-0.000 000 000 742 147 676 867 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 867(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
-0.000 000 000 742 147 676 867(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:

|-0.000 000 000 742 147 676 867| = 0.000 000 000 742 147 676 867


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: 0.000 000 000 742 147 676 867.

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) 0.000 000 000 742 147 676 867 × 2 = 0 + 0.000 000 001 484 295 353 734;
  • 2) 0.000 000 001 484 295 353 734 × 2 = 0 + 0.000 000 002 968 590 707 468;
  • 3) 0.000 000 002 968 590 707 468 × 2 = 0 + 0.000 000 005 937 181 414 936;
  • 4) 0.000 000 005 937 181 414 936 × 2 = 0 + 0.000 000 011 874 362 829 872;
  • 5) 0.000 000 011 874 362 829 872 × 2 = 0 + 0.000 000 023 748 725 659 744;
  • 6) 0.000 000 023 748 725 659 744 × 2 = 0 + 0.000 000 047 497 451 319 488;
  • 7) 0.000 000 047 497 451 319 488 × 2 = 0 + 0.000 000 094 994 902 638 976;
  • 8) 0.000 000 094 994 902 638 976 × 2 = 0 + 0.000 000 189 989 805 277 952;
  • 9) 0.000 000 189 989 805 277 952 × 2 = 0 + 0.000 000 379 979 610 555 904;
  • 10) 0.000 000 379 979 610 555 904 × 2 = 0 + 0.000 000 759 959 221 111 808;
  • 11) 0.000 000 759 959 221 111 808 × 2 = 0 + 0.000 001 519 918 442 223 616;
  • 12) 0.000 001 519 918 442 223 616 × 2 = 0 + 0.000 003 039 836 884 447 232;
  • 13) 0.000 003 039 836 884 447 232 × 2 = 0 + 0.000 006 079 673 768 894 464;
  • 14) 0.000 006 079 673 768 894 464 × 2 = 0 + 0.000 012 159 347 537 788 928;
  • 15) 0.000 012 159 347 537 788 928 × 2 = 0 + 0.000 024 318 695 075 577 856;
  • 16) 0.000 024 318 695 075 577 856 × 2 = 0 + 0.000 048 637 390 151 155 712;
  • 17) 0.000 048 637 390 151 155 712 × 2 = 0 + 0.000 097 274 780 302 311 424;
  • 18) 0.000 097 274 780 302 311 424 × 2 = 0 + 0.000 194 549 560 604 622 848;
  • 19) 0.000 194 549 560 604 622 848 × 2 = 0 + 0.000 389 099 121 209 245 696;
  • 20) 0.000 389 099 121 209 245 696 × 2 = 0 + 0.000 778 198 242 418 491 392;
  • 21) 0.000 778 198 242 418 491 392 × 2 = 0 + 0.001 556 396 484 836 982 784;
  • 22) 0.001 556 396 484 836 982 784 × 2 = 0 + 0.003 112 792 969 673 965 568;
  • 23) 0.003 112 792 969 673 965 568 × 2 = 0 + 0.006 225 585 939 347 931 136;
  • 24) 0.006 225 585 939 347 931 136 × 2 = 0 + 0.012 451 171 878 695 862 272;
  • 25) 0.012 451 171 878 695 862 272 × 2 = 0 + 0.024 902 343 757 391 724 544;
  • 26) 0.024 902 343 757 391 724 544 × 2 = 0 + 0.049 804 687 514 783 449 088;
  • 27) 0.049 804 687 514 783 449 088 × 2 = 0 + 0.099 609 375 029 566 898 176;
  • 28) 0.099 609 375 029 566 898 176 × 2 = 0 + 0.199 218 750 059 133 796 352;
  • 29) 0.199 218 750 059 133 796 352 × 2 = 0 + 0.398 437 500 118 267 592 704;
  • 30) 0.398 437 500 118 267 592 704 × 2 = 0 + 0.796 875 000 236 535 185 408;
  • 31) 0.796 875 000 236 535 185 408 × 2 = 1 + 0.593 750 000 473 070 370 816;
  • 32) 0.593 750 000 473 070 370 816 × 2 = 1 + 0.187 500 000 946 140 741 632;
  • 33) 0.187 500 000 946 140 741 632 × 2 = 0 + 0.375 000 001 892 281 483 264;
  • 34) 0.375 000 001 892 281 483 264 × 2 = 0 + 0.750 000 003 784 562 966 528;
  • 35) 0.750 000 003 784 562 966 528 × 2 = 1 + 0.500 000 007 569 125 933 056;
  • 36) 0.500 000 007 569 125 933 056 × 2 = 1 + 0.000 000 015 138 251 866 112;
  • 37) 0.000 000 015 138 251 866 112 × 2 = 0 + 0.000 000 030 276 503 732 224;
  • 38) 0.000 000 030 276 503 732 224 × 2 = 0 + 0.000 000 060 553 007 464 448;
  • 39) 0.000 000 060 553 007 464 448 × 2 = 0 + 0.000 000 121 106 014 928 896;
  • 40) 0.000 000 121 106 014 928 896 × 2 = 0 + 0.000 000 242 212 029 857 792;
  • 41) 0.000 000 242 212 029 857 792 × 2 = 0 + 0.000 000 484 424 059 715 584;
  • 42) 0.000 000 484 424 059 715 584 × 2 = 0 + 0.000 000 968 848 119 431 168;
  • 43) 0.000 000 968 848 119 431 168 × 2 = 0 + 0.000 001 937 696 238 862 336;
  • 44) 0.000 001 937 696 238 862 336 × 2 = 0 + 0.000 003 875 392 477 724 672;
  • 45) 0.000 003 875 392 477 724 672 × 2 = 0 + 0.000 007 750 784 955 449 344;
  • 46) 0.000 007 750 784 955 449 344 × 2 = 0 + 0.000 015 501 569 910 898 688;
  • 47) 0.000 015 501 569 910 898 688 × 2 = 0 + 0.000 031 003 139 821 797 376;
  • 48) 0.000 031 003 139 821 797 376 × 2 = 0 + 0.000 062 006 279 643 594 752;
  • 49) 0.000 062 006 279 643 594 752 × 2 = 0 + 0.000 124 012 559 287 189 504;
  • 50) 0.000 124 012 559 287 189 504 × 2 = 0 + 0.000 248 025 118 574 379 008;
  • 51) 0.000 248 025 118 574 379 008 × 2 = 0 + 0.000 496 050 237 148 758 016;
  • 52) 0.000 496 050 237 148 758 016 × 2 = 0 + 0.000 992 100 474 297 516 032;
  • 53) 0.000 992 100 474 297 516 032 × 2 = 0 + 0.001 984 200 948 595 032 064;
  • 54) 0.001 984 200 948 595 032 064 × 2 = 0 + 0.003 968 401 897 190 064 128;

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:


0.000 000 000 742 147 676 867(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 867(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:


0.000 000 000 742 147 676 867(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 -0.000 000 000 742 147 676 867 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