-0.000 000 000 742 147 34 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 34(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
-0.000 000 000 742 147 34(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:

|-0.000 000 000 742 147 34| = 0.000 000 000 742 147 34


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: 0.000 000 000 742 147 34.

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) 0.000 000 000 742 147 34 × 2 = 0 + 0.000 000 001 484 294 68;
  • 2) 0.000 000 001 484 294 68 × 2 = 0 + 0.000 000 002 968 589 36;
  • 3) 0.000 000 002 968 589 36 × 2 = 0 + 0.000 000 005 937 178 72;
  • 4) 0.000 000 005 937 178 72 × 2 = 0 + 0.000 000 011 874 357 44;
  • 5) 0.000 000 011 874 357 44 × 2 = 0 + 0.000 000 023 748 714 88;
  • 6) 0.000 000 023 748 714 88 × 2 = 0 + 0.000 000 047 497 429 76;
  • 7) 0.000 000 047 497 429 76 × 2 = 0 + 0.000 000 094 994 859 52;
  • 8) 0.000 000 094 994 859 52 × 2 = 0 + 0.000 000 189 989 719 04;
  • 9) 0.000 000 189 989 719 04 × 2 = 0 + 0.000 000 379 979 438 08;
  • 10) 0.000 000 379 979 438 08 × 2 = 0 + 0.000 000 759 958 876 16;
  • 11) 0.000 000 759 958 876 16 × 2 = 0 + 0.000 001 519 917 752 32;
  • 12) 0.000 001 519 917 752 32 × 2 = 0 + 0.000 003 039 835 504 64;
  • 13) 0.000 003 039 835 504 64 × 2 = 0 + 0.000 006 079 671 009 28;
  • 14) 0.000 006 079 671 009 28 × 2 = 0 + 0.000 012 159 342 018 56;
  • 15) 0.000 012 159 342 018 56 × 2 = 0 + 0.000 024 318 684 037 12;
  • 16) 0.000 024 318 684 037 12 × 2 = 0 + 0.000 048 637 368 074 24;
  • 17) 0.000 048 637 368 074 24 × 2 = 0 + 0.000 097 274 736 148 48;
  • 18) 0.000 097 274 736 148 48 × 2 = 0 + 0.000 194 549 472 296 96;
  • 19) 0.000 194 549 472 296 96 × 2 = 0 + 0.000 389 098 944 593 92;
  • 20) 0.000 389 098 944 593 92 × 2 = 0 + 0.000 778 197 889 187 84;
  • 21) 0.000 778 197 889 187 84 × 2 = 0 + 0.001 556 395 778 375 68;
  • 22) 0.001 556 395 778 375 68 × 2 = 0 + 0.003 112 791 556 751 36;
  • 23) 0.003 112 791 556 751 36 × 2 = 0 + 0.006 225 583 113 502 72;
  • 24) 0.006 225 583 113 502 72 × 2 = 0 + 0.012 451 166 227 005 44;
  • 25) 0.012 451 166 227 005 44 × 2 = 0 + 0.024 902 332 454 010 88;
  • 26) 0.024 902 332 454 010 88 × 2 = 0 + 0.049 804 664 908 021 76;
  • 27) 0.049 804 664 908 021 76 × 2 = 0 + 0.099 609 329 816 043 52;
  • 28) 0.099 609 329 816 043 52 × 2 = 0 + 0.199 218 659 632 087 04;
  • 29) 0.199 218 659 632 087 04 × 2 = 0 + 0.398 437 319 264 174 08;
  • 30) 0.398 437 319 264 174 08 × 2 = 0 + 0.796 874 638 528 348 16;
  • 31) 0.796 874 638 528 348 16 × 2 = 1 + 0.593 749 277 056 696 32;
  • 32) 0.593 749 277 056 696 32 × 2 = 1 + 0.187 498 554 113 392 64;
  • 33) 0.187 498 554 113 392 64 × 2 = 0 + 0.374 997 108 226 785 28;
  • 34) 0.374 997 108 226 785 28 × 2 = 0 + 0.749 994 216 453 570 56;
  • 35) 0.749 994 216 453 570 56 × 2 = 1 + 0.499 988 432 907 141 12;
  • 36) 0.499 988 432 907 141 12 × 2 = 0 + 0.999 976 865 814 282 24;
  • 37) 0.999 976 865 814 282 24 × 2 = 1 + 0.999 953 731 628 564 48;
  • 38) 0.999 953 731 628 564 48 × 2 = 1 + 0.999 907 463 257 128 96;
  • 39) 0.999 907 463 257 128 96 × 2 = 1 + 0.999 814 926 514 257 92;
  • 40) 0.999 814 926 514 257 92 × 2 = 1 + 0.999 629 853 028 515 84;
  • 41) 0.999 629 853 028 515 84 × 2 = 1 + 0.999 259 706 057 031 68;
  • 42) 0.999 259 706 057 031 68 × 2 = 1 + 0.998 519 412 114 063 36;
  • 43) 0.998 519 412 114 063 36 × 2 = 1 + 0.997 038 824 228 126 72;
  • 44) 0.997 038 824 228 126 72 × 2 = 1 + 0.994 077 648 456 253 44;
  • 45) 0.994 077 648 456 253 44 × 2 = 1 + 0.988 155 296 912 506 88;
  • 46) 0.988 155 296 912 506 88 × 2 = 1 + 0.976 310 593 825 013 76;
  • 47) 0.976 310 593 825 013 76 × 2 = 1 + 0.952 621 187 650 027 52;
  • 48) 0.952 621 187 650 027 52 × 2 = 1 + 0.905 242 375 300 055 04;
  • 49) 0.905 242 375 300 055 04 × 2 = 1 + 0.810 484 750 600 110 08;
  • 50) 0.810 484 750 600 110 08 × 2 = 1 + 0.620 969 501 200 220 16;
  • 51) 0.620 969 501 200 220 16 × 2 = 1 + 0.241 939 002 400 440 32;
  • 52) 0.241 939 002 400 440 32 × 2 = 0 + 0.483 878 004 800 880 64;
  • 53) 0.483 878 004 800 880 64 × 2 = 0 + 0.967 756 009 601 761 28;
  • 54) 0.967 756 009 601 761 28 × 2 = 1 + 0.935 512 019 203 522 56;

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:


0.000 000 000 742 147 34(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 34(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 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:


0.000 000 000 742 147 34(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2) × 20 =

1.1001 0111 1111 1111 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 0111 1111 1111 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 1011 1111 1111 1111 1001 =

100 1011 1111 1111 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 1011 1111 1111 1111 1001


Decimal number -0.000 000 000 742 147 34 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1001

Share

» Convert decimal -0.000 000 000 742 148 28 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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