-0.000 000 000 742 147 676 646 583 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 583(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
-0.000 000 000 742 147 676 646 583(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:

|-0.000 000 000 742 147 676 646 583| = 0.000 000 000 742 147 676 646 583


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: 0.000 000 000 742 147 676 646 583.

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) 0.000 000 000 742 147 676 646 583 × 2 = 0 + 0.000 000 001 484 295 353 293 166;
  • 2) 0.000 000 001 484 295 353 293 166 × 2 = 0 + 0.000 000 002 968 590 706 586 332;
  • 3) 0.000 000 002 968 590 706 586 332 × 2 = 0 + 0.000 000 005 937 181 413 172 664;
  • 4) 0.000 000 005 937 181 413 172 664 × 2 = 0 + 0.000 000 011 874 362 826 345 328;
  • 5) 0.000 000 011 874 362 826 345 328 × 2 = 0 + 0.000 000 023 748 725 652 690 656;
  • 6) 0.000 000 023 748 725 652 690 656 × 2 = 0 + 0.000 000 047 497 451 305 381 312;
  • 7) 0.000 000 047 497 451 305 381 312 × 2 = 0 + 0.000 000 094 994 902 610 762 624;
  • 8) 0.000 000 094 994 902 610 762 624 × 2 = 0 + 0.000 000 189 989 805 221 525 248;
  • 9) 0.000 000 189 989 805 221 525 248 × 2 = 0 + 0.000 000 379 979 610 443 050 496;
  • 10) 0.000 000 379 979 610 443 050 496 × 2 = 0 + 0.000 000 759 959 220 886 100 992;
  • 11) 0.000 000 759 959 220 886 100 992 × 2 = 0 + 0.000 001 519 918 441 772 201 984;
  • 12) 0.000 001 519 918 441 772 201 984 × 2 = 0 + 0.000 003 039 836 883 544 403 968;
  • 13) 0.000 003 039 836 883 544 403 968 × 2 = 0 + 0.000 006 079 673 767 088 807 936;
  • 14) 0.000 006 079 673 767 088 807 936 × 2 = 0 + 0.000 012 159 347 534 177 615 872;
  • 15) 0.000 012 159 347 534 177 615 872 × 2 = 0 + 0.000 024 318 695 068 355 231 744;
  • 16) 0.000 024 318 695 068 355 231 744 × 2 = 0 + 0.000 048 637 390 136 710 463 488;
  • 17) 0.000 048 637 390 136 710 463 488 × 2 = 0 + 0.000 097 274 780 273 420 926 976;
  • 18) 0.000 097 274 780 273 420 926 976 × 2 = 0 + 0.000 194 549 560 546 841 853 952;
  • 19) 0.000 194 549 560 546 841 853 952 × 2 = 0 + 0.000 389 099 121 093 683 707 904;
  • 20) 0.000 389 099 121 093 683 707 904 × 2 = 0 + 0.000 778 198 242 187 367 415 808;
  • 21) 0.000 778 198 242 187 367 415 808 × 2 = 0 + 0.001 556 396 484 374 734 831 616;
  • 22) 0.001 556 396 484 374 734 831 616 × 2 = 0 + 0.003 112 792 968 749 469 663 232;
  • 23) 0.003 112 792 968 749 469 663 232 × 2 = 0 + 0.006 225 585 937 498 939 326 464;
  • 24) 0.006 225 585 937 498 939 326 464 × 2 = 0 + 0.012 451 171 874 997 878 652 928;
  • 25) 0.012 451 171 874 997 878 652 928 × 2 = 0 + 0.024 902 343 749 995 757 305 856;
  • 26) 0.024 902 343 749 995 757 305 856 × 2 = 0 + 0.049 804 687 499 991 514 611 712;
  • 27) 0.049 804 687 499 991 514 611 712 × 2 = 0 + 0.099 609 374 999 983 029 223 424;
  • 28) 0.099 609 374 999 983 029 223 424 × 2 = 0 + 0.199 218 749 999 966 058 446 848;
  • 29) 0.199 218 749 999 966 058 446 848 × 2 = 0 + 0.398 437 499 999 932 116 893 696;
  • 30) 0.398 437 499 999 932 116 893 696 × 2 = 0 + 0.796 874 999 999 864 233 787 392;
  • 31) 0.796 874 999 999 864 233 787 392 × 2 = 1 + 0.593 749 999 999 728 467 574 784;
  • 32) 0.593 749 999 999 728 467 574 784 × 2 = 1 + 0.187 499 999 999 456 935 149 568;
  • 33) 0.187 499 999 999 456 935 149 568 × 2 = 0 + 0.374 999 999 998 913 870 299 136;
  • 34) 0.374 999 999 998 913 870 299 136 × 2 = 0 + 0.749 999 999 997 827 740 598 272;
  • 35) 0.749 999 999 997 827 740 598 272 × 2 = 1 + 0.499 999 999 995 655 481 196 544;
  • 36) 0.499 999 999 995 655 481 196 544 × 2 = 0 + 0.999 999 999 991 310 962 393 088;
  • 37) 0.999 999 999 991 310 962 393 088 × 2 = 1 + 0.999 999 999 982 621 924 786 176;
  • 38) 0.999 999 999 982 621 924 786 176 × 2 = 1 + 0.999 999 999 965 243 849 572 352;
  • 39) 0.999 999 999 965 243 849 572 352 × 2 = 1 + 0.999 999 999 930 487 699 144 704;
  • 40) 0.999 999 999 930 487 699 144 704 × 2 = 1 + 0.999 999 999 860 975 398 289 408;
  • 41) 0.999 999 999 860 975 398 289 408 × 2 = 1 + 0.999 999 999 721 950 796 578 816;
  • 42) 0.999 999 999 721 950 796 578 816 × 2 = 1 + 0.999 999 999 443 901 593 157 632;
  • 43) 0.999 999 999 443 901 593 157 632 × 2 = 1 + 0.999 999 998 887 803 186 315 264;
  • 44) 0.999 999 998 887 803 186 315 264 × 2 = 1 + 0.999 999 997 775 606 372 630 528;
  • 45) 0.999 999 997 775 606 372 630 528 × 2 = 1 + 0.999 999 995 551 212 745 261 056;
  • 46) 0.999 999 995 551 212 745 261 056 × 2 = 1 + 0.999 999 991 102 425 490 522 112;
  • 47) 0.999 999 991 102 425 490 522 112 × 2 = 1 + 0.999 999 982 204 850 981 044 224;
  • 48) 0.999 999 982 204 850 981 044 224 × 2 = 1 + 0.999 999 964 409 701 962 088 448;
  • 49) 0.999 999 964 409 701 962 088 448 × 2 = 1 + 0.999 999 928 819 403 924 176 896;
  • 50) 0.999 999 928 819 403 924 176 896 × 2 = 1 + 0.999 999 857 638 807 848 353 792;
  • 51) 0.999 999 857 638 807 848 353 792 × 2 = 1 + 0.999 999 715 277 615 696 707 584;
  • 52) 0.999 999 715 277 615 696 707 584 × 2 = 1 + 0.999 999 430 555 231 393 415 168;
  • 53) 0.999 999 430 555 231 393 415 168 × 2 = 1 + 0.999 998 861 110 462 786 830 336;
  • 54) 0.999 998 861 110 462 786 830 336 × 2 = 1 + 0.999 997 722 220 925 573 660 672;

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:


0.000 000 000 742 147 676 646 583(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 583(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:


0.000 000 000 742 147 676 646 583(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 -0.000 000 000 742 147 676 646 583 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 67 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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