-0.000 000 000 742 147 677 53 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 53(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
-0.000 000 000 742 147 677 53(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:

|-0.000 000 000 742 147 677 53| = 0.000 000 000 742 147 677 53


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: 0.000 000 000 742 147 677 53.

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) 0.000 000 000 742 147 677 53 × 2 = 0 + 0.000 000 001 484 295 355 06;
  • 2) 0.000 000 001 484 295 355 06 × 2 = 0 + 0.000 000 002 968 590 710 12;
  • 3) 0.000 000 002 968 590 710 12 × 2 = 0 + 0.000 000 005 937 181 420 24;
  • 4) 0.000 000 005 937 181 420 24 × 2 = 0 + 0.000 000 011 874 362 840 48;
  • 5) 0.000 000 011 874 362 840 48 × 2 = 0 + 0.000 000 023 748 725 680 96;
  • 6) 0.000 000 023 748 725 680 96 × 2 = 0 + 0.000 000 047 497 451 361 92;
  • 7) 0.000 000 047 497 451 361 92 × 2 = 0 + 0.000 000 094 994 902 723 84;
  • 8) 0.000 000 094 994 902 723 84 × 2 = 0 + 0.000 000 189 989 805 447 68;
  • 9) 0.000 000 189 989 805 447 68 × 2 = 0 + 0.000 000 379 979 610 895 36;
  • 10) 0.000 000 379 979 610 895 36 × 2 = 0 + 0.000 000 759 959 221 790 72;
  • 11) 0.000 000 759 959 221 790 72 × 2 = 0 + 0.000 001 519 918 443 581 44;
  • 12) 0.000 001 519 918 443 581 44 × 2 = 0 + 0.000 003 039 836 887 162 88;
  • 13) 0.000 003 039 836 887 162 88 × 2 = 0 + 0.000 006 079 673 774 325 76;
  • 14) 0.000 006 079 673 774 325 76 × 2 = 0 + 0.000 012 159 347 548 651 52;
  • 15) 0.000 012 159 347 548 651 52 × 2 = 0 + 0.000 024 318 695 097 303 04;
  • 16) 0.000 024 318 695 097 303 04 × 2 = 0 + 0.000 048 637 390 194 606 08;
  • 17) 0.000 048 637 390 194 606 08 × 2 = 0 + 0.000 097 274 780 389 212 16;
  • 18) 0.000 097 274 780 389 212 16 × 2 = 0 + 0.000 194 549 560 778 424 32;
  • 19) 0.000 194 549 560 778 424 32 × 2 = 0 + 0.000 389 099 121 556 848 64;
  • 20) 0.000 389 099 121 556 848 64 × 2 = 0 + 0.000 778 198 243 113 697 28;
  • 21) 0.000 778 198 243 113 697 28 × 2 = 0 + 0.001 556 396 486 227 394 56;
  • 22) 0.001 556 396 486 227 394 56 × 2 = 0 + 0.003 112 792 972 454 789 12;
  • 23) 0.003 112 792 972 454 789 12 × 2 = 0 + 0.006 225 585 944 909 578 24;
  • 24) 0.006 225 585 944 909 578 24 × 2 = 0 + 0.012 451 171 889 819 156 48;
  • 25) 0.012 451 171 889 819 156 48 × 2 = 0 + 0.024 902 343 779 638 312 96;
  • 26) 0.024 902 343 779 638 312 96 × 2 = 0 + 0.049 804 687 559 276 625 92;
  • 27) 0.049 804 687 559 276 625 92 × 2 = 0 + 0.099 609 375 118 553 251 84;
  • 28) 0.099 609 375 118 553 251 84 × 2 = 0 + 0.199 218 750 237 106 503 68;
  • 29) 0.199 218 750 237 106 503 68 × 2 = 0 + 0.398 437 500 474 213 007 36;
  • 30) 0.398 437 500 474 213 007 36 × 2 = 0 + 0.796 875 000 948 426 014 72;
  • 31) 0.796 875 000 948 426 014 72 × 2 = 1 + 0.593 750 001 896 852 029 44;
  • 32) 0.593 750 001 896 852 029 44 × 2 = 1 + 0.187 500 003 793 704 058 88;
  • 33) 0.187 500 003 793 704 058 88 × 2 = 0 + 0.375 000 007 587 408 117 76;
  • 34) 0.375 000 007 587 408 117 76 × 2 = 0 + 0.750 000 015 174 816 235 52;
  • 35) 0.750 000 015 174 816 235 52 × 2 = 1 + 0.500 000 030 349 632 471 04;
  • 36) 0.500 000 030 349 632 471 04 × 2 = 1 + 0.000 000 060 699 264 942 08;
  • 37) 0.000 000 060 699 264 942 08 × 2 = 0 + 0.000 000 121 398 529 884 16;
  • 38) 0.000 000 121 398 529 884 16 × 2 = 0 + 0.000 000 242 797 059 768 32;
  • 39) 0.000 000 242 797 059 768 32 × 2 = 0 + 0.000 000 485 594 119 536 64;
  • 40) 0.000 000 485 594 119 536 64 × 2 = 0 + 0.000 000 971 188 239 073 28;
  • 41) 0.000 000 971 188 239 073 28 × 2 = 0 + 0.000 001 942 376 478 146 56;
  • 42) 0.000 001 942 376 478 146 56 × 2 = 0 + 0.000 003 884 752 956 293 12;
  • 43) 0.000 003 884 752 956 293 12 × 2 = 0 + 0.000 007 769 505 912 586 24;
  • 44) 0.000 007 769 505 912 586 24 × 2 = 0 + 0.000 015 539 011 825 172 48;
  • 45) 0.000 015 539 011 825 172 48 × 2 = 0 + 0.000 031 078 023 650 344 96;
  • 46) 0.000 031 078 023 650 344 96 × 2 = 0 + 0.000 062 156 047 300 689 92;
  • 47) 0.000 062 156 047 300 689 92 × 2 = 0 + 0.000 124 312 094 601 379 84;
  • 48) 0.000 124 312 094 601 379 84 × 2 = 0 + 0.000 248 624 189 202 759 68;
  • 49) 0.000 248 624 189 202 759 68 × 2 = 0 + 0.000 497 248 378 405 519 36;
  • 50) 0.000 497 248 378 405 519 36 × 2 = 0 + 0.000 994 496 756 811 038 72;
  • 51) 0.000 994 496 756 811 038 72 × 2 = 0 + 0.001 988 993 513 622 077 44;
  • 52) 0.001 988 993 513 622 077 44 × 2 = 0 + 0.003 977 987 027 244 154 88;
  • 53) 0.003 977 987 027 244 154 88 × 2 = 0 + 0.007 955 974 054 488 309 76;
  • 54) 0.007 955 974 054 488 309 76 × 2 = 0 + 0.015 911 948 108 976 619 52;

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:


0.000 000 000 742 147 677 53(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 677 53(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 677 53(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 0000 0000 0000 000


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 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 677 53 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 678 21 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