 0.840 762 949 959 68 × 2 = 1 + 0.681 525 899 919 36;
  • 46) 0.681 525 899 919 36 × 2 = 1 + 0.363 051 799 838 72;
  • 47) 0.363 051 799 838 72 × 2 = 0 + 0.726 103 599 677 44;

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).


4. 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:


0.000 000 110 009 23(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 110(2)

5. Positive number before normalization:

0.000 000 110 009 23(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 009 23(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0110 110(2) × 20 =

1.1101 1000 0111 1100 0110 110(2) × 2-24


7. 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): -24

Mantissa (not normalized):
1.1101 1000 0111 1100 0110 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(10)


9. 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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

103(10) =

0110 0111(2)


11. 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 1100 0011 1110 0011 0110 =

110 1100 0011 1110 0011 0110


12. 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) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 1110 0011 0110


Decimal number 0.000 000 110 009 23 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0011 0110

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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