-0.000 000 000 742 147 676 545 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 545(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
-0.000 000 000 742 147 676 545(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:

|-0.000 000 000 742 147 676 545| = 0.000 000 000 742 147 676 545


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: 0.000 000 000 742 147 676 545.

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) 0.000 000 000 742 147 676 545 × 2 = 0 + 0.000 000 001 484 295 353 09;
  • 2) 0.000 000 001 484 295 353 09 × 2 = 0 + 0.000 000 002 968 590 706 18;
  • 3) 0.000 000 002 968 590 706 18 × 2 = 0 + 0.000 000 005 937 181 412 36;
  • 4) 0.000 000 005 937 181 412 36 × 2 = 0 + 0.000 000 011 874 362 824 72;
  • 5) 0.000 000 011 874 362 824 72 × 2 = 0 + 0.000 000 023 748 725 649 44;
  • 6) 0.000 000 023 748 725 649 44 × 2 = 0 + 0.000 000 047 497 451 298 88;
  • 7) 0.000 000 047 497 451 298 88 × 2 = 0 + 0.000 000 094 994 902 597 76;
  • 8) 0.000 000 094 994 902 597 76 × 2 = 0 + 0.000 000 189 989 805 195 52;
  • 9) 0.000 000 189 989 805 195 52 × 2 = 0 + 0.000 000 379 979 610 391 04;
  • 10) 0.000 000 379 979 610 391 04 × 2 = 0 + 0.000 000 759 959 220 782 08;
  • 11) 0.000 000 759 959 220 782 08 × 2 = 0 + 0.000 001 519 918 441 564 16;
  • 12) 0.000 001 519 918 441 564 16 × 2 = 0 + 0.000 003 039 836 883 128 32;
  • 13) 0.000 003 039 836 883 128 32 × 2 = 0 + 0.000 006 079 673 766 256 64;
  • 14) 0.000 006 079 673 766 256 64 × 2 = 0 + 0.000 012 159 347 532 513 28;
  • 15) 0.000 012 159 347 532 513 28 × 2 = 0 + 0.000 024 318 695 065 026 56;
  • 16) 0.000 024 318 695 065 026 56 × 2 = 0 + 0.000 048 637 390 130 053 12;
  • 17) 0.000 048 637 390 130 053 12 × 2 = 0 + 0.000 097 274 780 260 106 24;
  • 18) 0.000 097 274 780 260 106 24 × 2 = 0 + 0.000 194 549 560 520 212 48;
  • 19) 0.000 194 549 560 520 212 48 × 2 = 0 + 0.000 389 099 121 040 424 96;
  • 20) 0.000 389 099 121 040 424 96 × 2 = 0 + 0.000 778 198 242 080 849 92;
  • 21) 0.000 778 198 242 080 849 92 × 2 = 0 + 0.001 556 396 484 161 699 84;
  • 22) 0.001 556 396 484 161 699 84 × 2 = 0 + 0.003 112 792 968 323 399 68;
  • 23) 0.003 112 792 968 323 399 68 × 2 = 0 + 0.006 225 585 936 646 799 36;
  • 24) 0.006 225 585 936 646 799 36 × 2 = 0 + 0.012 451 171 873 293 598 72;
  • 25) 0.012 451 171 873 293 598 72 × 2 = 0 + 0.024 902 343 746 587 197 44;
  • 26) 0.024 902 343 746 587 197 44 × 2 = 0 + 0.049 804 687 493 174 394 88;
  • 27) 0.049 804 687 493 174 394 88 × 2 = 0 + 0.099 609 374 986 348 789 76;
  • 28) 0.099 609 374 986 348 789 76 × 2 = 0 + 0.199 218 749 972 697 579 52;
  • 29) 0.199 218 749 972 697 579 52 × 2 = 0 + 0.398 437 499 945 395 159 04;
  • 30) 0.398 437 499 945 395 159 04 × 2 = 0 + 0.796 874 999 890 790 318 08;
  • 31) 0.796 874 999 890 790 318 08 × 2 = 1 + 0.593 749 999 781 580 636 16;
  • 32) 0.593 749 999 781 580 636 16 × 2 = 1 + 0.187 499 999 563 161 272 32;
  • 33) 0.187 499 999 563 161 272 32 × 2 = 0 + 0.374 999 999 126 322 544 64;
  • 34) 0.374 999 999 126 322 544 64 × 2 = 0 + 0.749 999 998 252 645 089 28;
  • 35) 0.749 999 998 252 645 089 28 × 2 = 1 + 0.499 999 996 505 290 178 56;
  • 36) 0.499 999 996 505 290 178 56 × 2 = 0 + 0.999 999 993 010 580 357 12;
  • 37) 0.999 999 993 010 580 357 12 × 2 = 1 + 0.999 999 986 021 160 714 24;
  • 38) 0.999 999 986 021 160 714 24 × 2 = 1 + 0.999 999 972 042 321 428 48;
  • 39) 0.999 999 972 042 321 428 48 × 2 = 1 + 0.999 999 944 084 642 856 96;
  • 40) 0.999 999 944 084 642 856 96 × 2 = 1 + 0.999 999 888 169 285 713 92;
  • 41) 0.999 999 888 169 285 713 92 × 2 = 1 + 0.999 999 776 338 571 427 84;
  • 42) 0.999 999 776 338 571 427 84 × 2 = 1 + 0.999 999 552 677 142 855 68;
  • 43) 0.999 999 552 677 142 855 68 × 2 = 1 + 0.999 999 105 354 285 711 36;
  • 44) 0.999 999 105 354 285 711 36 × 2 = 1 + 0.999 998 210 708 571 422 72;
  • 45) 0.999 998 210 708 571 422 72 × 2 = 1 + 0.999 996 421 417 142 845 44;
  • 46) 0.999 996 421 417 142 845 44 × 2 = 1 + 0.999 992 842 834 285 690 88;
  • 47) 0.999 992 842 834 285 690 88 × 2 = 1 + 0.999 985 685 668 571 381 76;
  • 48) 0.999 985 685 668 571 381 76 × 2 = 1 + 0.999 971 371 337 142 763 52;
  • 49) 0.999 971 371 337 142 763 52 × 2 = 1 + 0.999 942 742 674 285 527 04;
  • 50) 0.999 942 742 674 285 527 04 × 2 = 1 + 0.999 885 485 348 571 054 08;
  • 51) 0.999 885 485 348 571 054 08 × 2 = 1 + 0.999 770 970 697 142 108 16;
  • 52) 0.999 770 970 697 142 108 16 × 2 = 1 + 0.999 541 941 394 284 216 32;
  • 53) 0.999 541 941 394 284 216 32 × 2 = 1 + 0.999 083 882 788 568 432 64;
  • 54) 0.999 083 882 788 568 432 64 × 2 = 1 + 0.998 167 765 577 136 865 28;

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:


0.000 000 000 742 147 676 545(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 545(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:


0.000 000 000 742 147 676 545(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 -0.000 000 000 742 147 676 545 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