-0.000 000 000 742 147 677 28 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 28(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 28(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 28| = 0.000 000 000 742 147 677 28


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 28.

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 28 × 2 = 0 + 0.000 000 001 484 295 354 56;
  • 2) 0.000 000 001 484 295 354 56 × 2 = 0 + 0.000 000 002 968 590 709 12;
  • 3) 0.000 000 002 968 590 709 12 × 2 = 0 + 0.000 000 005 937 181 418 24;
  • 4) 0.000 000 005 937 181 418 24 × 2 = 0 + 0.000 000 011 874 362 836 48;
  • 5) 0.000 000 011 874 362 836 48 × 2 = 0 + 0.000 000 023 748 725 672 96;
  • 6) 0.000 000 023 748 725 672 96 × 2 = 0 + 0.000 000 047 497 451 345 92;
  • 7) 0.000 000 047 497 451 345 92 × 2 = 0 + 0.000 000 094 994 902 691 84;
  • 8) 0.000 000 094 994 902 691 84 × 2 = 0 + 0.000 000 189 989 805 383 68;
  • 9) 0.000 000 189 989 805 383 68 × 2 = 0 + 0.000 000 379 979 610 767 36;
  • 10) 0.000 000 379 979 610 767 36 × 2 = 0 + 0.000 000 759 959 221 534 72;
  • 11) 0.000 000 759 959 221 534 72 × 2 = 0 + 0.000 001 519 918 443 069 44;
  • 12) 0.000 001 519 918 443 069 44 × 2 = 0 + 0.000 003 039 836 886 138 88;
  • 13) 0.000 003 039 836 886 138 88 × 2 = 0 + 0.000 006 079 673 772 277 76;
  • 14) 0.000 006 079 673 772 277 76 × 2 = 0 + 0.000 012 159 347 544 555 52;
  • 15) 0.000 012 159 347 544 555 52 × 2 = 0 + 0.000 024 318 695 089 111 04;
  • 16) 0.000 024 318 695 089 111 04 × 2 = 0 + 0.000 048 637 390 178 222 08;
  • 17) 0.000 048 637 390 178 222 08 × 2 = 0 + 0.000 097 274 780 356 444 16;
  • 18) 0.000 097 274 780 356 444 16 × 2 = 0 + 0.000 194 549 560 712 888 32;
  • 19) 0.000 194 549 560 712 888 32 × 2 = 0 + 0.000 389 099 121 425 776 64;
  • 20) 0.000 389 099 121 425 776 64 × 2 = 0 + 0.000 778 198 242 851 553 28;
  • 21) 0.000 778 198 242 851 553 28 × 2 = 0 + 0.001 556 396 485 703 106 56;
  • 22) 0.001 556 396 485 703 106 56 × 2 = 0 + 0.003 112 792 971 406 213 12;
  • 23) 0.003 112 792 971 406 213 12 × 2 = 0 + 0.006 225 585 942 812 426 24;
  • 24) 0.006 225 585 942 812 426 24 × 2 = 0 + 0.012 451 171 885 624 852 48;
  • 25) 0.012 451 171 885 624 852 48 × 2 = 0 + 0.024 902 343 771 249 704 96;
  • 26) 0.024 902 343 771 249 704 96 × 2 = 0 + 0.049 804 687 542 499 409 92;
  • 27) 0.049 804 687 542 499 409 92 × 2 = 0 + 0.099 609 375 084 998 819 84;
  • 28) 0.099 609 375 084 998 819 84 × 2 = 0 + 0.199 218 750 169 997 639 68;
  • 29) 0.199 218 750 169 997 639 68 × 2 = 0 + 0.398 437 500 339 995 279 36;
  • 30) 0.398 437 500 339 995 279 36 × 2 = 0 + 0.796 875 000 679 990 558 72;
  • 31) 0.796 875 000 679 990 558 72 × 2 = 1 + 0.593 750 001 359 981 117 44;
  • 32) 0.593 750 001 359 981 117 44 × 2 = 1 + 0.187 500 002 719 962 234 88;
  • 33) 0.187 500 002 719 962 234 88 × 2 = 0 + 0.375 000 005 439 924 469 76;
  • 34) 0.375 000 005 439 924 469 76 × 2 = 0 + 0.750 000 010 879 848 939 52;
  • 35) 0.750 000 010 879 848 939 52 × 2 = 1 + 0.500 000 021 759 697 879 04;
  • 36) 0.500 000 021 759 697 879 04 × 2 = 1 + 0.000 000 043 519 395 758 08;
  • 37) 0.000 000 043 519 395 758 08 × 2 = 0 + 0.000 000 087 038 791 516 16;
  • 38) 0.000 000 087 038 791 516 16 × 2 = 0 + 0.000 000 174 077 583 032 32;
  • 39) 0.000 000 174 077 583 032 32 × 2 = 0 + 0.000 000 348 155 166 064 64;
  • 40) 0.000 000 348 155 166 064 64 × 2 = 0 + 0.000 000 696 310 332 129 28;
  • 41) 0.000 000 696 310 332 129 28 × 2 = 0 + 0.000 001 392 620 664 258 56;
  • 42) 0.000 001 392 620 664 258 56 × 2 = 0 + 0.000 002 785 241 328 517 12;
  • 43) 0.000 002 785 241 328 517 12 × 2 = 0 + 0.000 005 570 482 657 034 24;
  • 44) 0.000 005 570 482 657 034 24 × 2 = 0 + 0.000 011 140 965 314 068 48;
  • 45) 0.000 011 140 965 314 068 48 × 2 = 0 + 0.000 022 281 930 628 136 96;
  • 46) 0.000 022 281 930 628 136 96 × 2 = 0 + 0.000 044 563 861 256 273 92;
  • 47) 0.000 044 563 861 256 273 92 × 2 = 0 + 0.000 089 127 722 512 547 84;
  • 48) 0.000 089 127 722 512 547 84 × 2 = 0 + 0.000 178 255 445 025 095 68;
  • 49) 0.000 178 255 445 025 095 68 × 2 = 0 + 0.000 356 510 890 050 191 36;
  • 50) 0.000 356 510 890 050 191 36 × 2 = 0 + 0.000 713 021 780 100 382 72;
  • 51) 0.000 713 021 780 100 382 72 × 2 = 0 + 0.001 426 043 560 200 765 44;
  • 52) 0.001 426 043 560 200 765 44 × 2 = 0 + 0.002 852 087 120 401 530 88;
  • 53) 0.002 852 087 120 401 530 88 × 2 = 0 + 0.005 704 174 240 803 061 76;
  • 54) 0.005 704 174 240 803 061 76 × 2 = 0 + 0.011 408 348 481 606 123 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 28(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 28(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 28(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 28 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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