-0.000 000 000 845 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 845(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
-0.000 000 000 845(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:

|-0.000 000 000 845| = 0.000 000 000 845


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: 0.000 000 000 845.

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) 0.000 000 000 845 × 2 = 0 + 0.000 000 001 69;
  • 2) 0.000 000 001 69 × 2 = 0 + 0.000 000 003 38;
  • 3) 0.000 000 003 38 × 2 = 0 + 0.000 000 006 76;
  • 4) 0.000 000 006 76 × 2 = 0 + 0.000 000 013 52;
  • 5) 0.000 000 013 52 × 2 = 0 + 0.000 000 027 04;
  • 6) 0.000 000 027 04 × 2 = 0 + 0.000 000 054 08;
  • 7) 0.000 000 054 08 × 2 = 0 + 0.000 000 108 16;
  • 8) 0.000 000 108 16 × 2 = 0 + 0.000 000 216 32;
  • 9) 0.000 000 216 32 × 2 = 0 + 0.000 000 432 64;
  • 10) 0.000 000 432 64 × 2 = 0 + 0.000 000 865 28;
  • 11) 0.000 000 865 28 × 2 = 0 + 0.000 001 730 56;
  • 12) 0.000 001 730 56 × 2 = 0 + 0.000 003 461 12;
  • 13) 0.000 003 461 12 × 2 = 0 + 0.000 006 922 24;
  • 14) 0.000 006 922 24 × 2 = 0 + 0.000 013 844 48;
  • 15) 0.000 013 844 48 × 2 = 0 + 0.000 027 688 96;
  • 16) 0.000 027 688 96 × 2 = 0 + 0.000 055 377 92;
  • 17) 0.000 055 377 92 × 2 = 0 + 0.000 110 755 84;
  • 18) 0.000 110 755 84 × 2 = 0 + 0.000 221 511 68;
  • 19) 0.000 221 511 68 × 2 = 0 + 0.000 443 023 36;
  • 20) 0.000 443 023 36 × 2 = 0 + 0.000 886 046 72;
  • 21) 0.000 886 046 72 × 2 = 0 + 0.001 772 093 44;
  • 22) 0.001 772 093 44 × 2 = 0 + 0.003 544 186 88;
  • 23) 0.003 544 186 88 × 2 = 0 + 0.007 088 373 76;
  • 24) 0.007 088 373 76 × 2 = 0 + 0.014 176 747 52;
  • 25) 0.014 176 747 52 × 2 = 0 + 0.028 353 495 04;
  • 26) 0.028 353 495 04 × 2 = 0 + 0.056 706 990 08;
  • 27) 0.056 706 990 08 × 2 = 0 + 0.113 413 980 16;
  • 28) 0.113 413 980 16 × 2 = 0 + 0.226 827 960 32;
  • 29) 0.226 827 960 32 × 2 = 0 + 0.453 655 920 64;
  • 30) 0.453 655 920 64 × 2 = 0 + 0.907 311 841 28;
  • 31) 0.907 311 841 28 × 2 = 1 + 0.814 623 682 56;
  • 32) 0.814 623 682 56 × 2 = 1 + 0.629 247 365 12;
  • 33) 0.629 247 365 12 × 2 = 1 + 0.258 494 730 24;
  • 34) 0.258 494 730 24 × 2 = 0 + 0.516 989 460 48;
  • 35) 0.516 989 460 48 × 2 = 1 + 0.033 978 920 96;
  • 36) 0.033 978 920 96 × 2 = 0 + 0.067 957 841 92;
  • 37) 0.067 957 841 92 × 2 = 0 + 0.135 915 683 84;
  • 38) 0.135 915 683 84 × 2 = 0 + 0.271 831 367 68;
  • 39) 0.271 831 367 68 × 2 = 0 + 0.543 662 735 36;
  • 40) 0.543 662 735 36 × 2 = 1 + 0.087 325 470 72;
  • 41) 0.087 325 470 72 × 2 = 0 + 0.174 650 941 44;
  • 42) 0.174 650 941 44 × 2 = 0 + 0.349 301 882 88;
  • 43) 0.349 301 882 88 × 2 = 0 + 0.698 603 765 76;
  • 44) 0.698 603 765 76 × 2 = 1 + 0.397 207 531 52;
  • 45) 0.397 207 531 52 × 2 = 0 + 0.794 415 063 04;
  • 46) 0.794 415 063 04 × 2 = 1 + 0.588 830 126 08;
  • 47) 0.588 830 126 08 × 2 = 1 + 0.177 660 252 16;
  • 48) 0.177 660 252 16 × 2 = 0 + 0.355 320 504 32;
  • 49) 0.355 320 504 32 × 2 = 0 + 0.710 641 008 64;
  • 50) 0.710 641 008 64 × 2 = 1 + 0.421 282 017 28;
  • 51) 0.421 282 017 28 × 2 = 0 + 0.842 564 034 56;
  • 52) 0.842 564 034 56 × 2 = 1 + 0.685 128 069 12;
  • 53) 0.685 128 069 12 × 2 = 1 + 0.370 256 138 24;
  • 54) 0.370 256 138 24 × 2 = 0 + 0.740 512 276 48;

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:


0.000 000 000 845(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0001 0001 0110 0101 10(2)

6. Positive number before normalization:

0.000 000 000 845(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0001 0001 0110 0101 10(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:


0.000 000 000 845(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0001 0001 0110 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 0001 0001 0110 0101 10(2) × 20 =

1.1101 0000 1000 1011 0010 110(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 0000 1000 1011 0010 110


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 1000 0100 0101 1001 0110 =

110 1000 0100 0101 1001 0110


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 1000 0100 0101 1001 0110


Decimal number -0.000 000 000 845 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 1000 0100 0101 1001 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