-0.000 000 000 742 147 676 646 686 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 686 3(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
-0.000 000 000 742 147 676 646 686 3(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:

|-0.000 000 000 742 147 676 646 686 3| = 0.000 000 000 742 147 676 646 686 3


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: 0.000 000 000 742 147 676 646 686 3.

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) 0.000 000 000 742 147 676 646 686 3 × 2 = 0 + 0.000 000 001 484 295 353 293 372 6;
  • 2) 0.000 000 001 484 295 353 293 372 6 × 2 = 0 + 0.000 000 002 968 590 706 586 745 2;
  • 3) 0.000 000 002 968 590 706 586 745 2 × 2 = 0 + 0.000 000 005 937 181 413 173 490 4;
  • 4) 0.000 000 005 937 181 413 173 490 4 × 2 = 0 + 0.000 000 011 874 362 826 346 980 8;
  • 5) 0.000 000 011 874 362 826 346 980 8 × 2 = 0 + 0.000 000 023 748 725 652 693 961 6;
  • 6) 0.000 000 023 748 725 652 693 961 6 × 2 = 0 + 0.000 000 047 497 451 305 387 923 2;
  • 7) 0.000 000 047 497 451 305 387 923 2 × 2 = 0 + 0.000 000 094 994 902 610 775 846 4;
  • 8) 0.000 000 094 994 902 610 775 846 4 × 2 = 0 + 0.000 000 189 989 805 221 551 692 8;
  • 9) 0.000 000 189 989 805 221 551 692 8 × 2 = 0 + 0.000 000 379 979 610 443 103 385 6;
  • 10) 0.000 000 379 979 610 443 103 385 6 × 2 = 0 + 0.000 000 759 959 220 886 206 771 2;
  • 11) 0.000 000 759 959 220 886 206 771 2 × 2 = 0 + 0.000 001 519 918 441 772 413 542 4;
  • 12) 0.000 001 519 918 441 772 413 542 4 × 2 = 0 + 0.000 003 039 836 883 544 827 084 8;
  • 13) 0.000 003 039 836 883 544 827 084 8 × 2 = 0 + 0.000 006 079 673 767 089 654 169 6;
  • 14) 0.000 006 079 673 767 089 654 169 6 × 2 = 0 + 0.000 012 159 347 534 179 308 339 2;
  • 15) 0.000 012 159 347 534 179 308 339 2 × 2 = 0 + 0.000 024 318 695 068 358 616 678 4;
  • 16) 0.000 024 318 695 068 358 616 678 4 × 2 = 0 + 0.000 048 637 390 136 717 233 356 8;
  • 17) 0.000 048 637 390 136 717 233 356 8 × 2 = 0 + 0.000 097 274 780 273 434 466 713 6;
  • 18) 0.000 097 274 780 273 434 466 713 6 × 2 = 0 + 0.000 194 549 560 546 868 933 427 2;
  • 19) 0.000 194 549 560 546 868 933 427 2 × 2 = 0 + 0.000 389 099 121 093 737 866 854 4;
  • 20) 0.000 389 099 121 093 737 866 854 4 × 2 = 0 + 0.000 778 198 242 187 475 733 708 8;
  • 21) 0.000 778 198 242 187 475 733 708 8 × 2 = 0 + 0.001 556 396 484 374 951 467 417 6;
  • 22) 0.001 556 396 484 374 951 467 417 6 × 2 = 0 + 0.003 112 792 968 749 902 934 835 2;
  • 23) 0.003 112 792 968 749 902 934 835 2 × 2 = 0 + 0.006 225 585 937 499 805 869 670 4;
  • 24) 0.006 225 585 937 499 805 869 670 4 × 2 = 0 + 0.012 451 171 874 999 611 739 340 8;
  • 25) 0.012 451 171 874 999 611 739 340 8 × 2 = 0 + 0.024 902 343 749 999 223 478 681 6;
  • 26) 0.024 902 343 749 999 223 478 681 6 × 2 = 0 + 0.049 804 687 499 998 446 957 363 2;
  • 27) 0.049 804 687 499 998 446 957 363 2 × 2 = 0 + 0.099 609 374 999 996 893 914 726 4;
  • 28) 0.099 609 374 999 996 893 914 726 4 × 2 = 0 + 0.199 218 749 999 993 787 829 452 8;
  • 29) 0.199 218 749 999 993 787 829 452 8 × 2 = 0 + 0.398 437 499 999 987 575 658 905 6;
  • 30) 0.398 437 499 999 987 575 658 905 6 × 2 = 0 + 0.796 874 999 999 975 151 317 811 2;
  • 31) 0.796 874 999 999 975 151 317 811 2 × 2 = 1 + 0.593 749 999 999 950 302 635 622 4;
  • 32) 0.593 749 999 999 950 302 635 622 4 × 2 = 1 + 0.187 499 999 999 900 605 271 244 8;
  • 33) 0.187 499 999 999 900 605 271 244 8 × 2 = 0 + 0.374 999 999 999 801 210 542 489 6;
  • 34) 0.374 999 999 999 801 210 542 489 6 × 2 = 0 + 0.749 999 999 999 602 421 084 979 2;
  • 35) 0.749 999 999 999 602 421 084 979 2 × 2 = 1 + 0.499 999 999 999 204 842 169 958 4;
  • 36) 0.499 999 999 999 204 842 169 958 4 × 2 = 0 + 0.999 999 999 998 409 684 339 916 8;
  • 37) 0.999 999 999 998 409 684 339 916 8 × 2 = 1 + 0.999 999 999 996 819 368 679 833 6;
  • 38) 0.999 999 999 996 819 368 679 833 6 × 2 = 1 + 0.999 999 999 993 638 737 359 667 2;
  • 39) 0.999 999 999 993 638 737 359 667 2 × 2 = 1 + 0.999 999 999 987 277 474 719 334 4;
  • 40) 0.999 999 999 987 277 474 719 334 4 × 2 = 1 + 0.999 999 999 974 554 949 438 668 8;
  • 41) 0.999 999 999 974 554 949 438 668 8 × 2 = 1 + 0.999 999 999 949 109 898 877 337 6;
  • 42) 0.999 999 999 949 109 898 877 337 6 × 2 = 1 + 0.999 999 999 898 219 797 754 675 2;
  • 43) 0.999 999 999 898 219 797 754 675 2 × 2 = 1 + 0.999 999 999 796 439 595 509 350 4;
  • 44) 0.999 999 999 796 439 595 509 350 4 × 2 = 1 + 0.999 999 999 592 879 191 018 700 8;
  • 45) 0.999 999 999 592 879 191 018 700 8 × 2 = 1 + 0.999 999 999 185 758 382 037 401 6;
  • 46) 0.999 999 999 185 758 382 037 401 6 × 2 = 1 + 0.999 999 998 371 516 764 074 803 2;
  • 47) 0.999 999 998 371 516 764 074 803 2 × 2 = 1 + 0.999 999 996 743 033 528 149 606 4;
  • 48) 0.999 999 996 743 033 528 149 606 4 × 2 = 1 + 0.999 999 993 486 067 056 299 212 8;
  • 49) 0.999 999 993 486 067 056 299 212 8 × 2 = 1 + 0.999 999 986 972 134 112 598 425 6;
  • 50) 0.999 999 986 972 134 112 598 425 6 × 2 = 1 + 0.999 999 973 944 268 225 196 851 2;
  • 51) 0.999 999 973 944 268 225 196 851 2 × 2 = 1 + 0.999 999 947 888 536 450 393 702 4;
  • 52) 0.999 999 947 888 536 450 393 702 4 × 2 = 1 + 0.999 999 895 777 072 900 787 404 8;
  • 53) 0.999 999 895 777 072 900 787 404 8 × 2 = 1 + 0.999 999 791 554 145 801 574 809 6;
  • 54) 0.999 999 791 554 145 801 574 809 6 × 2 = 1 + 0.999 999 583 108 291 603 149 619 2;

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:


0.000 000 000 742 147 676 646 686 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 686 3(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:


0.000 000 000 742 147 676 646 686 3(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 -0.000 000 000 742 147 676 646 686 3 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