-0.000 000 000 735 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 735(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
-0.000 000 000 735(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:

|-0.000 000 000 735| = 0.000 000 000 735


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: 0.000 000 000 735.

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) 0.000 000 000 735 × 2 = 0 + 0.000 000 001 47;
  • 2) 0.000 000 001 47 × 2 = 0 + 0.000 000 002 94;
  • 3) 0.000 000 002 94 × 2 = 0 + 0.000 000 005 88;
  • 4) 0.000 000 005 88 × 2 = 0 + 0.000 000 011 76;
  • 5) 0.000 000 011 76 × 2 = 0 + 0.000 000 023 52;
  • 6) 0.000 000 023 52 × 2 = 0 + 0.000 000 047 04;
  • 7) 0.000 000 047 04 × 2 = 0 + 0.000 000 094 08;
  • 8) 0.000 000 094 08 × 2 = 0 + 0.000 000 188 16;
  • 9) 0.000 000 188 16 × 2 = 0 + 0.000 000 376 32;
  • 10) 0.000 000 376 32 × 2 = 0 + 0.000 000 752 64;
  • 11) 0.000 000 752 64 × 2 = 0 + 0.000 001 505 28;
  • 12) 0.000 001 505 28 × 2 = 0 + 0.000 003 010 56;
  • 13) 0.000 003 010 56 × 2 = 0 + 0.000 006 021 12;
  • 14) 0.000 006 021 12 × 2 = 0 + 0.000 012 042 24;
  • 15) 0.000 012 042 24 × 2 = 0 + 0.000 024 084 48;
  • 16) 0.000 024 084 48 × 2 = 0 + 0.000 048 168 96;
  • 17) 0.000 048 168 96 × 2 = 0 + 0.000 096 337 92;
  • 18) 0.000 096 337 92 × 2 = 0 + 0.000 192 675 84;
  • 19) 0.000 192 675 84 × 2 = 0 + 0.000 385 351 68;
  • 20) 0.000 385 351 68 × 2 = 0 + 0.000 770 703 36;
  • 21) 0.000 770 703 36 × 2 = 0 + 0.001 541 406 72;
  • 22) 0.001 541 406 72 × 2 = 0 + 0.003 082 813 44;
  • 23) 0.003 082 813 44 × 2 = 0 + 0.006 165 626 88;
  • 24) 0.006 165 626 88 × 2 = 0 + 0.012 331 253 76;
  • 25) 0.012 331 253 76 × 2 = 0 + 0.024 662 507 52;
  • 26) 0.024 662 507 52 × 2 = 0 + 0.049 325 015 04;
  • 27) 0.049 325 015 04 × 2 = 0 + 0.098 650 030 08;
  • 28) 0.098 650 030 08 × 2 = 0 + 0.197 300 060 16;
  • 29) 0.197 300 060 16 × 2 = 0 + 0.394 600 120 32;
  • 30) 0.394 600 120 32 × 2 = 0 + 0.789 200 240 64;
  • 31) 0.789 200 240 64 × 2 = 1 + 0.578 400 481 28;
  • 32) 0.578 400 481 28 × 2 = 1 + 0.156 800 962 56;
  • 33) 0.156 800 962 56 × 2 = 0 + 0.313 601 925 12;
  • 34) 0.313 601 925 12 × 2 = 0 + 0.627 203 850 24;
  • 35) 0.627 203 850 24 × 2 = 1 + 0.254 407 700 48;
  • 36) 0.254 407 700 48 × 2 = 0 + 0.508 815 400 96;
  • 37) 0.508 815 400 96 × 2 = 1 + 0.017 630 801 92;
  • 38) 0.017 630 801 92 × 2 = 0 + 0.035 261 603 84;
  • 39) 0.035 261 603 84 × 2 = 0 + 0.070 523 207 68;
  • 40) 0.070 523 207 68 × 2 = 0 + 0.141 046 415 36;
  • 41) 0.141 046 415 36 × 2 = 0 + 0.282 092 830 72;
  • 42) 0.282 092 830 72 × 2 = 0 + 0.564 185 661 44;
  • 43) 0.564 185 661 44 × 2 = 1 + 0.128 371 322 88;
  • 44) 0.128 371 322 88 × 2 = 0 + 0.256 742 645 76;
  • 45) 0.256 742 645 76 × 2 = 0 + 0.513 485 291 52;
  • 46) 0.513 485 291 52 × 2 = 1 + 0.026 970 583 04;
  • 47) 0.026 970 583 04 × 2 = 0 + 0.053 941 166 08;
  • 48) 0.053 941 166 08 × 2 = 0 + 0.107 882 332 16;
  • 49) 0.107 882 332 16 × 2 = 0 + 0.215 764 664 32;
  • 50) 0.215 764 664 32 × 2 = 0 + 0.431 529 328 64;
  • 51) 0.431 529 328 64 × 2 = 0 + 0.863 058 657 28;
  • 52) 0.863 058 657 28 × 2 = 1 + 0.726 117 314 56;
  • 53) 0.726 117 314 56 × 2 = 1 + 0.452 234 629 12;
  • 54) 0.452 234 629 12 × 2 = 0 + 0.904 469 258 24;

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:


0.000 000 000 735(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0010 0100 0001 10(2)

6. Positive number before normalization:

0.000 000 000 735(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0010 0100 0001 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:


0.000 000 000 735(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0010 0100 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1000 0010 0100 0001 10(2) × 20 =

1.1001 0100 0001 0010 0000 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):
1.1001 0100 0001 0010 0000 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. 100 1010 0000 1001 0000 0110 =

100 1010 0000 1001 0000 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) =
100 1010 0000 1001 0000 0110


Decimal number -0.000 000 000 735 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1010 0000 1001 0000 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