-0.000 000 000 742 147 676 692 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 692(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
-0.000 000 000 742 147 676 692(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:

|-0.000 000 000 742 147 676 692| = 0.000 000 000 742 147 676 692


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: 0.000 000 000 742 147 676 692.

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) 0.000 000 000 742 147 676 692 × 2 = 0 + 0.000 000 001 484 295 353 384;
  • 2) 0.000 000 001 484 295 353 384 × 2 = 0 + 0.000 000 002 968 590 706 768;
  • 3) 0.000 000 002 968 590 706 768 × 2 = 0 + 0.000 000 005 937 181 413 536;
  • 4) 0.000 000 005 937 181 413 536 × 2 = 0 + 0.000 000 011 874 362 827 072;
  • 5) 0.000 000 011 874 362 827 072 × 2 = 0 + 0.000 000 023 748 725 654 144;
  • 6) 0.000 000 023 748 725 654 144 × 2 = 0 + 0.000 000 047 497 451 308 288;
  • 7) 0.000 000 047 497 451 308 288 × 2 = 0 + 0.000 000 094 994 902 616 576;
  • 8) 0.000 000 094 994 902 616 576 × 2 = 0 + 0.000 000 189 989 805 233 152;
  • 9) 0.000 000 189 989 805 233 152 × 2 = 0 + 0.000 000 379 979 610 466 304;
  • 10) 0.000 000 379 979 610 466 304 × 2 = 0 + 0.000 000 759 959 220 932 608;
  • 11) 0.000 000 759 959 220 932 608 × 2 = 0 + 0.000 001 519 918 441 865 216;
  • 12) 0.000 001 519 918 441 865 216 × 2 = 0 + 0.000 003 039 836 883 730 432;
  • 13) 0.000 003 039 836 883 730 432 × 2 = 0 + 0.000 006 079 673 767 460 864;
  • 14) 0.000 006 079 673 767 460 864 × 2 = 0 + 0.000 012 159 347 534 921 728;
  • 15) 0.000 012 159 347 534 921 728 × 2 = 0 + 0.000 024 318 695 069 843 456;
  • 16) 0.000 024 318 695 069 843 456 × 2 = 0 + 0.000 048 637 390 139 686 912;
  • 17) 0.000 048 637 390 139 686 912 × 2 = 0 + 0.000 097 274 780 279 373 824;
  • 18) 0.000 097 274 780 279 373 824 × 2 = 0 + 0.000 194 549 560 558 747 648;
  • 19) 0.000 194 549 560 558 747 648 × 2 = 0 + 0.000 389 099 121 117 495 296;
  • 20) 0.000 389 099 121 117 495 296 × 2 = 0 + 0.000 778 198 242 234 990 592;
  • 21) 0.000 778 198 242 234 990 592 × 2 = 0 + 0.001 556 396 484 469 981 184;
  • 22) 0.001 556 396 484 469 981 184 × 2 = 0 + 0.003 112 792 968 939 962 368;
  • 23) 0.003 112 792 968 939 962 368 × 2 = 0 + 0.006 225 585 937 879 924 736;
  • 24) 0.006 225 585 937 879 924 736 × 2 = 0 + 0.012 451 171 875 759 849 472;
  • 25) 0.012 451 171 875 759 849 472 × 2 = 0 + 0.024 902 343 751 519 698 944;
  • 26) 0.024 902 343 751 519 698 944 × 2 = 0 + 0.049 804 687 503 039 397 888;
  • 27) 0.049 804 687 503 039 397 888 × 2 = 0 + 0.099 609 375 006 078 795 776;
  • 28) 0.099 609 375 006 078 795 776 × 2 = 0 + 0.199 218 750 012 157 591 552;
  • 29) 0.199 218 750 012 157 591 552 × 2 = 0 + 0.398 437 500 024 315 183 104;
  • 30) 0.398 437 500 024 315 183 104 × 2 = 0 + 0.796 875 000 048 630 366 208;
  • 31) 0.796 875 000 048 630 366 208 × 2 = 1 + 0.593 750 000 097 260 732 416;
  • 32) 0.593 750 000 097 260 732 416 × 2 = 1 + 0.187 500 000 194 521 464 832;
  • 33) 0.187 500 000 194 521 464 832 × 2 = 0 + 0.375 000 000 389 042 929 664;
  • 34) 0.375 000 000 389 042 929 664 × 2 = 0 + 0.750 000 000 778 085 859 328;
  • 35) 0.750 000 000 778 085 859 328 × 2 = 1 + 0.500 000 001 556 171 718 656;
  • 36) 0.500 000 001 556 171 718 656 × 2 = 1 + 0.000 000 003 112 343 437 312;
  • 37) 0.000 000 003 112 343 437 312 × 2 = 0 + 0.000 000 006 224 686 874 624;
  • 38) 0.000 000 006 224 686 874 624 × 2 = 0 + 0.000 000 012 449 373 749 248;
  • 39) 0.000 000 012 449 373 749 248 × 2 = 0 + 0.000 000 024 898 747 498 496;
  • 40) 0.000 000 024 898 747 498 496 × 2 = 0 + 0.000 000 049 797 494 996 992;
  • 41) 0.000 000 049 797 494 996 992 × 2 = 0 + 0.000 000 099 594 989 993 984;
  • 42) 0.000 000 099 594 989 993 984 × 2 = 0 + 0.000 000 199 189 979 987 968;
  • 43) 0.000 000 199 189 979 987 968 × 2 = 0 + 0.000 000 398 379 959 975 936;
  • 44) 0.000 000 398 379 959 975 936 × 2 = 0 + 0.000 000 796 759 919 951 872;
  • 45) 0.000 000 796 759 919 951 872 × 2 = 0 + 0.000 001 593 519 839 903 744;
  • 46) 0.000 001 593 519 839 903 744 × 2 = 0 + 0.000 003 187 039 679 807 488;
  • 47) 0.000 003 187 039 679 807 488 × 2 = 0 + 0.000 006 374 079 359 614 976;
  • 48) 0.000 006 374 079 359 614 976 × 2 = 0 + 0.000 012 748 158 719 229 952;
  • 49) 0.000 012 748 158 719 229 952 × 2 = 0 + 0.000 025 496 317 438 459 904;
  • 50) 0.000 025 496 317 438 459 904 × 2 = 0 + 0.000 050 992 634 876 919 808;
  • 51) 0.000 050 992 634 876 919 808 × 2 = 0 + 0.000 101 985 269 753 839 616;
  • 52) 0.000 101 985 269 753 839 616 × 2 = 0 + 0.000 203 970 539 507 679 232;
  • 53) 0.000 203 970 539 507 679 232 × 2 = 0 + 0.000 407 941 079 015 358 464;
  • 54) 0.000 407 941 079 015 358 464 × 2 = 0 + 0.000 815 882 158 030 716 928;

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:


0.000 000 000 742 147 676 692(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 692(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:


0.000 000 000 742 147 676 692(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 -0.000 000 000 742 147 676 692 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