-0.000 000 000 742 147 676 832 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 832(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
-0.000 000 000 742 147 676 832(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:

|-0.000 000 000 742 147 676 832| = 0.000 000 000 742 147 676 832


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: 0.000 000 000 742 147 676 832.

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) 0.000 000 000 742 147 676 832 × 2 = 0 + 0.000 000 001 484 295 353 664;
  • 2) 0.000 000 001 484 295 353 664 × 2 = 0 + 0.000 000 002 968 590 707 328;
  • 3) 0.000 000 002 968 590 707 328 × 2 = 0 + 0.000 000 005 937 181 414 656;
  • 4) 0.000 000 005 937 181 414 656 × 2 = 0 + 0.000 000 011 874 362 829 312;
  • 5) 0.000 000 011 874 362 829 312 × 2 = 0 + 0.000 000 023 748 725 658 624;
  • 6) 0.000 000 023 748 725 658 624 × 2 = 0 + 0.000 000 047 497 451 317 248;
  • 7) 0.000 000 047 497 451 317 248 × 2 = 0 + 0.000 000 094 994 902 634 496;
  • 8) 0.000 000 094 994 902 634 496 × 2 = 0 + 0.000 000 189 989 805 268 992;
  • 9) 0.000 000 189 989 805 268 992 × 2 = 0 + 0.000 000 379 979 610 537 984;
  • 10) 0.000 000 379 979 610 537 984 × 2 = 0 + 0.000 000 759 959 221 075 968;
  • 11) 0.000 000 759 959 221 075 968 × 2 = 0 + 0.000 001 519 918 442 151 936;
  • 12) 0.000 001 519 918 442 151 936 × 2 = 0 + 0.000 003 039 836 884 303 872;
  • 13) 0.000 003 039 836 884 303 872 × 2 = 0 + 0.000 006 079 673 768 607 744;
  • 14) 0.000 006 079 673 768 607 744 × 2 = 0 + 0.000 012 159 347 537 215 488;
  • 15) 0.000 012 159 347 537 215 488 × 2 = 0 + 0.000 024 318 695 074 430 976;
  • 16) 0.000 024 318 695 074 430 976 × 2 = 0 + 0.000 048 637 390 148 861 952;
  • 17) 0.000 048 637 390 148 861 952 × 2 = 0 + 0.000 097 274 780 297 723 904;
  • 18) 0.000 097 274 780 297 723 904 × 2 = 0 + 0.000 194 549 560 595 447 808;
  • 19) 0.000 194 549 560 595 447 808 × 2 = 0 + 0.000 389 099 121 190 895 616;
  • 20) 0.000 389 099 121 190 895 616 × 2 = 0 + 0.000 778 198 242 381 791 232;
  • 21) 0.000 778 198 242 381 791 232 × 2 = 0 + 0.001 556 396 484 763 582 464;
  • 22) 0.001 556 396 484 763 582 464 × 2 = 0 + 0.003 112 792 969 527 164 928;
  • 23) 0.003 112 792 969 527 164 928 × 2 = 0 + 0.006 225 585 939 054 329 856;
  • 24) 0.006 225 585 939 054 329 856 × 2 = 0 + 0.012 451 171 878 108 659 712;
  • 25) 0.012 451 171 878 108 659 712 × 2 = 0 + 0.024 902 343 756 217 319 424;
  • 26) 0.024 902 343 756 217 319 424 × 2 = 0 + 0.049 804 687 512 434 638 848;
  • 27) 0.049 804 687 512 434 638 848 × 2 = 0 + 0.099 609 375 024 869 277 696;
  • 28) 0.099 609 375 024 869 277 696 × 2 = 0 + 0.199 218 750 049 738 555 392;
  • 29) 0.199 218 750 049 738 555 392 × 2 = 0 + 0.398 437 500 099 477 110 784;
  • 30) 0.398 437 500 099 477 110 784 × 2 = 0 + 0.796 875 000 198 954 221 568;
  • 31) 0.796 875 000 198 954 221 568 × 2 = 1 + 0.593 750 000 397 908 443 136;
  • 32) 0.593 750 000 397 908 443 136 × 2 = 1 + 0.187 500 000 795 816 886 272;
  • 33) 0.187 500 000 795 816 886 272 × 2 = 0 + 0.375 000 001 591 633 772 544;
  • 34) 0.375 000 001 591 633 772 544 × 2 = 0 + 0.750 000 003 183 267 545 088;
  • 35) 0.750 000 003 183 267 545 088 × 2 = 1 + 0.500 000 006 366 535 090 176;
  • 36) 0.500 000 006 366 535 090 176 × 2 = 1 + 0.000 000 012 733 070 180 352;
  • 37) 0.000 000 012 733 070 180 352 × 2 = 0 + 0.000 000 025 466 140 360 704;
  • 38) 0.000 000 025 466 140 360 704 × 2 = 0 + 0.000 000 050 932 280 721 408;
  • 39) 0.000 000 050 932 280 721 408 × 2 = 0 + 0.000 000 101 864 561 442 816;
  • 40) 0.000 000 101 864 561 442 816 × 2 = 0 + 0.000 000 203 729 122 885 632;
  • 41) 0.000 000 203 729 122 885 632 × 2 = 0 + 0.000 000 407 458 245 771 264;
  • 42) 0.000 000 407 458 245 771 264 × 2 = 0 + 0.000 000 814 916 491 542 528;
  • 43) 0.000 000 814 916 491 542 528 × 2 = 0 + 0.000 001 629 832 983 085 056;
  • 44) 0.000 001 629 832 983 085 056 × 2 = 0 + 0.000 003 259 665 966 170 112;
  • 45) 0.000 003 259 665 966 170 112 × 2 = 0 + 0.000 006 519 331 932 340 224;
  • 46) 0.000 006 519 331 932 340 224 × 2 = 0 + 0.000 013 038 663 864 680 448;
  • 47) 0.000 013 038 663 864 680 448 × 2 = 0 + 0.000 026 077 327 729 360 896;
  • 48) 0.000 026 077 327 729 360 896 × 2 = 0 + 0.000 052 154 655 458 721 792;
  • 49) 0.000 052 154 655 458 721 792 × 2 = 0 + 0.000 104 309 310 917 443 584;
  • 50) 0.000 104 309 310 917 443 584 × 2 = 0 + 0.000 208 618 621 834 887 168;
  • 51) 0.000 208 618 621 834 887 168 × 2 = 0 + 0.000 417 237 243 669 774 336;
  • 52) 0.000 417 237 243 669 774 336 × 2 = 0 + 0.000 834 474 487 339 548 672;
  • 53) 0.000 834 474 487 339 548 672 × 2 = 0 + 0.001 668 948 974 679 097 344;
  • 54) 0.001 668 948 974 679 097 344 × 2 = 0 + 0.003 337 897 949 358 194 688;

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:


0.000 000 000 742 147 676 832(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 832(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 832(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 832 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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