-0.000 000 000 742 147 676 249 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 249(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
-0.000 000 000 742 147 676 249(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:

|-0.000 000 000 742 147 676 249| = 0.000 000 000 742 147 676 249


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: 0.000 000 000 742 147 676 249.

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) 0.000 000 000 742 147 676 249 × 2 = 0 + 0.000 000 001 484 295 352 498;
  • 2) 0.000 000 001 484 295 352 498 × 2 = 0 + 0.000 000 002 968 590 704 996;
  • 3) 0.000 000 002 968 590 704 996 × 2 = 0 + 0.000 000 005 937 181 409 992;
  • 4) 0.000 000 005 937 181 409 992 × 2 = 0 + 0.000 000 011 874 362 819 984;
  • 5) 0.000 000 011 874 362 819 984 × 2 = 0 + 0.000 000 023 748 725 639 968;
  • 6) 0.000 000 023 748 725 639 968 × 2 = 0 + 0.000 000 047 497 451 279 936;
  • 7) 0.000 000 047 497 451 279 936 × 2 = 0 + 0.000 000 094 994 902 559 872;
  • 8) 0.000 000 094 994 902 559 872 × 2 = 0 + 0.000 000 189 989 805 119 744;
  • 9) 0.000 000 189 989 805 119 744 × 2 = 0 + 0.000 000 379 979 610 239 488;
  • 10) 0.000 000 379 979 610 239 488 × 2 = 0 + 0.000 000 759 959 220 478 976;
  • 11) 0.000 000 759 959 220 478 976 × 2 = 0 + 0.000 001 519 918 440 957 952;
  • 12) 0.000 001 519 918 440 957 952 × 2 = 0 + 0.000 003 039 836 881 915 904;
  • 13) 0.000 003 039 836 881 915 904 × 2 = 0 + 0.000 006 079 673 763 831 808;
  • 14) 0.000 006 079 673 763 831 808 × 2 = 0 + 0.000 012 159 347 527 663 616;
  • 15) 0.000 012 159 347 527 663 616 × 2 = 0 + 0.000 024 318 695 055 327 232;
  • 16) 0.000 024 318 695 055 327 232 × 2 = 0 + 0.000 048 637 390 110 654 464;
  • 17) 0.000 048 637 390 110 654 464 × 2 = 0 + 0.000 097 274 780 221 308 928;
  • 18) 0.000 097 274 780 221 308 928 × 2 = 0 + 0.000 194 549 560 442 617 856;
  • 19) 0.000 194 549 560 442 617 856 × 2 = 0 + 0.000 389 099 120 885 235 712;
  • 20) 0.000 389 099 120 885 235 712 × 2 = 0 + 0.000 778 198 241 770 471 424;
  • 21) 0.000 778 198 241 770 471 424 × 2 = 0 + 0.001 556 396 483 540 942 848;
  • 22) 0.001 556 396 483 540 942 848 × 2 = 0 + 0.003 112 792 967 081 885 696;
  • 23) 0.003 112 792 967 081 885 696 × 2 = 0 + 0.006 225 585 934 163 771 392;
  • 24) 0.006 225 585 934 163 771 392 × 2 = 0 + 0.012 451 171 868 327 542 784;
  • 25) 0.012 451 171 868 327 542 784 × 2 = 0 + 0.024 902 343 736 655 085 568;
  • 26) 0.024 902 343 736 655 085 568 × 2 = 0 + 0.049 804 687 473 310 171 136;
  • 27) 0.049 804 687 473 310 171 136 × 2 = 0 + 0.099 609 374 946 620 342 272;
  • 28) 0.099 609 374 946 620 342 272 × 2 = 0 + 0.199 218 749 893 240 684 544;
  • 29) 0.199 218 749 893 240 684 544 × 2 = 0 + 0.398 437 499 786 481 369 088;
  • 30) 0.398 437 499 786 481 369 088 × 2 = 0 + 0.796 874 999 572 962 738 176;
  • 31) 0.796 874 999 572 962 738 176 × 2 = 1 + 0.593 749 999 145 925 476 352;
  • 32) 0.593 749 999 145 925 476 352 × 2 = 1 + 0.187 499 998 291 850 952 704;
  • 33) 0.187 499 998 291 850 952 704 × 2 = 0 + 0.374 999 996 583 701 905 408;
  • 34) 0.374 999 996 583 701 905 408 × 2 = 0 + 0.749 999 993 167 403 810 816;
  • 35) 0.749 999 993 167 403 810 816 × 2 = 1 + 0.499 999 986 334 807 621 632;
  • 36) 0.499 999 986 334 807 621 632 × 2 = 0 + 0.999 999 972 669 615 243 264;
  • 37) 0.999 999 972 669 615 243 264 × 2 = 1 + 0.999 999 945 339 230 486 528;
  • 38) 0.999 999 945 339 230 486 528 × 2 = 1 + 0.999 999 890 678 460 973 056;
  • 39) 0.999 999 890 678 460 973 056 × 2 = 1 + 0.999 999 781 356 921 946 112;
  • 40) 0.999 999 781 356 921 946 112 × 2 = 1 + 0.999 999 562 713 843 892 224;
  • 41) 0.999 999 562 713 843 892 224 × 2 = 1 + 0.999 999 125 427 687 784 448;
  • 42) 0.999 999 125 427 687 784 448 × 2 = 1 + 0.999 998 250 855 375 568 896;
  • 43) 0.999 998 250 855 375 568 896 × 2 = 1 + 0.999 996 501 710 751 137 792;
  • 44) 0.999 996 501 710 751 137 792 × 2 = 1 + 0.999 993 003 421 502 275 584;
  • 45) 0.999 993 003 421 502 275 584 × 2 = 1 + 0.999 986 006 843 004 551 168;
  • 46) 0.999 986 006 843 004 551 168 × 2 = 1 + 0.999 972 013 686 009 102 336;
  • 47) 0.999 972 013 686 009 102 336 × 2 = 1 + 0.999 944 027 372 018 204 672;
  • 48) 0.999 944 027 372 018 204 672 × 2 = 1 + 0.999 888 054 744 036 409 344;
  • 49) 0.999 888 054 744 036 409 344 × 2 = 1 + 0.999 776 109 488 072 818 688;
  • 50) 0.999 776 109 488 072 818 688 × 2 = 1 + 0.999 552 218 976 145 637 376;
  • 51) 0.999 552 218 976 145 637 376 × 2 = 1 + 0.999 104 437 952 291 274 752;
  • 52) 0.999 104 437 952 291 274 752 × 2 = 1 + 0.998 208 875 904 582 549 504;
  • 53) 0.998 208 875 904 582 549 504 × 2 = 1 + 0.996 417 751 809 165 099 008;
  • 54) 0.996 417 751 809 165 099 008 × 2 = 1 + 0.992 835 503 618 330 198 016;

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:


0.000 000 000 742 147 676 249(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 249(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:


0.000 000 000 742 147 676 249(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 -0.000 000 000 742 147 676 249 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