-0.000 000 000 742 147 676 646 535 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 535(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
-0.000 000 000 742 147 676 646 535(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:

|-0.000 000 000 742 147 676 646 535| = 0.000 000 000 742 147 676 646 535


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: 0.000 000 000 742 147 676 646 535.

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) 0.000 000 000 742 147 676 646 535 × 2 = 0 + 0.000 000 001 484 295 353 293 07;
  • 2) 0.000 000 001 484 295 353 293 07 × 2 = 0 + 0.000 000 002 968 590 706 586 14;
  • 3) 0.000 000 002 968 590 706 586 14 × 2 = 0 + 0.000 000 005 937 181 413 172 28;
  • 4) 0.000 000 005 937 181 413 172 28 × 2 = 0 + 0.000 000 011 874 362 826 344 56;
  • 5) 0.000 000 011 874 362 826 344 56 × 2 = 0 + 0.000 000 023 748 725 652 689 12;
  • 6) 0.000 000 023 748 725 652 689 12 × 2 = 0 + 0.000 000 047 497 451 305 378 24;
  • 7) 0.000 000 047 497 451 305 378 24 × 2 = 0 + 0.000 000 094 994 902 610 756 48;
  • 8) 0.000 000 094 994 902 610 756 48 × 2 = 0 + 0.000 000 189 989 805 221 512 96;
  • 9) 0.000 000 189 989 805 221 512 96 × 2 = 0 + 0.000 000 379 979 610 443 025 92;
  • 10) 0.000 000 379 979 610 443 025 92 × 2 = 0 + 0.000 000 759 959 220 886 051 84;
  • 11) 0.000 000 759 959 220 886 051 84 × 2 = 0 + 0.000 001 519 918 441 772 103 68;
  • 12) 0.000 001 519 918 441 772 103 68 × 2 = 0 + 0.000 003 039 836 883 544 207 36;
  • 13) 0.000 003 039 836 883 544 207 36 × 2 = 0 + 0.000 006 079 673 767 088 414 72;
  • 14) 0.000 006 079 673 767 088 414 72 × 2 = 0 + 0.000 012 159 347 534 176 829 44;
  • 15) 0.000 012 159 347 534 176 829 44 × 2 = 0 + 0.000 024 318 695 068 353 658 88;
  • 16) 0.000 024 318 695 068 353 658 88 × 2 = 0 + 0.000 048 637 390 136 707 317 76;
  • 17) 0.000 048 637 390 136 707 317 76 × 2 = 0 + 0.000 097 274 780 273 414 635 52;
  • 18) 0.000 097 274 780 273 414 635 52 × 2 = 0 + 0.000 194 549 560 546 829 271 04;
  • 19) 0.000 194 549 560 546 829 271 04 × 2 = 0 + 0.000 389 099 121 093 658 542 08;
  • 20) 0.000 389 099 121 093 658 542 08 × 2 = 0 + 0.000 778 198 242 187 317 084 16;
  • 21) 0.000 778 198 242 187 317 084 16 × 2 = 0 + 0.001 556 396 484 374 634 168 32;
  • 22) 0.001 556 396 484 374 634 168 32 × 2 = 0 + 0.003 112 792 968 749 268 336 64;
  • 23) 0.003 112 792 968 749 268 336 64 × 2 = 0 + 0.006 225 585 937 498 536 673 28;
  • 24) 0.006 225 585 937 498 536 673 28 × 2 = 0 + 0.012 451 171 874 997 073 346 56;
  • 25) 0.012 451 171 874 997 073 346 56 × 2 = 0 + 0.024 902 343 749 994 146 693 12;
  • 26) 0.024 902 343 749 994 146 693 12 × 2 = 0 + 0.049 804 687 499 988 293 386 24;
  • 27) 0.049 804 687 499 988 293 386 24 × 2 = 0 + 0.099 609 374 999 976 586 772 48;
  • 28) 0.099 609 374 999 976 586 772 48 × 2 = 0 + 0.199 218 749 999 953 173 544 96;
  • 29) 0.199 218 749 999 953 173 544 96 × 2 = 0 + 0.398 437 499 999 906 347 089 92;
  • 30) 0.398 437 499 999 906 347 089 92 × 2 = 0 + 0.796 874 999 999 812 694 179 84;
  • 31) 0.796 874 999 999 812 694 179 84 × 2 = 1 + 0.593 749 999 999 625 388 359 68;
  • 32) 0.593 749 999 999 625 388 359 68 × 2 = 1 + 0.187 499 999 999 250 776 719 36;
  • 33) 0.187 499 999 999 250 776 719 36 × 2 = 0 + 0.374 999 999 998 501 553 438 72;
  • 34) 0.374 999 999 998 501 553 438 72 × 2 = 0 + 0.749 999 999 997 003 106 877 44;
  • 35) 0.749 999 999 997 003 106 877 44 × 2 = 1 + 0.499 999 999 994 006 213 754 88;
  • 36) 0.499 999 999 994 006 213 754 88 × 2 = 0 + 0.999 999 999 988 012 427 509 76;
  • 37) 0.999 999 999 988 012 427 509 76 × 2 = 1 + 0.999 999 999 976 024 855 019 52;
  • 38) 0.999 999 999 976 024 855 019 52 × 2 = 1 + 0.999 999 999 952 049 710 039 04;
  • 39) 0.999 999 999 952 049 710 039 04 × 2 = 1 + 0.999 999 999 904 099 420 078 08;
  • 40) 0.999 999 999 904 099 420 078 08 × 2 = 1 + 0.999 999 999 808 198 840 156 16;
  • 41) 0.999 999 999 808 198 840 156 16 × 2 = 1 + 0.999 999 999 616 397 680 312 32;
  • 42) 0.999 999 999 616 397 680 312 32 × 2 = 1 + 0.999 999 999 232 795 360 624 64;
  • 43) 0.999 999 999 232 795 360 624 64 × 2 = 1 + 0.999 999 998 465 590 721 249 28;
  • 44) 0.999 999 998 465 590 721 249 28 × 2 = 1 + 0.999 999 996 931 181 442 498 56;
  • 45) 0.999 999 996 931 181 442 498 56 × 2 = 1 + 0.999 999 993 862 362 884 997 12;
  • 46) 0.999 999 993 862 362 884 997 12 × 2 = 1 + 0.999 999 987 724 725 769 994 24;
  • 47) 0.999 999 987 724 725 769 994 24 × 2 = 1 + 0.999 999 975 449 451 539 988 48;
  • 48) 0.999 999 975 449 451 539 988 48 × 2 = 1 + 0.999 999 950 898 903 079 976 96;
  • 49) 0.999 999 950 898 903 079 976 96 × 2 = 1 + 0.999 999 901 797 806 159 953 92;
  • 50) 0.999 999 901 797 806 159 953 92 × 2 = 1 + 0.999 999 803 595 612 319 907 84;
  • 51) 0.999 999 803 595 612 319 907 84 × 2 = 1 + 0.999 999 607 191 224 639 815 68;
  • 52) 0.999 999 607 191 224 639 815 68 × 2 = 1 + 0.999 999 214 382 449 279 631 36;
  • 53) 0.999 999 214 382 449 279 631 36 × 2 = 1 + 0.999 998 428 764 898 559 262 72;
  • 54) 0.999 998 428 764 898 559 262 72 × 2 = 1 + 0.999 996 857 529 797 118 525 44;

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:


0.000 000 000 742 147 676 646 535(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 535(10) =

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


0.000 000 000 742 147 676 646 535(10) =

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

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

1.1001 0111 1111 1111 1111 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 1111 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 1111 1111 =

100 1011 1111 1111 1111 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 1111 1111


Decimal number -0.000 000 000 742 147 676 646 535 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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