-0.000 000 000 742 147 676 239 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 239(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
-0.000 000 000 742 147 676 239(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:

|-0.000 000 000 742 147 676 239| = 0.000 000 000 742 147 676 239


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: 0.000 000 000 742 147 676 239.

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) 0.000 000 000 742 147 676 239 × 2 = 0 + 0.000 000 001 484 295 352 478;
  • 2) 0.000 000 001 484 295 352 478 × 2 = 0 + 0.000 000 002 968 590 704 956;
  • 3) 0.000 000 002 968 590 704 956 × 2 = 0 + 0.000 000 005 937 181 409 912;
  • 4) 0.000 000 005 937 181 409 912 × 2 = 0 + 0.000 000 011 874 362 819 824;
  • 5) 0.000 000 011 874 362 819 824 × 2 = 0 + 0.000 000 023 748 725 639 648;
  • 6) 0.000 000 023 748 725 639 648 × 2 = 0 + 0.000 000 047 497 451 279 296;
  • 7) 0.000 000 047 497 451 279 296 × 2 = 0 + 0.000 000 094 994 902 558 592;
  • 8) 0.000 000 094 994 902 558 592 × 2 = 0 + 0.000 000 189 989 805 117 184;
  • 9) 0.000 000 189 989 805 117 184 × 2 = 0 + 0.000 000 379 979 610 234 368;
  • 10) 0.000 000 379 979 610 234 368 × 2 = 0 + 0.000 000 759 959 220 468 736;
  • 11) 0.000 000 759 959 220 468 736 × 2 = 0 + 0.000 001 519 918 440 937 472;
  • 12) 0.000 001 519 918 440 937 472 × 2 = 0 + 0.000 003 039 836 881 874 944;
  • 13) 0.000 003 039 836 881 874 944 × 2 = 0 + 0.000 006 079 673 763 749 888;
  • 14) 0.000 006 079 673 763 749 888 × 2 = 0 + 0.000 012 159 347 527 499 776;
  • 15) 0.000 012 159 347 527 499 776 × 2 = 0 + 0.000 024 318 695 054 999 552;
  • 16) 0.000 024 318 695 054 999 552 × 2 = 0 + 0.000 048 637 390 109 999 104;
  • 17) 0.000 048 637 390 109 999 104 × 2 = 0 + 0.000 097 274 780 219 998 208;
  • 18) 0.000 097 274 780 219 998 208 × 2 = 0 + 0.000 194 549 560 439 996 416;
  • 19) 0.000 194 549 560 439 996 416 × 2 = 0 + 0.000 389 099 120 879 992 832;
  • 20) 0.000 389 099 120 879 992 832 × 2 = 0 + 0.000 778 198 241 759 985 664;
  • 21) 0.000 778 198 241 759 985 664 × 2 = 0 + 0.001 556 396 483 519 971 328;
  • 22) 0.001 556 396 483 519 971 328 × 2 = 0 + 0.003 112 792 967 039 942 656;
  • 23) 0.003 112 792 967 039 942 656 × 2 = 0 + 0.006 225 585 934 079 885 312;
  • 24) 0.006 225 585 934 079 885 312 × 2 = 0 + 0.012 451 171 868 159 770 624;
  • 25) 0.012 451 171 868 159 770 624 × 2 = 0 + 0.024 902 343 736 319 541 248;
  • 26) 0.024 902 343 736 319 541 248 × 2 = 0 + 0.049 804 687 472 639 082 496;
  • 27) 0.049 804 687 472 639 082 496 × 2 = 0 + 0.099 609 374 945 278 164 992;
  • 28) 0.099 609 374 945 278 164 992 × 2 = 0 + 0.199 218 749 890 556 329 984;
  • 29) 0.199 218 749 890 556 329 984 × 2 = 0 + 0.398 437 499 781 112 659 968;
  • 30) 0.398 437 499 781 112 659 968 × 2 = 0 + 0.796 874 999 562 225 319 936;
  • 31) 0.796 874 999 562 225 319 936 × 2 = 1 + 0.593 749 999 124 450 639 872;
  • 32) 0.593 749 999 124 450 639 872 × 2 = 1 + 0.187 499 998 248 901 279 744;
  • 33) 0.187 499 998 248 901 279 744 × 2 = 0 + 0.374 999 996 497 802 559 488;
  • 34) 0.374 999 996 497 802 559 488 × 2 = 0 + 0.749 999 992 995 605 118 976;
  • 35) 0.749 999 992 995 605 118 976 × 2 = 1 + 0.499 999 985 991 210 237 952;
  • 36) 0.499 999 985 991 210 237 952 × 2 = 0 + 0.999 999 971 982 420 475 904;
  • 37) 0.999 999 971 982 420 475 904 × 2 = 1 + 0.999 999 943 964 840 951 808;
  • 38) 0.999 999 943 964 840 951 808 × 2 = 1 + 0.999 999 887 929 681 903 616;
  • 39) 0.999 999 887 929 681 903 616 × 2 = 1 + 0.999 999 775 859 363 807 232;
  • 40) 0.999 999 775 859 363 807 232 × 2 = 1 + 0.999 999 551 718 727 614 464;
  • 41) 0.999 999 551 718 727 614 464 × 2 = 1 + 0.999 999 103 437 455 228 928;
  • 42) 0.999 999 103 437 455 228 928 × 2 = 1 + 0.999 998 206 874 910 457 856;
  • 43) 0.999 998 206 874 910 457 856 × 2 = 1 + 0.999 996 413 749 820 915 712;
  • 44) 0.999 996 413 749 820 915 712 × 2 = 1 + 0.999 992 827 499 641 831 424;
  • 45) 0.999 992 827 499 641 831 424 × 2 = 1 + 0.999 985 654 999 283 662 848;
  • 46) 0.999 985 654 999 283 662 848 × 2 = 1 + 0.999 971 309 998 567 325 696;
  • 47) 0.999 971 309 998 567 325 696 × 2 = 1 + 0.999 942 619 997 134 651 392;
  • 48) 0.999 942 619 997 134 651 392 × 2 = 1 + 0.999 885 239 994 269 302 784;
  • 49) 0.999 885 239 994 269 302 784 × 2 = 1 + 0.999 770 479 988 538 605 568;
  • 50) 0.999 770 479 988 538 605 568 × 2 = 1 + 0.999 540 959 977 077 211 136;
  • 51) 0.999 540 959 977 077 211 136 × 2 = 1 + 0.999 081 919 954 154 422 272;
  • 52) 0.999 081 919 954 154 422 272 × 2 = 1 + 0.998 163 839 908 308 844 544;
  • 53) 0.998 163 839 908 308 844 544 × 2 = 1 + 0.996 327 679 816 617 689 088;
  • 54) 0.996 327 679 816 617 689 088 × 2 = 1 + 0.992 655 359 633 235 378 176;

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:


0.000 000 000 742 147 676 239(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 239(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:


0.000 000 000 742 147 676 239(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 -0.000 000 000 742 147 676 239 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