0.000 000 110 008 69 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 008 69(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
0.000 000 110 008 69(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: 0.000 000 110 008 69.

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) 0.000 000 110 008 69 × 2 = 0 + 0.000 000 220 017 38;
  • 2) 0.000 000 220 017 38 × 2 = 0 + 0.000 000 440 034 76;
  • 3) 0.000 000 440 034 76 × 2 = 0 + 0.000 000 880 069 52;
  • 4) 0.000 000 880 069 52 × 2 = 0 + 0.000 001 760 139 04;
  • 5) 0.000 001 760 139 04 × 2 = 0 + 0.000 003 520 278 08;
  • 6) 0.000 003 520 278 08 × 2 = 0 + 0.000 007 040 556 16;
  • 7) 0.000 007 040 556 16 × 2 = 0 + 0.000 014 081 112 32;
  • 8) 0.000 014 081 112 32 × 2 = 0 + 0.000 028 162 224 64;
  • 9) 0.000 028 162 224 64 × 2 = 0 + 0.000 056 324 449 28;
  • 10) 0.000 056 324 449 28 × 2 = 0 + 0.000 112 648 898 56;
  • 11) 0.000 112 648 898 56 × 2 = 0 + 0.000 225 297 797 12;
  • 12) 0.000 225 297 797 12 × 2 = 0 + 0.000 450 595 594 24;
  • 13) 0.000 450 595 594 24 × 2 = 0 + 0.000 901 191 188 48;
  • 14) 0.000 901 191 188 48 × 2 = 0 + 0.001 802 382 376 96;
  • 15) 0.001 802 382 376 96 × 2 = 0 + 0.003 604 764 753 92;
  • 16) 0.003 604 764 753 92 × 2 = 0 + 0.007 209 529 507 84;
  • 17) 0.007 209 529 507 84 × 2 = 0 + 0.014 419 059 015 68;
  • 18) 0.014 419 059 015 68 × 2 = 0 + 0.028 838 118 031 36;
  • 19) 0.028 838 118 031 36 × 2 = 0 + 0.057 676 236 062 72;
  • 20) 0.057 676 236 062 72 × 2 = 0 + 0.115 352 472 125 44;
  • 21) 0.115 352 472 125 44 × 2 = 0 + 0.230 704 944 250 88;
  • 22) 0.230 704 944 250 88 × 2 = 0 + 0.461 409 888 501 76;
  • 23) 0.461 409 888 501 76 × 2 = 0 + 0.922 819 777 003 52;
  • 24) 0.922 819 777 003 52 × 2 = 1 + 0.845 639 554 007 04;
  • 25) 0.845 639 554 007 04 × 2 = 1 + 0.691 279 108 014 08;
  • 26) 0.691 279 108 014 08 × 2 = 1 + 0.382 558 216 028 16;
  • 27) 0.382 558 216 028 16 × 2 = 0 + 0.765 116 432 056 32;
  • 28) 0.765 116 432 056 32 × 2 = 1 + 0.530 232 864 112 64;
  • 29) 0.530 232 864 112 64 × 2 = 1 + 0.060 465 728 225 28;
  • 30) 0.060 465 728 225 28 × 2 = 0 + 0.120 931 456 450 56;
  • 31) 0.120 931 456 450 56 × 2 = 0 + 0.241 862 912 901 12;
  • 32) 0.241 862 912 901 12 × 2 = 0 + 0.483 725 825 802 24;
  • 33) 0.483 725 825 802 24 × 2 = 0 + 0.967 451 651 604 48;
  • 34) 0.967 451 651 604 48 × 2 = 1 + 0.934 903 303 208 96;
  • 35) 0.934 903 303 208 96 × 2 = 1 + 0.869 806 606 417 92;
  • 36) 0.869 806 606 417 92 × 2 = 1 + 0.739 613 212 835 84;
  • 37) 0.739 613 212 835 84 × 2 = 1 + 0.479 226 425 671 68;
  • 38) 0.479 226 425 671 68 × 2 = 0 + 0.958 452 851 343 36;
  • 39) 0.958 452 851 343 36 × 2 = 1 + 0.916 905 702 686 72;
  • 40) 0.916 905 702 686 72 × 2 = 1 + 0.833 811 405 373 44;
  • 41) 0.833 811 405 373 44 × 2 = 1 + 0.667 622 810 746 88;
  • 42) 0.667 622 810 746 88 × 2 = 1 + 0.335 245 621 493 76;
  • 43) 0.335 245 621 493 76 × 2 = 0 + 0.670 491 242 987 52;
  • 44) 0.670 491 242 987 52 × 2 = 1 + 0.340 982 485 975 04;
  • 45) 0.340 982 485 975 04 × 2 = 0 + 0.681 964 971 950 08;
  • 46) 0.681 964 971 950 08 × 2 = 1 + 0.363 929 943 900 16;
  • 47) 0.363 929 943 900 16 × 2 = 0 + 0.727 859 887 800 32;

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:


0.000 000 110 008 69(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1101 010(2)

5. Positive number before normalization:

0.000 000 110 008 69(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1101 010(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:


0.000 000 110 008 69(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1101 010(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1101 010(2) × 20 =

1.1101 1000 0111 1011 1101 010(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 1011 1101 010


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 1101 1110 1010 =

110 1100 0011 1101 1110 1010


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 1101 1110 1010


Decimal number 0.000 000 110 008 69 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1101 1110 1010

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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