-0.000 000 000 742 147 676 646 679 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 679 1(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
-0.000 000 000 742 147 676 646 679 1(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:

|-0.000 000 000 742 147 676 646 679 1| = 0.000 000 000 742 147 676 646 679 1


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: 0.000 000 000 742 147 676 646 679 1.

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) 0.000 000 000 742 147 676 646 679 1 × 2 = 0 + 0.000 000 001 484 295 353 293 358 2;
  • 2) 0.000 000 001 484 295 353 293 358 2 × 2 = 0 + 0.000 000 002 968 590 706 586 716 4;
  • 3) 0.000 000 002 968 590 706 586 716 4 × 2 = 0 + 0.000 000 005 937 181 413 173 432 8;
  • 4) 0.000 000 005 937 181 413 173 432 8 × 2 = 0 + 0.000 000 011 874 362 826 346 865 6;
  • 5) 0.000 000 011 874 362 826 346 865 6 × 2 = 0 + 0.000 000 023 748 725 652 693 731 2;
  • 6) 0.000 000 023 748 725 652 693 731 2 × 2 = 0 + 0.000 000 047 497 451 305 387 462 4;
  • 7) 0.000 000 047 497 451 305 387 462 4 × 2 = 0 + 0.000 000 094 994 902 610 774 924 8;
  • 8) 0.000 000 094 994 902 610 774 924 8 × 2 = 0 + 0.000 000 189 989 805 221 549 849 6;
  • 9) 0.000 000 189 989 805 221 549 849 6 × 2 = 0 + 0.000 000 379 979 610 443 099 699 2;
  • 10) 0.000 000 379 979 610 443 099 699 2 × 2 = 0 + 0.000 000 759 959 220 886 199 398 4;
  • 11) 0.000 000 759 959 220 886 199 398 4 × 2 = 0 + 0.000 001 519 918 441 772 398 796 8;
  • 12) 0.000 001 519 918 441 772 398 796 8 × 2 = 0 + 0.000 003 039 836 883 544 797 593 6;
  • 13) 0.000 003 039 836 883 544 797 593 6 × 2 = 0 + 0.000 006 079 673 767 089 595 187 2;
  • 14) 0.000 006 079 673 767 089 595 187 2 × 2 = 0 + 0.000 012 159 347 534 179 190 374 4;
  • 15) 0.000 012 159 347 534 179 190 374 4 × 2 = 0 + 0.000 024 318 695 068 358 380 748 8;
  • 16) 0.000 024 318 695 068 358 380 748 8 × 2 = 0 + 0.000 048 637 390 136 716 761 497 6;
  • 17) 0.000 048 637 390 136 716 761 497 6 × 2 = 0 + 0.000 097 274 780 273 433 522 995 2;
  • 18) 0.000 097 274 780 273 433 522 995 2 × 2 = 0 + 0.000 194 549 560 546 867 045 990 4;
  • 19) 0.000 194 549 560 546 867 045 990 4 × 2 = 0 + 0.000 389 099 121 093 734 091 980 8;
  • 20) 0.000 389 099 121 093 734 091 980 8 × 2 = 0 + 0.000 778 198 242 187 468 183 961 6;
  • 21) 0.000 778 198 242 187 468 183 961 6 × 2 = 0 + 0.001 556 396 484 374 936 367 923 2;
  • 22) 0.001 556 396 484 374 936 367 923 2 × 2 = 0 + 0.003 112 792 968 749 872 735 846 4;
  • 23) 0.003 112 792 968 749 872 735 846 4 × 2 = 0 + 0.006 225 585 937 499 745 471 692 8;
  • 24) 0.006 225 585 937 499 745 471 692 8 × 2 = 0 + 0.012 451 171 874 999 490 943 385 6;
  • 25) 0.012 451 171 874 999 490 943 385 6 × 2 = 0 + 0.024 902 343 749 998 981 886 771 2;
  • 26) 0.024 902 343 749 998 981 886 771 2 × 2 = 0 + 0.049 804 687 499 997 963 773 542 4;
  • 27) 0.049 804 687 499 997 963 773 542 4 × 2 = 0 + 0.099 609 374 999 995 927 547 084 8;
  • 28) 0.099 609 374 999 995 927 547 084 8 × 2 = 0 + 0.199 218 749 999 991 855 094 169 6;
  • 29) 0.199 218 749 999 991 855 094 169 6 × 2 = 0 + 0.398 437 499 999 983 710 188 339 2;
  • 30) 0.398 437 499 999 983 710 188 339 2 × 2 = 0 + 0.796 874 999 999 967 420 376 678 4;
  • 31) 0.796 874 999 999 967 420 376 678 4 × 2 = 1 + 0.593 749 999 999 934 840 753 356 8;
  • 32) 0.593 749 999 999 934 840 753 356 8 × 2 = 1 + 0.187 499 999 999 869 681 506 713 6;
  • 33) 0.187 499 999 999 869 681 506 713 6 × 2 = 0 + 0.374 999 999 999 739 363 013 427 2;
  • 34) 0.374 999 999 999 739 363 013 427 2 × 2 = 0 + 0.749 999 999 999 478 726 026 854 4;
  • 35) 0.749 999 999 999 478 726 026 854 4 × 2 = 1 + 0.499 999 999 998 957 452 053 708 8;
  • 36) 0.499 999 999 998 957 452 053 708 8 × 2 = 0 + 0.999 999 999 997 914 904 107 417 6;
  • 37) 0.999 999 999 997 914 904 107 417 6 × 2 = 1 + 0.999 999 999 995 829 808 214 835 2;
  • 38) 0.999 999 999 995 829 808 214 835 2 × 2 = 1 + 0.999 999 999 991 659 616 429 670 4;
  • 39) 0.999 999 999 991 659 616 429 670 4 × 2 = 1 + 0.999 999 999 983 319 232 859 340 8;
  • 40) 0.999 999 999 983 319 232 859 340 8 × 2 = 1 + 0.999 999 999 966 638 465 718 681 6;
  • 41) 0.999 999 999 966 638 465 718 681 6 × 2 = 1 + 0.999 999 999 933 276 931 437 363 2;
  • 42) 0.999 999 999 933 276 931 437 363 2 × 2 = 1 + 0.999 999 999 866 553 862 874 726 4;
  • 43) 0.999 999 999 866 553 862 874 726 4 × 2 = 1 + 0.999 999 999 733 107 725 749 452 8;
  • 44) 0.999 999 999 733 107 725 749 452 8 × 2 = 1 + 0.999 999 999 466 215 451 498 905 6;
  • 45) 0.999 999 999 466 215 451 498 905 6 × 2 = 1 + 0.999 999 998 932 430 902 997 811 2;
  • 46) 0.999 999 998 932 430 902 997 811 2 × 2 = 1 + 0.999 999 997 864 861 805 995 622 4;
  • 47) 0.999 999 997 864 861 805 995 622 4 × 2 = 1 + 0.999 999 995 729 723 611 991 244 8;
  • 48) 0.999 999 995 729 723 611 991 244 8 × 2 = 1 + 0.999 999 991 459 447 223 982 489 6;
  • 49) 0.999 999 991 459 447 223 982 489 6 × 2 = 1 + 0.999 999 982 918 894 447 964 979 2;
  • 50) 0.999 999 982 918 894 447 964 979 2 × 2 = 1 + 0.999 999 965 837 788 895 929 958 4;
  • 51) 0.999 999 965 837 788 895 929 958 4 × 2 = 1 + 0.999 999 931 675 577 791 859 916 8;
  • 52) 0.999 999 931 675 577 791 859 916 8 × 2 = 1 + 0.999 999 863 351 155 583 719 833 6;
  • 53) 0.999 999 863 351 155 583 719 833 6 × 2 = 1 + 0.999 999 726 702 311 167 439 667 2;
  • 54) 0.999 999 726 702 311 167 439 667 2 × 2 = 1 + 0.999 999 453 404 622 334 879 334 4;

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:


0.000 000 000 742 147 676 646 679 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 679 1(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:


0.000 000 000 742 147 676 646 679 1(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 -0.000 000 000 742 147 676 646 679 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 671 5 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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