-0.000 000 000 742 145 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 145(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
-0.000 000 000 742 145(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:

|-0.000 000 000 742 145| = 0.000 000 000 742 145


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: 0.000 000 000 742 145.

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) 0.000 000 000 742 145 × 2 = 0 + 0.000 000 001 484 29;
  • 2) 0.000 000 001 484 29 × 2 = 0 + 0.000 000 002 968 58;
  • 3) 0.000 000 002 968 58 × 2 = 0 + 0.000 000 005 937 16;
  • 4) 0.000 000 005 937 16 × 2 = 0 + 0.000 000 011 874 32;
  • 5) 0.000 000 011 874 32 × 2 = 0 + 0.000 000 023 748 64;
  • 6) 0.000 000 023 748 64 × 2 = 0 + 0.000 000 047 497 28;
  • 7) 0.000 000 047 497 28 × 2 = 0 + 0.000 000 094 994 56;
  • 8) 0.000 000 094 994 56 × 2 = 0 + 0.000 000 189 989 12;
  • 9) 0.000 000 189 989 12 × 2 = 0 + 0.000 000 379 978 24;
  • 10) 0.000 000 379 978 24 × 2 = 0 + 0.000 000 759 956 48;
  • 11) 0.000 000 759 956 48 × 2 = 0 + 0.000 001 519 912 96;
  • 12) 0.000 001 519 912 96 × 2 = 0 + 0.000 003 039 825 92;
  • 13) 0.000 003 039 825 92 × 2 = 0 + 0.000 006 079 651 84;
  • 14) 0.000 006 079 651 84 × 2 = 0 + 0.000 012 159 303 68;
  • 15) 0.000 012 159 303 68 × 2 = 0 + 0.000 024 318 607 36;
  • 16) 0.000 024 318 607 36 × 2 = 0 + 0.000 048 637 214 72;
  • 17) 0.000 048 637 214 72 × 2 = 0 + 0.000 097 274 429 44;
  • 18) 0.000 097 274 429 44 × 2 = 0 + 0.000 194 548 858 88;
  • 19) 0.000 194 548 858 88 × 2 = 0 + 0.000 389 097 717 76;
  • 20) 0.000 389 097 717 76 × 2 = 0 + 0.000 778 195 435 52;
  • 21) 0.000 778 195 435 52 × 2 = 0 + 0.001 556 390 871 04;
  • 22) 0.001 556 390 871 04 × 2 = 0 + 0.003 112 781 742 08;
  • 23) 0.003 112 781 742 08 × 2 = 0 + 0.006 225 563 484 16;
  • 24) 0.006 225 563 484 16 × 2 = 0 + 0.012 451 126 968 32;
  • 25) 0.012 451 126 968 32 × 2 = 0 + 0.024 902 253 936 64;
  • 26) 0.024 902 253 936 64 × 2 = 0 + 0.049 804 507 873 28;
  • 27) 0.049 804 507 873 28 × 2 = 0 + 0.099 609 015 746 56;
  • 28) 0.099 609 015 746 56 × 2 = 0 + 0.199 218 031 493 12;
  • 29) 0.199 218 031 493 12 × 2 = 0 + 0.398 436 062 986 24;
  • 30) 0.398 436 062 986 24 × 2 = 0 + 0.796 872 125 972 48;
  • 31) 0.796 872 125 972 48 × 2 = 1 + 0.593 744 251 944 96;
  • 32) 0.593 744 251 944 96 × 2 = 1 + 0.187 488 503 889 92;
  • 33) 0.187 488 503 889 92 × 2 = 0 + 0.374 977 007 779 84;
  • 34) 0.374 977 007 779 84 × 2 = 0 + 0.749 954 015 559 68;
  • 35) 0.749 954 015 559 68 × 2 = 1 + 0.499 908 031 119 36;
  • 36) 0.499 908 031 119 36 × 2 = 0 + 0.999 816 062 238 72;
  • 37) 0.999 816 062 238 72 × 2 = 1 + 0.999 632 124 477 44;
  • 38) 0.999 632 124 477 44 × 2 = 1 + 0.999 264 248 954 88;
  • 39) 0.999 264 248 954 88 × 2 = 1 + 0.998 528 497 909 76;
  • 40) 0.998 528 497 909 76 × 2 = 1 + 0.997 056 995 819 52;
  • 41) 0.997 056 995 819 52 × 2 = 1 + 0.994 113 991 639 04;
  • 42) 0.994 113 991 639 04 × 2 = 1 + 0.988 227 983 278 08;
  • 43) 0.988 227 983 278 08 × 2 = 1 + 0.976 455 966 556 16;
  • 44) 0.976 455 966 556 16 × 2 = 1 + 0.952 911 933 112 32;
  • 45) 0.952 911 933 112 32 × 2 = 1 + 0.905 823 866 224 64;
  • 46) 0.905 823 866 224 64 × 2 = 1 + 0.811 647 732 449 28;
  • 47) 0.811 647 732 449 28 × 2 = 1 + 0.623 295 464 898 56;
  • 48) 0.623 295 464 898 56 × 2 = 1 + 0.246 590 929 797 12;
  • 49) 0.246 590 929 797 12 × 2 = 0 + 0.493 181 859 594 24;
  • 50) 0.493 181 859 594 24 × 2 = 0 + 0.986 363 719 188 48;
  • 51) 0.986 363 719 188 48 × 2 = 1 + 0.972 727 438 376 96;
  • 52) 0.972 727 438 376 96 × 2 = 1 + 0.945 454 876 753 92;
  • 53) 0.945 454 876 753 92 × 2 = 1 + 0.890 909 753 507 84;
  • 54) 0.890 909 753 507 84 × 2 = 1 + 0.781 819 507 015 68;

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:


0.000 000 000 742 145(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0011 11(2)

6. Positive number before normalization:

0.000 000 000 742 145(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0011 11(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:


0.000 000 000 742 145(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 0011 11(2) × 20 =

1.1001 0111 1111 1111 1001 111(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 0111 1111 1111 1001 111


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 1011 1111 1111 1100 1111 =

100 1011 1111 1111 1100 1111


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 1011 1111 1111 1100 1111


Decimal number -0.000 000 000 742 145 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1100 1111

Share

» Convert decimal -0.000 000 000 742 227 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: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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