-0.000 000 000 742 147 676 101 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 101(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
-0.000 000 000 742 147 676 101(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:

|-0.000 000 000 742 147 676 101| = 0.000 000 000 742 147 676 101


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: 0.000 000 000 742 147 676 101.

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) 0.000 000 000 742 147 676 101 × 2 = 0 + 0.000 000 001 484 295 352 202;
  • 2) 0.000 000 001 484 295 352 202 × 2 = 0 + 0.000 000 002 968 590 704 404;
  • 3) 0.000 000 002 968 590 704 404 × 2 = 0 + 0.000 000 005 937 181 408 808;
  • 4) 0.000 000 005 937 181 408 808 × 2 = 0 + 0.000 000 011 874 362 817 616;
  • 5) 0.000 000 011 874 362 817 616 × 2 = 0 + 0.000 000 023 748 725 635 232;
  • 6) 0.000 000 023 748 725 635 232 × 2 = 0 + 0.000 000 047 497 451 270 464;
  • 7) 0.000 000 047 497 451 270 464 × 2 = 0 + 0.000 000 094 994 902 540 928;
  • 8) 0.000 000 094 994 902 540 928 × 2 = 0 + 0.000 000 189 989 805 081 856;
  • 9) 0.000 000 189 989 805 081 856 × 2 = 0 + 0.000 000 379 979 610 163 712;
  • 10) 0.000 000 379 979 610 163 712 × 2 = 0 + 0.000 000 759 959 220 327 424;
  • 11) 0.000 000 759 959 220 327 424 × 2 = 0 + 0.000 001 519 918 440 654 848;
  • 12) 0.000 001 519 918 440 654 848 × 2 = 0 + 0.000 003 039 836 881 309 696;
  • 13) 0.000 003 039 836 881 309 696 × 2 = 0 + 0.000 006 079 673 762 619 392;
  • 14) 0.000 006 079 673 762 619 392 × 2 = 0 + 0.000 012 159 347 525 238 784;
  • 15) 0.000 012 159 347 525 238 784 × 2 = 0 + 0.000 024 318 695 050 477 568;
  • 16) 0.000 024 318 695 050 477 568 × 2 = 0 + 0.000 048 637 390 100 955 136;
  • 17) 0.000 048 637 390 100 955 136 × 2 = 0 + 0.000 097 274 780 201 910 272;
  • 18) 0.000 097 274 780 201 910 272 × 2 = 0 + 0.000 194 549 560 403 820 544;
  • 19) 0.000 194 549 560 403 820 544 × 2 = 0 + 0.000 389 099 120 807 641 088;
  • 20) 0.000 389 099 120 807 641 088 × 2 = 0 + 0.000 778 198 241 615 282 176;
  • 21) 0.000 778 198 241 615 282 176 × 2 = 0 + 0.001 556 396 483 230 564 352;
  • 22) 0.001 556 396 483 230 564 352 × 2 = 0 + 0.003 112 792 966 461 128 704;
  • 23) 0.003 112 792 966 461 128 704 × 2 = 0 + 0.006 225 585 932 922 257 408;
  • 24) 0.006 225 585 932 922 257 408 × 2 = 0 + 0.012 451 171 865 844 514 816;
  • 25) 0.012 451 171 865 844 514 816 × 2 = 0 + 0.024 902 343 731 689 029 632;
  • 26) 0.024 902 343 731 689 029 632 × 2 = 0 + 0.049 804 687 463 378 059 264;
  • 27) 0.049 804 687 463 378 059 264 × 2 = 0 + 0.099 609 374 926 756 118 528;
  • 28) 0.099 609 374 926 756 118 528 × 2 = 0 + 0.199 218 749 853 512 237 056;
  • 29) 0.199 218 749 853 512 237 056 × 2 = 0 + 0.398 437 499 707 024 474 112;
  • 30) 0.398 437 499 707 024 474 112 × 2 = 0 + 0.796 874 999 414 048 948 224;
  • 31) 0.796 874 999 414 048 948 224 × 2 = 1 + 0.593 749 998 828 097 896 448;
  • 32) 0.593 749 998 828 097 896 448 × 2 = 1 + 0.187 499 997 656 195 792 896;
  • 33) 0.187 499 997 656 195 792 896 × 2 = 0 + 0.374 999 995 312 391 585 792;
  • 34) 0.374 999 995 312 391 585 792 × 2 = 0 + 0.749 999 990 624 783 171 584;
  • 35) 0.749 999 990 624 783 171 584 × 2 = 1 + 0.499 999 981 249 566 343 168;
  • 36) 0.499 999 981 249 566 343 168 × 2 = 0 + 0.999 999 962 499 132 686 336;
  • 37) 0.999 999 962 499 132 686 336 × 2 = 1 + 0.999 999 924 998 265 372 672;
  • 38) 0.999 999 924 998 265 372 672 × 2 = 1 + 0.999 999 849 996 530 745 344;
  • 39) 0.999 999 849 996 530 745 344 × 2 = 1 + 0.999 999 699 993 061 490 688;
  • 40) 0.999 999 699 993 061 490 688 × 2 = 1 + 0.999 999 399 986 122 981 376;
  • 41) 0.999 999 399 986 122 981 376 × 2 = 1 + 0.999 998 799 972 245 962 752;
  • 42) 0.999 998 799 972 245 962 752 × 2 = 1 + 0.999 997 599 944 491 925 504;
  • 43) 0.999 997 599 944 491 925 504 × 2 = 1 + 0.999 995 199 888 983 851 008;
  • 44) 0.999 995 199 888 983 851 008 × 2 = 1 + 0.999 990 399 777 967 702 016;
  • 45) 0.999 990 399 777 967 702 016 × 2 = 1 + 0.999 980 799 555 935 404 032;
  • 46) 0.999 980 799 555 935 404 032 × 2 = 1 + 0.999 961 599 111 870 808 064;
  • 47) 0.999 961 599 111 870 808 064 × 2 = 1 + 0.999 923 198 223 741 616 128;
  • 48) 0.999 923 198 223 741 616 128 × 2 = 1 + 0.999 846 396 447 483 232 256;
  • 49) 0.999 846 396 447 483 232 256 × 2 = 1 + 0.999 692 792 894 966 464 512;
  • 50) 0.999 692 792 894 966 464 512 × 2 = 1 + 0.999 385 585 789 932 929 024;
  • 51) 0.999 385 585 789 932 929 024 × 2 = 1 + 0.998 771 171 579 865 858 048;
  • 52) 0.998 771 171 579 865 858 048 × 2 = 1 + 0.997 542 343 159 731 716 096;
  • 53) 0.997 542 343 159 731 716 096 × 2 = 1 + 0.995 084 686 319 463 432 192;
  • 54) 0.995 084 686 319 463 432 192 × 2 = 1 + 0.990 169 372 638 926 864 384;

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:


0.000 000 000 742 147 676 101(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 101(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:


0.000 000 000 742 147 676 101(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 -0.000 000 000 742 147 676 101 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