-0.000 000 000 742 147 676 144 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 144(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
-0.000 000 000 742 147 676 144(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:

|-0.000 000 000 742 147 676 144| = 0.000 000 000 742 147 676 144


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: 0.000 000 000 742 147 676 144.

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) 0.000 000 000 742 147 676 144 × 2 = 0 + 0.000 000 001 484 295 352 288;
  • 2) 0.000 000 001 484 295 352 288 × 2 = 0 + 0.000 000 002 968 590 704 576;
  • 3) 0.000 000 002 968 590 704 576 × 2 = 0 + 0.000 000 005 937 181 409 152;
  • 4) 0.000 000 005 937 181 409 152 × 2 = 0 + 0.000 000 011 874 362 818 304;
  • 5) 0.000 000 011 874 362 818 304 × 2 = 0 + 0.000 000 023 748 725 636 608;
  • 6) 0.000 000 023 748 725 636 608 × 2 = 0 + 0.000 000 047 497 451 273 216;
  • 7) 0.000 000 047 497 451 273 216 × 2 = 0 + 0.000 000 094 994 902 546 432;
  • 8) 0.000 000 094 994 902 546 432 × 2 = 0 + 0.000 000 189 989 805 092 864;
  • 9) 0.000 000 189 989 805 092 864 × 2 = 0 + 0.000 000 379 979 610 185 728;
  • 10) 0.000 000 379 979 610 185 728 × 2 = 0 + 0.000 000 759 959 220 371 456;
  • 11) 0.000 000 759 959 220 371 456 × 2 = 0 + 0.000 001 519 918 440 742 912;
  • 12) 0.000 001 519 918 440 742 912 × 2 = 0 + 0.000 003 039 836 881 485 824;
  • 13) 0.000 003 039 836 881 485 824 × 2 = 0 + 0.000 006 079 673 762 971 648;
  • 14) 0.000 006 079 673 762 971 648 × 2 = 0 + 0.000 012 159 347 525 943 296;
  • 15) 0.000 012 159 347 525 943 296 × 2 = 0 + 0.000 024 318 695 051 886 592;
  • 16) 0.000 024 318 695 051 886 592 × 2 = 0 + 0.000 048 637 390 103 773 184;
  • 17) 0.000 048 637 390 103 773 184 × 2 = 0 + 0.000 097 274 780 207 546 368;
  • 18) 0.000 097 274 780 207 546 368 × 2 = 0 + 0.000 194 549 560 415 092 736;
  • 19) 0.000 194 549 560 415 092 736 × 2 = 0 + 0.000 389 099 120 830 185 472;
  • 20) 0.000 389 099 120 830 185 472 × 2 = 0 + 0.000 778 198 241 660 370 944;
  • 21) 0.000 778 198 241 660 370 944 × 2 = 0 + 0.001 556 396 483 320 741 888;
  • 22) 0.001 556 396 483 320 741 888 × 2 = 0 + 0.003 112 792 966 641 483 776;
  • 23) 0.003 112 792 966 641 483 776 × 2 = 0 + 0.006 225 585 933 282 967 552;
  • 24) 0.006 225 585 933 282 967 552 × 2 = 0 + 0.012 451 171 866 565 935 104;
  • 25) 0.012 451 171 866 565 935 104 × 2 = 0 + 0.024 902 343 733 131 870 208;
  • 26) 0.024 902 343 733 131 870 208 × 2 = 0 + 0.049 804 687 466 263 740 416;
  • 27) 0.049 804 687 466 263 740 416 × 2 = 0 + 0.099 609 374 932 527 480 832;
  • 28) 0.099 609 374 932 527 480 832 × 2 = 0 + 0.199 218 749 865 054 961 664;
  • 29) 0.199 218 749 865 054 961 664 × 2 = 0 + 0.398 437 499 730 109 923 328;
  • 30) 0.398 437 499 730 109 923 328 × 2 = 0 + 0.796 874 999 460 219 846 656;
  • 31) 0.796 874 999 460 219 846 656 × 2 = 1 + 0.593 749 998 920 439 693 312;
  • 32) 0.593 749 998 920 439 693 312 × 2 = 1 + 0.187 499 997 840 879 386 624;
  • 33) 0.187 499 997 840 879 386 624 × 2 = 0 + 0.374 999 995 681 758 773 248;
  • 34) 0.374 999 995 681 758 773 248 × 2 = 0 + 0.749 999 991 363 517 546 496;
  • 35) 0.749 999 991 363 517 546 496 × 2 = 1 + 0.499 999 982 727 035 092 992;
  • 36) 0.499 999 982 727 035 092 992 × 2 = 0 + 0.999 999 965 454 070 185 984;
  • 37) 0.999 999 965 454 070 185 984 × 2 = 1 + 0.999 999 930 908 140 371 968;
  • 38) 0.999 999 930 908 140 371 968 × 2 = 1 + 0.999 999 861 816 280 743 936;
  • 39) 0.999 999 861 816 280 743 936 × 2 = 1 + 0.999 999 723 632 561 487 872;
  • 40) 0.999 999 723 632 561 487 872 × 2 = 1 + 0.999 999 447 265 122 975 744;
  • 41) 0.999 999 447 265 122 975 744 × 2 = 1 + 0.999 998 894 530 245 951 488;
  • 42) 0.999 998 894 530 245 951 488 × 2 = 1 + 0.999 997 789 060 491 902 976;
  • 43) 0.999 997 789 060 491 902 976 × 2 = 1 + 0.999 995 578 120 983 805 952;
  • 44) 0.999 995 578 120 983 805 952 × 2 = 1 + 0.999 991 156 241 967 611 904;
  • 45) 0.999 991 156 241 967 611 904 × 2 = 1 + 0.999 982 312 483 935 223 808;
  • 46) 0.999 982 312 483 935 223 808 × 2 = 1 + 0.999 964 624 967 870 447 616;
  • 47) 0.999 964 624 967 870 447 616 × 2 = 1 + 0.999 929 249 935 740 895 232;
  • 48) 0.999 929 249 935 740 895 232 × 2 = 1 + 0.999 858 499 871 481 790 464;
  • 49) 0.999 858 499 871 481 790 464 × 2 = 1 + 0.999 716 999 742 963 580 928;
  • 50) 0.999 716 999 742 963 580 928 × 2 = 1 + 0.999 433 999 485 927 161 856;
  • 51) 0.999 433 999 485 927 161 856 × 2 = 1 + 0.998 867 998 971 854 323 712;
  • 52) 0.998 867 998 971 854 323 712 × 2 = 1 + 0.997 735 997 943 708 647 424;
  • 53) 0.997 735 997 943 708 647 424 × 2 = 1 + 0.995 471 995 887 417 294 848;
  • 54) 0.995 471 995 887 417 294 848 × 2 = 1 + 0.990 943 991 774 834 589 696;

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:


0.000 000 000 742 147 676 144(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 144(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:


0.000 000 000 742 147 676 144(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 -0.000 000 000 742 147 676 144 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