-0.000 000 000 742 147 676 646 705 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 705 2(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
-0.000 000 000 742 147 676 646 705 2(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:

|-0.000 000 000 742 147 676 646 705 2| = 0.000 000 000 742 147 676 646 705 2


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: 0.000 000 000 742 147 676 646 705 2.

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) 0.000 000 000 742 147 676 646 705 2 × 2 = 0 + 0.000 000 001 484 295 353 293 410 4;
  • 2) 0.000 000 001 484 295 353 293 410 4 × 2 = 0 + 0.000 000 002 968 590 706 586 820 8;
  • 3) 0.000 000 002 968 590 706 586 820 8 × 2 = 0 + 0.000 000 005 937 181 413 173 641 6;
  • 4) 0.000 000 005 937 181 413 173 641 6 × 2 = 0 + 0.000 000 011 874 362 826 347 283 2;
  • 5) 0.000 000 011 874 362 826 347 283 2 × 2 = 0 + 0.000 000 023 748 725 652 694 566 4;
  • 6) 0.000 000 023 748 725 652 694 566 4 × 2 = 0 + 0.000 000 047 497 451 305 389 132 8;
  • 7) 0.000 000 047 497 451 305 389 132 8 × 2 = 0 + 0.000 000 094 994 902 610 778 265 6;
  • 8) 0.000 000 094 994 902 610 778 265 6 × 2 = 0 + 0.000 000 189 989 805 221 556 531 2;
  • 9) 0.000 000 189 989 805 221 556 531 2 × 2 = 0 + 0.000 000 379 979 610 443 113 062 4;
  • 10) 0.000 000 379 979 610 443 113 062 4 × 2 = 0 + 0.000 000 759 959 220 886 226 124 8;
  • 11) 0.000 000 759 959 220 886 226 124 8 × 2 = 0 + 0.000 001 519 918 441 772 452 249 6;
  • 12) 0.000 001 519 918 441 772 452 249 6 × 2 = 0 + 0.000 003 039 836 883 544 904 499 2;
  • 13) 0.000 003 039 836 883 544 904 499 2 × 2 = 0 + 0.000 006 079 673 767 089 808 998 4;
  • 14) 0.000 006 079 673 767 089 808 998 4 × 2 = 0 + 0.000 012 159 347 534 179 617 996 8;
  • 15) 0.000 012 159 347 534 179 617 996 8 × 2 = 0 + 0.000 024 318 695 068 359 235 993 6;
  • 16) 0.000 024 318 695 068 359 235 993 6 × 2 = 0 + 0.000 048 637 390 136 718 471 987 2;
  • 17) 0.000 048 637 390 136 718 471 987 2 × 2 = 0 + 0.000 097 274 780 273 436 943 974 4;
  • 18) 0.000 097 274 780 273 436 943 974 4 × 2 = 0 + 0.000 194 549 560 546 873 887 948 8;
  • 19) 0.000 194 549 560 546 873 887 948 8 × 2 = 0 + 0.000 389 099 121 093 747 775 897 6;
  • 20) 0.000 389 099 121 093 747 775 897 6 × 2 = 0 + 0.000 778 198 242 187 495 551 795 2;
  • 21) 0.000 778 198 242 187 495 551 795 2 × 2 = 0 + 0.001 556 396 484 374 991 103 590 4;
  • 22) 0.001 556 396 484 374 991 103 590 4 × 2 = 0 + 0.003 112 792 968 749 982 207 180 8;
  • 23) 0.003 112 792 968 749 982 207 180 8 × 2 = 0 + 0.006 225 585 937 499 964 414 361 6;
  • 24) 0.006 225 585 937 499 964 414 361 6 × 2 = 0 + 0.012 451 171 874 999 928 828 723 2;
  • 25) 0.012 451 171 874 999 928 828 723 2 × 2 = 0 + 0.024 902 343 749 999 857 657 446 4;
  • 26) 0.024 902 343 749 999 857 657 446 4 × 2 = 0 + 0.049 804 687 499 999 715 314 892 8;
  • 27) 0.049 804 687 499 999 715 314 892 8 × 2 = 0 + 0.099 609 374 999 999 430 629 785 6;
  • 28) 0.099 609 374 999 999 430 629 785 6 × 2 = 0 + 0.199 218 749 999 998 861 259 571 2;
  • 29) 0.199 218 749 999 998 861 259 571 2 × 2 = 0 + 0.398 437 499 999 997 722 519 142 4;
  • 30) 0.398 437 499 999 997 722 519 142 4 × 2 = 0 + 0.796 874 999 999 995 445 038 284 8;
  • 31) 0.796 874 999 999 995 445 038 284 8 × 2 = 1 + 0.593 749 999 999 990 890 076 569 6;
  • 32) 0.593 749 999 999 990 890 076 569 6 × 2 = 1 + 0.187 499 999 999 981 780 153 139 2;
  • 33) 0.187 499 999 999 981 780 153 139 2 × 2 = 0 + 0.374 999 999 999 963 560 306 278 4;
  • 34) 0.374 999 999 999 963 560 306 278 4 × 2 = 0 + 0.749 999 999 999 927 120 612 556 8;
  • 35) 0.749 999 999 999 927 120 612 556 8 × 2 = 1 + 0.499 999 999 999 854 241 225 113 6;
  • 36) 0.499 999 999 999 854 241 225 113 6 × 2 = 0 + 0.999 999 999 999 708 482 450 227 2;
  • 37) 0.999 999 999 999 708 482 450 227 2 × 2 = 1 + 0.999 999 999 999 416 964 900 454 4;
  • 38) 0.999 999 999 999 416 964 900 454 4 × 2 = 1 + 0.999 999 999 998 833 929 800 908 8;
  • 39) 0.999 999 999 998 833 929 800 908 8 × 2 = 1 + 0.999 999 999 997 667 859 601 817 6;
  • 40) 0.999 999 999 997 667 859 601 817 6 × 2 = 1 + 0.999 999 999 995 335 719 203 635 2;
  • 41) 0.999 999 999 995 335 719 203 635 2 × 2 = 1 + 0.999 999 999 990 671 438 407 270 4;
  • 42) 0.999 999 999 990 671 438 407 270 4 × 2 = 1 + 0.999 999 999 981 342 876 814 540 8;
  • 43) 0.999 999 999 981 342 876 814 540 8 × 2 = 1 + 0.999 999 999 962 685 753 629 081 6;
  • 44) 0.999 999 999 962 685 753 629 081 6 × 2 = 1 + 0.999 999 999 925 371 507 258 163 2;
  • 45) 0.999 999 999 925 371 507 258 163 2 × 2 = 1 + 0.999 999 999 850 743 014 516 326 4;
  • 46) 0.999 999 999 850 743 014 516 326 4 × 2 = 1 + 0.999 999 999 701 486 029 032 652 8;
  • 47) 0.999 999 999 701 486 029 032 652 8 × 2 = 1 + 0.999 999 999 402 972 058 065 305 6;
  • 48) 0.999 999 999 402 972 058 065 305 6 × 2 = 1 + 0.999 999 998 805 944 116 130 611 2;
  • 49) 0.999 999 998 805 944 116 130 611 2 × 2 = 1 + 0.999 999 997 611 888 232 261 222 4;
  • 50) 0.999 999 997 611 888 232 261 222 4 × 2 = 1 + 0.999 999 995 223 776 464 522 444 8;
  • 51) 0.999 999 995 223 776 464 522 444 8 × 2 = 1 + 0.999 999 990 447 552 929 044 889 6;
  • 52) 0.999 999 990 447 552 929 044 889 6 × 2 = 1 + 0.999 999 980 895 105 858 089 779 2;
  • 53) 0.999 999 980 895 105 858 089 779 2 × 2 = 1 + 0.999 999 961 790 211 716 179 558 4;
  • 54) 0.999 999 961 790 211 716 179 558 4 × 2 = 1 + 0.999 999 923 580 423 432 359 116 8;

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:


0.000 000 000 742 147 676 646 705 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 705 2(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:


0.000 000 000 742 147 676 646 705 2(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 -0.000 000 000 742 147 676 646 705 2 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