-0.000 000 000 742 147 676 341 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 341(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
-0.000 000 000 742 147 676 341(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:

|-0.000 000 000 742 147 676 341| = 0.000 000 000 742 147 676 341


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: 0.000 000 000 742 147 676 341.

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) 0.000 000 000 742 147 676 341 × 2 = 0 + 0.000 000 001 484 295 352 682;
  • 2) 0.000 000 001 484 295 352 682 × 2 = 0 + 0.000 000 002 968 590 705 364;
  • 3) 0.000 000 002 968 590 705 364 × 2 = 0 + 0.000 000 005 937 181 410 728;
  • 4) 0.000 000 005 937 181 410 728 × 2 = 0 + 0.000 000 011 874 362 821 456;
  • 5) 0.000 000 011 874 362 821 456 × 2 = 0 + 0.000 000 023 748 725 642 912;
  • 6) 0.000 000 023 748 725 642 912 × 2 = 0 + 0.000 000 047 497 451 285 824;
  • 7) 0.000 000 047 497 451 285 824 × 2 = 0 + 0.000 000 094 994 902 571 648;
  • 8) 0.000 000 094 994 902 571 648 × 2 = 0 + 0.000 000 189 989 805 143 296;
  • 9) 0.000 000 189 989 805 143 296 × 2 = 0 + 0.000 000 379 979 610 286 592;
  • 10) 0.000 000 379 979 610 286 592 × 2 = 0 + 0.000 000 759 959 220 573 184;
  • 11) 0.000 000 759 959 220 573 184 × 2 = 0 + 0.000 001 519 918 441 146 368;
  • 12) 0.000 001 519 918 441 146 368 × 2 = 0 + 0.000 003 039 836 882 292 736;
  • 13) 0.000 003 039 836 882 292 736 × 2 = 0 + 0.000 006 079 673 764 585 472;
  • 14) 0.000 006 079 673 764 585 472 × 2 = 0 + 0.000 012 159 347 529 170 944;
  • 15) 0.000 012 159 347 529 170 944 × 2 = 0 + 0.000 024 318 695 058 341 888;
  • 16) 0.000 024 318 695 058 341 888 × 2 = 0 + 0.000 048 637 390 116 683 776;
  • 17) 0.000 048 637 390 116 683 776 × 2 = 0 + 0.000 097 274 780 233 367 552;
  • 18) 0.000 097 274 780 233 367 552 × 2 = 0 + 0.000 194 549 560 466 735 104;
  • 19) 0.000 194 549 560 466 735 104 × 2 = 0 + 0.000 389 099 120 933 470 208;
  • 20) 0.000 389 099 120 933 470 208 × 2 = 0 + 0.000 778 198 241 866 940 416;
  • 21) 0.000 778 198 241 866 940 416 × 2 = 0 + 0.001 556 396 483 733 880 832;
  • 22) 0.001 556 396 483 733 880 832 × 2 = 0 + 0.003 112 792 967 467 761 664;
  • 23) 0.003 112 792 967 467 761 664 × 2 = 0 + 0.006 225 585 934 935 523 328;
  • 24) 0.006 225 585 934 935 523 328 × 2 = 0 + 0.012 451 171 869 871 046 656;
  • 25) 0.012 451 171 869 871 046 656 × 2 = 0 + 0.024 902 343 739 742 093 312;
  • 26) 0.024 902 343 739 742 093 312 × 2 = 0 + 0.049 804 687 479 484 186 624;
  • 27) 0.049 804 687 479 484 186 624 × 2 = 0 + 0.099 609 374 958 968 373 248;
  • 28) 0.099 609 374 958 968 373 248 × 2 = 0 + 0.199 218 749 917 936 746 496;
  • 29) 0.199 218 749 917 936 746 496 × 2 = 0 + 0.398 437 499 835 873 492 992;
  • 30) 0.398 437 499 835 873 492 992 × 2 = 0 + 0.796 874 999 671 746 985 984;
  • 31) 0.796 874 999 671 746 985 984 × 2 = 1 + 0.593 749 999 343 493 971 968;
  • 32) 0.593 749 999 343 493 971 968 × 2 = 1 + 0.187 499 998 686 987 943 936;
  • 33) 0.187 499 998 686 987 943 936 × 2 = 0 + 0.374 999 997 373 975 887 872;
  • 34) 0.374 999 997 373 975 887 872 × 2 = 0 + 0.749 999 994 747 951 775 744;
  • 35) 0.749 999 994 747 951 775 744 × 2 = 1 + 0.499 999 989 495 903 551 488;
  • 36) 0.499 999 989 495 903 551 488 × 2 = 0 + 0.999 999 978 991 807 102 976;
  • 37) 0.999 999 978 991 807 102 976 × 2 = 1 + 0.999 999 957 983 614 205 952;
  • 38) 0.999 999 957 983 614 205 952 × 2 = 1 + 0.999 999 915 967 228 411 904;
  • 39) 0.999 999 915 967 228 411 904 × 2 = 1 + 0.999 999 831 934 456 823 808;
  • 40) 0.999 999 831 934 456 823 808 × 2 = 1 + 0.999 999 663 868 913 647 616;
  • 41) 0.999 999 663 868 913 647 616 × 2 = 1 + 0.999 999 327 737 827 295 232;
  • 42) 0.999 999 327 737 827 295 232 × 2 = 1 + 0.999 998 655 475 654 590 464;
  • 43) 0.999 998 655 475 654 590 464 × 2 = 1 + 0.999 997 310 951 309 180 928;
  • 44) 0.999 997 310 951 309 180 928 × 2 = 1 + 0.999 994 621 902 618 361 856;
  • 45) 0.999 994 621 902 618 361 856 × 2 = 1 + 0.999 989 243 805 236 723 712;
  • 46) 0.999 989 243 805 236 723 712 × 2 = 1 + 0.999 978 487 610 473 447 424;
  • 47) 0.999 978 487 610 473 447 424 × 2 = 1 + 0.999 956 975 220 946 894 848;
  • 48) 0.999 956 975 220 946 894 848 × 2 = 1 + 0.999 913 950 441 893 789 696;
  • 49) 0.999 913 950 441 893 789 696 × 2 = 1 + 0.999 827 900 883 787 579 392;
  • 50) 0.999 827 900 883 787 579 392 × 2 = 1 + 0.999 655 801 767 575 158 784;
  • 51) 0.999 655 801 767 575 158 784 × 2 = 1 + 0.999 311 603 535 150 317 568;
  • 52) 0.999 311 603 535 150 317 568 × 2 = 1 + 0.998 623 207 070 300 635 136;
  • 53) 0.998 623 207 070 300 635 136 × 2 = 1 + 0.997 246 414 140 601 270 272;
  • 54) 0.997 246 414 140 601 270 272 × 2 = 1 + 0.994 492 828 281 202 540 544;

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:


0.000 000 000 742 147 676 341(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 341(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:


0.000 000 000 742 147 676 341(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 -0.000 000 000 742 147 676 341 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