-0.000 000 000 742 147 676 287 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 287(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
-0.000 000 000 742 147 676 287(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:

|-0.000 000 000 742 147 676 287| = 0.000 000 000 742 147 676 287


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: 0.000 000 000 742 147 676 287.

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) 0.000 000 000 742 147 676 287 × 2 = 0 + 0.000 000 001 484 295 352 574;
  • 2) 0.000 000 001 484 295 352 574 × 2 = 0 + 0.000 000 002 968 590 705 148;
  • 3) 0.000 000 002 968 590 705 148 × 2 = 0 + 0.000 000 005 937 181 410 296;
  • 4) 0.000 000 005 937 181 410 296 × 2 = 0 + 0.000 000 011 874 362 820 592;
  • 5) 0.000 000 011 874 362 820 592 × 2 = 0 + 0.000 000 023 748 725 641 184;
  • 6) 0.000 000 023 748 725 641 184 × 2 = 0 + 0.000 000 047 497 451 282 368;
  • 7) 0.000 000 047 497 451 282 368 × 2 = 0 + 0.000 000 094 994 902 564 736;
  • 8) 0.000 000 094 994 902 564 736 × 2 = 0 + 0.000 000 189 989 805 129 472;
  • 9) 0.000 000 189 989 805 129 472 × 2 = 0 + 0.000 000 379 979 610 258 944;
  • 10) 0.000 000 379 979 610 258 944 × 2 = 0 + 0.000 000 759 959 220 517 888;
  • 11) 0.000 000 759 959 220 517 888 × 2 = 0 + 0.000 001 519 918 441 035 776;
  • 12) 0.000 001 519 918 441 035 776 × 2 = 0 + 0.000 003 039 836 882 071 552;
  • 13) 0.000 003 039 836 882 071 552 × 2 = 0 + 0.000 006 079 673 764 143 104;
  • 14) 0.000 006 079 673 764 143 104 × 2 = 0 + 0.000 012 159 347 528 286 208;
  • 15) 0.000 012 159 347 528 286 208 × 2 = 0 + 0.000 024 318 695 056 572 416;
  • 16) 0.000 024 318 695 056 572 416 × 2 = 0 + 0.000 048 637 390 113 144 832;
  • 17) 0.000 048 637 390 113 144 832 × 2 = 0 + 0.000 097 274 780 226 289 664;
  • 18) 0.000 097 274 780 226 289 664 × 2 = 0 + 0.000 194 549 560 452 579 328;
  • 19) 0.000 194 549 560 452 579 328 × 2 = 0 + 0.000 389 099 120 905 158 656;
  • 20) 0.000 389 099 120 905 158 656 × 2 = 0 + 0.000 778 198 241 810 317 312;
  • 21) 0.000 778 198 241 810 317 312 × 2 = 0 + 0.001 556 396 483 620 634 624;
  • 22) 0.001 556 396 483 620 634 624 × 2 = 0 + 0.003 112 792 967 241 269 248;
  • 23) 0.003 112 792 967 241 269 248 × 2 = 0 + 0.006 225 585 934 482 538 496;
  • 24) 0.006 225 585 934 482 538 496 × 2 = 0 + 0.012 451 171 868 965 076 992;
  • 25) 0.012 451 171 868 965 076 992 × 2 = 0 + 0.024 902 343 737 930 153 984;
  • 26) 0.024 902 343 737 930 153 984 × 2 = 0 + 0.049 804 687 475 860 307 968;
  • 27) 0.049 804 687 475 860 307 968 × 2 = 0 + 0.099 609 374 951 720 615 936;
  • 28) 0.099 609 374 951 720 615 936 × 2 = 0 + 0.199 218 749 903 441 231 872;
  • 29) 0.199 218 749 903 441 231 872 × 2 = 0 + 0.398 437 499 806 882 463 744;
  • 30) 0.398 437 499 806 882 463 744 × 2 = 0 + 0.796 874 999 613 764 927 488;
  • 31) 0.796 874 999 613 764 927 488 × 2 = 1 + 0.593 749 999 227 529 854 976;
  • 32) 0.593 749 999 227 529 854 976 × 2 = 1 + 0.187 499 998 455 059 709 952;
  • 33) 0.187 499 998 455 059 709 952 × 2 = 0 + 0.374 999 996 910 119 419 904;
  • 34) 0.374 999 996 910 119 419 904 × 2 = 0 + 0.749 999 993 820 238 839 808;
  • 35) 0.749 999 993 820 238 839 808 × 2 = 1 + 0.499 999 987 640 477 679 616;
  • 36) 0.499 999 987 640 477 679 616 × 2 = 0 + 0.999 999 975 280 955 359 232;
  • 37) 0.999 999 975 280 955 359 232 × 2 = 1 + 0.999 999 950 561 910 718 464;
  • 38) 0.999 999 950 561 910 718 464 × 2 = 1 + 0.999 999 901 123 821 436 928;
  • 39) 0.999 999 901 123 821 436 928 × 2 = 1 + 0.999 999 802 247 642 873 856;
  • 40) 0.999 999 802 247 642 873 856 × 2 = 1 + 0.999 999 604 495 285 747 712;
  • 41) 0.999 999 604 495 285 747 712 × 2 = 1 + 0.999 999 208 990 571 495 424;
  • 42) 0.999 999 208 990 571 495 424 × 2 = 1 + 0.999 998 417 981 142 990 848;
  • 43) 0.999 998 417 981 142 990 848 × 2 = 1 + 0.999 996 835 962 285 981 696;
  • 44) 0.999 996 835 962 285 981 696 × 2 = 1 + 0.999 993 671 924 571 963 392;
  • 45) 0.999 993 671 924 571 963 392 × 2 = 1 + 0.999 987 343 849 143 926 784;
  • 46) 0.999 987 343 849 143 926 784 × 2 = 1 + 0.999 974 687 698 287 853 568;
  • 47) 0.999 974 687 698 287 853 568 × 2 = 1 + 0.999 949 375 396 575 707 136;
  • 48) 0.999 949 375 396 575 707 136 × 2 = 1 + 0.999 898 750 793 151 414 272;
  • 49) 0.999 898 750 793 151 414 272 × 2 = 1 + 0.999 797 501 586 302 828 544;
  • 50) 0.999 797 501 586 302 828 544 × 2 = 1 + 0.999 595 003 172 605 657 088;
  • 51) 0.999 595 003 172 605 657 088 × 2 = 1 + 0.999 190 006 345 211 314 176;
  • 52) 0.999 190 006 345 211 314 176 × 2 = 1 + 0.998 380 012 690 422 628 352;
  • 53) 0.998 380 012 690 422 628 352 × 2 = 1 + 0.996 760 025 380 845 256 704;
  • 54) 0.996 760 025 380 845 256 704 × 2 = 1 + 0.993 520 050 761 690 513 408;

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:


0.000 000 000 742 147 676 287(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 287(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:


0.000 000 000 742 147 676 287(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 -0.000 000 000 742 147 676 287 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