-0.000 000 099 985 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 099 985(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
-0.000 000 099 985(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:

|-0.000 000 099 985| = 0.000 000 099 985


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: 0.000 000 099 985.

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) 0.000 000 099 985 × 2 = 0 + 0.000 000 199 97;
  • 2) 0.000 000 199 97 × 2 = 0 + 0.000 000 399 94;
  • 3) 0.000 000 399 94 × 2 = 0 + 0.000 000 799 88;
  • 4) 0.000 000 799 88 × 2 = 0 + 0.000 001 599 76;
  • 5) 0.000 001 599 76 × 2 = 0 + 0.000 003 199 52;
  • 6) 0.000 003 199 52 × 2 = 0 + 0.000 006 399 04;
  • 7) 0.000 006 399 04 × 2 = 0 + 0.000 012 798 08;
  • 8) 0.000 012 798 08 × 2 = 0 + 0.000 025 596 16;
  • 9) 0.000 025 596 16 × 2 = 0 + 0.000 051 192 32;
  • 10) 0.000 051 192 32 × 2 = 0 + 0.000 102 384 64;
  • 11) 0.000 102 384 64 × 2 = 0 + 0.000 204 769 28;
  • 12) 0.000 204 769 28 × 2 = 0 + 0.000 409 538 56;
  • 13) 0.000 409 538 56 × 2 = 0 + 0.000 819 077 12;
  • 14) 0.000 819 077 12 × 2 = 0 + 0.001 638 154 24;
  • 15) 0.001 638 154 24 × 2 = 0 + 0.003 276 308 48;
  • 16) 0.003 276 308 48 × 2 = 0 + 0.006 552 616 96;
  • 17) 0.006 552 616 96 × 2 = 0 + 0.013 105 233 92;
  • 18) 0.013 105 233 92 × 2 = 0 + 0.026 210 467 84;
  • 19) 0.026 210 467 84 × 2 = 0 + 0.052 420 935 68;
  • 20) 0.052 420 935 68 × 2 = 0 + 0.104 841 871 36;
  • 21) 0.104 841 871 36 × 2 = 0 + 0.209 683 742 72;
  • 22) 0.209 683 742 72 × 2 = 0 + 0.419 367 485 44;
  • 23) 0.419 367 485 44 × 2 = 0 + 0.838 734 970 88;
  • 24) 0.838 734 970 88 × 2 = 1 + 0.677 469 941 76;
  • 25) 0.677 469 941 76 × 2 = 1 + 0.354 939 883 52;
  • 26) 0.354 939 883 52 × 2 = 0 + 0.709 879 767 04;
  • 27) 0.709 879 767 04 × 2 = 1 + 0.419 759 534 08;
  • 28) 0.419 759 534 08 × 2 = 0 + 0.839 519 068 16;
  • 29) 0.839 519 068 16 × 2 = 1 + 0.679 038 136 32;
  • 30) 0.679 038 136 32 × 2 = 1 + 0.358 076 272 64;
  • 31) 0.358 076 272 64 × 2 = 0 + 0.716 152 545 28;
  • 32) 0.716 152 545 28 × 2 = 1 + 0.432 305 090 56;
  • 33) 0.432 305 090 56 × 2 = 0 + 0.864 610 181 12;
  • 34) 0.864 610 181 12 × 2 = 1 + 0.729 220 362 24;
  • 35) 0.729 220 362 24 × 2 = 1 + 0.458 440 724 48;
  • 36) 0.458 440 724 48 × 2 = 0 + 0.916 881 448 96;
  • 37) 0.916 881 448 96 × 2 = 1 + 0.833 762 897 92;
  • 38) 0.833 762 897 92 × 2 = 1 + 0.667 525 795 84;
  • 39) 0.667 525 795 84 × 2 = 1 + 0.335 051 591 68;
  • 40) 0.335 051 591 68 × 2 = 0 + 0.670 103 183 36;
  • 41) 0.670 103 183 36 × 2 = 1 + 0.340 206 366 72;
  • 42) 0.340 206 366 72 × 2 = 0 + 0.680 412 733 44;
  • 43) 0.680 412 733 44 × 2 = 1 + 0.360 825 466 88;
  • 44) 0.360 825 466 88 × 2 = 0 + 0.721 650 933 76;
  • 45) 0.721 650 933 76 × 2 = 1 + 0.443 301 867 52;
  • 46) 0.443 301 867 52 × 2 = 0 + 0.886 603 735 04;
  • 47) 0.886 603 735 04 × 2 = 1 + 0.773 207 470 08;

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:


0.000 000 099 985(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0110 1110 1010 101(2)

6. Positive number before normalization:

0.000 000 099 985(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0110 1110 1010 101(2)

7. 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:


0.000 000 099 985(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0110 1110 1010 101(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0110 1110 1010 101(2) × 20 =

1.1010 1101 0110 1110 1010 101(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1101 0110 1110 1010 101


9. 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)


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;
  • 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;

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) =

103(10) =

0110 0111(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. 101 0110 1011 0111 0101 0101 =

101 0110 1011 0111 0101 0101


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 0111

Mantissa (23 bits) =
101 0110 1011 0111 0101 0101


Decimal number -0.000 000 099 985 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1011 0111 0101 0101

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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