-0.000 000 000 742 147 676 630 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 630 5(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
-0.000 000 000 742 147 676 630 5(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:

|-0.000 000 000 742 147 676 630 5| = 0.000 000 000 742 147 676 630 5


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: 0.000 000 000 742 147 676 630 5.

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) 0.000 000 000 742 147 676 630 5 × 2 = 0 + 0.000 000 001 484 295 353 261;
  • 2) 0.000 000 001 484 295 353 261 × 2 = 0 + 0.000 000 002 968 590 706 522;
  • 3) 0.000 000 002 968 590 706 522 × 2 = 0 + 0.000 000 005 937 181 413 044;
  • 4) 0.000 000 005 937 181 413 044 × 2 = 0 + 0.000 000 011 874 362 826 088;
  • 5) 0.000 000 011 874 362 826 088 × 2 = 0 + 0.000 000 023 748 725 652 176;
  • 6) 0.000 000 023 748 725 652 176 × 2 = 0 + 0.000 000 047 497 451 304 352;
  • 7) 0.000 000 047 497 451 304 352 × 2 = 0 + 0.000 000 094 994 902 608 704;
  • 8) 0.000 000 094 994 902 608 704 × 2 = 0 + 0.000 000 189 989 805 217 408;
  • 9) 0.000 000 189 989 805 217 408 × 2 = 0 + 0.000 000 379 979 610 434 816;
  • 10) 0.000 000 379 979 610 434 816 × 2 = 0 + 0.000 000 759 959 220 869 632;
  • 11) 0.000 000 759 959 220 869 632 × 2 = 0 + 0.000 001 519 918 441 739 264;
  • 12) 0.000 001 519 918 441 739 264 × 2 = 0 + 0.000 003 039 836 883 478 528;
  • 13) 0.000 003 039 836 883 478 528 × 2 = 0 + 0.000 006 079 673 766 957 056;
  • 14) 0.000 006 079 673 766 957 056 × 2 = 0 + 0.000 012 159 347 533 914 112;
  • 15) 0.000 012 159 347 533 914 112 × 2 = 0 + 0.000 024 318 695 067 828 224;
  • 16) 0.000 024 318 695 067 828 224 × 2 = 0 + 0.000 048 637 390 135 656 448;
  • 17) 0.000 048 637 390 135 656 448 × 2 = 0 + 0.000 097 274 780 271 312 896;
  • 18) 0.000 097 274 780 271 312 896 × 2 = 0 + 0.000 194 549 560 542 625 792;
  • 19) 0.000 194 549 560 542 625 792 × 2 = 0 + 0.000 389 099 121 085 251 584;
  • 20) 0.000 389 099 121 085 251 584 × 2 = 0 + 0.000 778 198 242 170 503 168;
  • 21) 0.000 778 198 242 170 503 168 × 2 = 0 + 0.001 556 396 484 341 006 336;
  • 22) 0.001 556 396 484 341 006 336 × 2 = 0 + 0.003 112 792 968 682 012 672;
  • 23) 0.003 112 792 968 682 012 672 × 2 = 0 + 0.006 225 585 937 364 025 344;
  • 24) 0.006 225 585 937 364 025 344 × 2 = 0 + 0.012 451 171 874 728 050 688;
  • 25) 0.012 451 171 874 728 050 688 × 2 = 0 + 0.024 902 343 749 456 101 376;
  • 26) 0.024 902 343 749 456 101 376 × 2 = 0 + 0.049 804 687 498 912 202 752;
  • 27) 0.049 804 687 498 912 202 752 × 2 = 0 + 0.099 609 374 997 824 405 504;
  • 28) 0.099 609 374 997 824 405 504 × 2 = 0 + 0.199 218 749 995 648 811 008;
  • 29) 0.199 218 749 995 648 811 008 × 2 = 0 + 0.398 437 499 991 297 622 016;
  • 30) 0.398 437 499 991 297 622 016 × 2 = 0 + 0.796 874 999 982 595 244 032;
  • 31) 0.796 874 999 982 595 244 032 × 2 = 1 + 0.593 749 999 965 190 488 064;
  • 32) 0.593 749 999 965 190 488 064 × 2 = 1 + 0.187 499 999 930 380 976 128;
  • 33) 0.187 499 999 930 380 976 128 × 2 = 0 + 0.374 999 999 860 761 952 256;
  • 34) 0.374 999 999 860 761 952 256 × 2 = 0 + 0.749 999 999 721 523 904 512;
  • 35) 0.749 999 999 721 523 904 512 × 2 = 1 + 0.499 999 999 443 047 809 024;
  • 36) 0.499 999 999 443 047 809 024 × 2 = 0 + 0.999 999 998 886 095 618 048;
  • 37) 0.999 999 998 886 095 618 048 × 2 = 1 + 0.999 999 997 772 191 236 096;
  • 38) 0.999 999 997 772 191 236 096 × 2 = 1 + 0.999 999 995 544 382 472 192;
  • 39) 0.999 999 995 544 382 472 192 × 2 = 1 + 0.999 999 991 088 764 944 384;
  • 40) 0.999 999 991 088 764 944 384 × 2 = 1 + 0.999 999 982 177 529 888 768;
  • 41) 0.999 999 982 177 529 888 768 × 2 = 1 + 0.999 999 964 355 059 777 536;
  • 42) 0.999 999 964 355 059 777 536 × 2 = 1 + 0.999 999 928 710 119 555 072;
  • 43) 0.999 999 928 710 119 555 072 × 2 = 1 + 0.999 999 857 420 239 110 144;
  • 44) 0.999 999 857 420 239 110 144 × 2 = 1 + 0.999 999 714 840 478 220 288;
  • 45) 0.999 999 714 840 478 220 288 × 2 = 1 + 0.999 999 429 680 956 440 576;
  • 46) 0.999 999 429 680 956 440 576 × 2 = 1 + 0.999 998 859 361 912 881 152;
  • 47) 0.999 998 859 361 912 881 152 × 2 = 1 + 0.999 997 718 723 825 762 304;
  • 48) 0.999 997 718 723 825 762 304 × 2 = 1 + 0.999 995 437 447 651 524 608;
  • 49) 0.999 995 437 447 651 524 608 × 2 = 1 + 0.999 990 874 895 303 049 216;
  • 50) 0.999 990 874 895 303 049 216 × 2 = 1 + 0.999 981 749 790 606 098 432;
  • 51) 0.999 981 749 790 606 098 432 × 2 = 1 + 0.999 963 499 581 212 196 864;
  • 52) 0.999 963 499 581 212 196 864 × 2 = 1 + 0.999 926 999 162 424 393 728;
  • 53) 0.999 926 999 162 424 393 728 × 2 = 1 + 0.999 853 998 324 848 787 456;
  • 54) 0.999 853 998 324 848 787 456 × 2 = 1 + 0.999 707 996 649 697 574 912;

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:


0.000 000 000 742 147 676 630 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 630 5(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:


0.000 000 000 742 147 676 630 5(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 -0.000 000 000 742 147 676 630 5 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