-0.000 000 000 743 81 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 743 81(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
-0.000 000 000 743 81(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:

|-0.000 000 000 743 81| = 0.000 000 000 743 81


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: 0.000 000 000 743 81.

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) 0.000 000 000 743 81 × 2 = 0 + 0.000 000 001 487 62;
  • 2) 0.000 000 001 487 62 × 2 = 0 + 0.000 000 002 975 24;
  • 3) 0.000 000 002 975 24 × 2 = 0 + 0.000 000 005 950 48;
  • 4) 0.000 000 005 950 48 × 2 = 0 + 0.000 000 011 900 96;
  • 5) 0.000 000 011 900 96 × 2 = 0 + 0.000 000 023 801 92;
  • 6) 0.000 000 023 801 92 × 2 = 0 + 0.000 000 047 603 84;
  • 7) 0.000 000 047 603 84 × 2 = 0 + 0.000 000 095 207 68;
  • 8) 0.000 000 095 207 68 × 2 = 0 + 0.000 000 190 415 36;
  • 9) 0.000 000 190 415 36 × 2 = 0 + 0.000 000 380 830 72;
  • 10) 0.000 000 380 830 72 × 2 = 0 + 0.000 000 761 661 44;
  • 11) 0.000 000 761 661 44 × 2 = 0 + 0.000 001 523 322 88;
  • 12) 0.000 001 523 322 88 × 2 = 0 + 0.000 003 046 645 76;
  • 13) 0.000 003 046 645 76 × 2 = 0 + 0.000 006 093 291 52;
  • 14) 0.000 006 093 291 52 × 2 = 0 + 0.000 012 186 583 04;
  • 15) 0.000 012 186 583 04 × 2 = 0 + 0.000 024 373 166 08;
  • 16) 0.000 024 373 166 08 × 2 = 0 + 0.000 048 746 332 16;
  • 17) 0.000 048 746 332 16 × 2 = 0 + 0.000 097 492 664 32;
  • 18) 0.000 097 492 664 32 × 2 = 0 + 0.000 194 985 328 64;
  • 19) 0.000 194 985 328 64 × 2 = 0 + 0.000 389 970 657 28;
  • 20) 0.000 389 970 657 28 × 2 = 0 + 0.000 779 941 314 56;
  • 21) 0.000 779 941 314 56 × 2 = 0 + 0.001 559 882 629 12;
  • 22) 0.001 559 882 629 12 × 2 = 0 + 0.003 119 765 258 24;
  • 23) 0.003 119 765 258 24 × 2 = 0 + 0.006 239 530 516 48;
  • 24) 0.006 239 530 516 48 × 2 = 0 + 0.012 479 061 032 96;
  • 25) 0.012 479 061 032 96 × 2 = 0 + 0.024 958 122 065 92;
  • 26) 0.024 958 122 065 92 × 2 = 0 + 0.049 916 244 131 84;
  • 27) 0.049 916 244 131 84 × 2 = 0 + 0.099 832 488 263 68;
  • 28) 0.099 832 488 263 68 × 2 = 0 + 0.199 664 976 527 36;
  • 29) 0.199 664 976 527 36 × 2 = 0 + 0.399 329 953 054 72;
  • 30) 0.399 329 953 054 72 × 2 = 0 + 0.798 659 906 109 44;
  • 31) 0.798 659 906 109 44 × 2 = 1 + 0.597 319 812 218 88;
  • 32) 0.597 319 812 218 88 × 2 = 1 + 0.194 639 624 437 76;
  • 33) 0.194 639 624 437 76 × 2 = 0 + 0.389 279 248 875 52;
  • 34) 0.389 279 248 875 52 × 2 = 0 + 0.778 558 497 751 04;
  • 35) 0.778 558 497 751 04 × 2 = 1 + 0.557 116 995 502 08;
  • 36) 0.557 116 995 502 08 × 2 = 1 + 0.114 233 991 004 16;
  • 37) 0.114 233 991 004 16 × 2 = 0 + 0.228 467 982 008 32;
  • 38) 0.228 467 982 008 32 × 2 = 0 + 0.456 935 964 016 64;
  • 39) 0.456 935 964 016 64 × 2 = 0 + 0.913 871 928 033 28;
  • 40) 0.913 871 928 033 28 × 2 = 1 + 0.827 743 856 066 56;
  • 41) 0.827 743 856 066 56 × 2 = 1 + 0.655 487 712 133 12;
  • 42) 0.655 487 712 133 12 × 2 = 1 + 0.310 975 424 266 24;
  • 43) 0.310 975 424 266 24 × 2 = 0 + 0.621 950 848 532 48;
  • 44) 0.621 950 848 532 48 × 2 = 1 + 0.243 901 697 064 96;
  • 45) 0.243 901 697 064 96 × 2 = 0 + 0.487 803 394 129 92;
  • 46) 0.487 803 394 129 92 × 2 = 0 + 0.975 606 788 259 84;
  • 47) 0.975 606 788 259 84 × 2 = 1 + 0.951 213 576 519 68;
  • 48) 0.951 213 576 519 68 × 2 = 1 + 0.902 427 153 039 36;
  • 49) 0.902 427 153 039 36 × 2 = 1 + 0.804 854 306 078 72;
  • 50) 0.804 854 306 078 72 × 2 = 1 + 0.609 708 612 157 44;
  • 51) 0.609 708 612 157 44 × 2 = 1 + 0.219 417 224 314 88;
  • 52) 0.219 417 224 314 88 × 2 = 0 + 0.438 834 448 629 76;
  • 53) 0.438 834 448 629 76 × 2 = 0 + 0.877 668 897 259 52;
  • 54) 0.877 668 897 259 52 × 2 = 1 + 0.755 337 794 519 04;

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:


0.000 000 000 743 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1101 0011 1110 01(2)

6. Positive number before normalization:

0.000 000 000 743 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1101 0011 1110 01(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:


0.000 000 000 743 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1101 0011 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0001 1101 0011 1110 01(2) × 20 =

1.1001 1000 1110 1001 1111 001(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 1000 1110 1001 1111 001


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 1100 0111 0100 1111 1001 =

100 1100 0111 0100 1111 1001


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 1100 0111 0100 1111 1001


Decimal number -0.000 000 000 743 81 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0111 0100 1111 1001

Share

» Convert decimal -0.000 000 000 744 68 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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