-0.000 000 000 742 147 676 646 701 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 701 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
-0.000 000 000 742 147 676 646 701 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:

|-0.000 000 000 742 147 676 646 701 3| = 0.000 000 000 742 147 676 646 701 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: 0.000 000 000 742 147 676 646 701 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) 0.000 000 000 742 147 676 646 701 3 × 2 = 0 + 0.000 000 001 484 295 353 293 402 6;
  • 2) 0.000 000 001 484 295 353 293 402 6 × 2 = 0 + 0.000 000 002 968 590 706 586 805 2;
  • 3) 0.000 000 002 968 590 706 586 805 2 × 2 = 0 + 0.000 000 005 937 181 413 173 610 4;
  • 4) 0.000 000 005 937 181 413 173 610 4 × 2 = 0 + 0.000 000 011 874 362 826 347 220 8;
  • 5) 0.000 000 011 874 362 826 347 220 8 × 2 = 0 + 0.000 000 023 748 725 652 694 441 6;
  • 6) 0.000 000 023 748 725 652 694 441 6 × 2 = 0 + 0.000 000 047 497 451 305 388 883 2;
  • 7) 0.000 000 047 497 451 305 388 883 2 × 2 = 0 + 0.000 000 094 994 902 610 777 766 4;
  • 8) 0.000 000 094 994 902 610 777 766 4 × 2 = 0 + 0.000 000 189 989 805 221 555 532 8;
  • 9) 0.000 000 189 989 805 221 555 532 8 × 2 = 0 + 0.000 000 379 979 610 443 111 065 6;
  • 10) 0.000 000 379 979 610 443 111 065 6 × 2 = 0 + 0.000 000 759 959 220 886 222 131 2;
  • 11) 0.000 000 759 959 220 886 222 131 2 × 2 = 0 + 0.000 001 519 918 441 772 444 262 4;
  • 12) 0.000 001 519 918 441 772 444 262 4 × 2 = 0 + 0.000 003 039 836 883 544 888 524 8;
  • 13) 0.000 003 039 836 883 544 888 524 8 × 2 = 0 + 0.000 006 079 673 767 089 777 049 6;
  • 14) 0.000 006 079 673 767 089 777 049 6 × 2 = 0 + 0.000 012 159 347 534 179 554 099 2;
  • 15) 0.000 012 159 347 534 179 554 099 2 × 2 = 0 + 0.000 024 318 695 068 359 108 198 4;
  • 16) 0.000 024 318 695 068 359 108 198 4 × 2 = 0 + 0.000 048 637 390 136 718 216 396 8;
  • 17) 0.000 048 637 390 136 718 216 396 8 × 2 = 0 + 0.000 097 274 780 273 436 432 793 6;
  • 18) 0.000 097 274 780 273 436 432 793 6 × 2 = 0 + 0.000 194 549 560 546 872 865 587 2;
  • 19) 0.000 194 549 560 546 872 865 587 2 × 2 = 0 + 0.000 389 099 121 093 745 731 174 4;
  • 20) 0.000 389 099 121 093 745 731 174 4 × 2 = 0 + 0.000 778 198 242 187 491 462 348 8;
  • 21) 0.000 778 198 242 187 491 462 348 8 × 2 = 0 + 0.001 556 396 484 374 982 924 697 6;
  • 22) 0.001 556 396 484 374 982 924 697 6 × 2 = 0 + 0.003 112 792 968 749 965 849 395 2;
  • 23) 0.003 112 792 968 749 965 849 395 2 × 2 = 0 + 0.006 225 585 937 499 931 698 790 4;
  • 24) 0.006 225 585 937 499 931 698 790 4 × 2 = 0 + 0.012 451 171 874 999 863 397 580 8;
  • 25) 0.012 451 171 874 999 863 397 580 8 × 2 = 0 + 0.024 902 343 749 999 726 795 161 6;
  • 26) 0.024 902 343 749 999 726 795 161 6 × 2 = 0 + 0.049 804 687 499 999 453 590 323 2;
  • 27) 0.049 804 687 499 999 453 590 323 2 × 2 = 0 + 0.099 609 374 999 998 907 180 646 4;
  • 28) 0.099 609 374 999 998 907 180 646 4 × 2 = 0 + 0.199 218 749 999 997 814 361 292 8;
  • 29) 0.199 218 749 999 997 814 361 292 8 × 2 = 0 + 0.398 437 499 999 995 628 722 585 6;
  • 30) 0.398 437 499 999 995 628 722 585 6 × 2 = 0 + 0.796 874 999 999 991 257 445 171 2;
  • 31) 0.796 874 999 999 991 257 445 171 2 × 2 = 1 + 0.593 749 999 999 982 514 890 342 4;
  • 32) 0.593 749 999 999 982 514 890 342 4 × 2 = 1 + 0.187 499 999 999 965 029 780 684 8;
  • 33) 0.187 499 999 999 965 029 780 684 8 × 2 = 0 + 0.374 999 999 999 930 059 561 369 6;
  • 34) 0.374 999 999 999 930 059 561 369 6 × 2 = 0 + 0.749 999 999 999 860 119 122 739 2;
  • 35) 0.749 999 999 999 860 119 122 739 2 × 2 = 1 + 0.499 999 999 999 720 238 245 478 4;
  • 36) 0.499 999 999 999 720 238 245 478 4 × 2 = 0 + 0.999 999 999 999 440 476 490 956 8;
  • 37) 0.999 999 999 999 440 476 490 956 8 × 2 = 1 + 0.999 999 999 998 880 952 981 913 6;
  • 38) 0.999 999 999 998 880 952 981 913 6 × 2 = 1 + 0.999 999 999 997 761 905 963 827 2;
  • 39) 0.999 999 999 997 761 905 963 827 2 × 2 = 1 + 0.999 999 999 995 523 811 927 654 4;
  • 40) 0.999 999 999 995 523 811 927 654 4 × 2 = 1 + 0.999 999 999 991 047 623 855 308 8;
  • 41) 0.999 999 999 991 047 623 855 308 8 × 2 = 1 + 0.999 999 999 982 095 247 710 617 6;
  • 42) 0.999 999 999 982 095 247 710 617 6 × 2 = 1 + 0.999 999 999 964 190 495 421 235 2;
  • 43) 0.999 999 999 964 190 495 421 235 2 × 2 = 1 + 0.999 999 999 928 380 990 842 470 4;
  • 44) 0.999 999 999 928 380 990 842 470 4 × 2 = 1 + 0.999 999 999 856 761 981 684 940 8;
  • 45) 0.999 999 999 856 761 981 684 940 8 × 2 = 1 + 0.999 999 999 713 523 963 369 881 6;
  • 46) 0.999 999 999 713 523 963 369 881 6 × 2 = 1 + 0.999 999 999 427 047 926 739 763 2;
  • 47) 0.999 999 999 427 047 926 739 763 2 × 2 = 1 + 0.999 999 998 854 095 853 479 526 4;
  • 48) 0.999 999 998 854 095 853 479 526 4 × 2 = 1 + 0.999 999 997 708 191 706 959 052 8;
  • 49) 0.999 999 997 708 191 706 959 052 8 × 2 = 1 + 0.999 999 995 416 383 413 918 105 6;
  • 50) 0.999 999 995 416 383 413 918 105 6 × 2 = 1 + 0.999 999 990 832 766 827 836 211 2;
  • 51) 0.999 999 990 832 766 827 836 211 2 × 2 = 1 + 0.999 999 981 665 533 655 672 422 4;
  • 52) 0.999 999 981 665 533 655 672 422 4 × 2 = 1 + 0.999 999 963 331 067 311 344 844 8;
  • 53) 0.999 999 963 331 067 311 344 844 8 × 2 = 1 + 0.999 999 926 662 134 622 689 689 6;
  • 54) 0.999 999 926 662 134 622 689 689 6 × 2 = 1 + 0.999 999 853 324 269 245 379 379 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:


0.000 000 000 742 147 676 646 701 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 701 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:


0.000 000 000 742 147 676 646 701 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 -0.000 000 000 742 147 676 646 701 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