-0.000 000 000 742 147 683 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 683(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
-0.000 000 000 742 147 683(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:

|-0.000 000 000 742 147 683| = 0.000 000 000 742 147 683


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: 0.000 000 000 742 147 683.

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) 0.000 000 000 742 147 683 × 2 = 0 + 0.000 000 001 484 295 366;
  • 2) 0.000 000 001 484 295 366 × 2 = 0 + 0.000 000 002 968 590 732;
  • 3) 0.000 000 002 968 590 732 × 2 = 0 + 0.000 000 005 937 181 464;
  • 4) 0.000 000 005 937 181 464 × 2 = 0 + 0.000 000 011 874 362 928;
  • 5) 0.000 000 011 874 362 928 × 2 = 0 + 0.000 000 023 748 725 856;
  • 6) 0.000 000 023 748 725 856 × 2 = 0 + 0.000 000 047 497 451 712;
  • 7) 0.000 000 047 497 451 712 × 2 = 0 + 0.000 000 094 994 903 424;
  • 8) 0.000 000 094 994 903 424 × 2 = 0 + 0.000 000 189 989 806 848;
  • 9) 0.000 000 189 989 806 848 × 2 = 0 + 0.000 000 379 979 613 696;
  • 10) 0.000 000 379 979 613 696 × 2 = 0 + 0.000 000 759 959 227 392;
  • 11) 0.000 000 759 959 227 392 × 2 = 0 + 0.000 001 519 918 454 784;
  • 12) 0.000 001 519 918 454 784 × 2 = 0 + 0.000 003 039 836 909 568;
  • 13) 0.000 003 039 836 909 568 × 2 = 0 + 0.000 006 079 673 819 136;
  • 14) 0.000 006 079 673 819 136 × 2 = 0 + 0.000 012 159 347 638 272;
  • 15) 0.000 012 159 347 638 272 × 2 = 0 + 0.000 024 318 695 276 544;
  • 16) 0.000 024 318 695 276 544 × 2 = 0 + 0.000 048 637 390 553 088;
  • 17) 0.000 048 637 390 553 088 × 2 = 0 + 0.000 097 274 781 106 176;
  • 18) 0.000 097 274 781 106 176 × 2 = 0 + 0.000 194 549 562 212 352;
  • 19) 0.000 194 549 562 212 352 × 2 = 0 + 0.000 389 099 124 424 704;
  • 20) 0.000 389 099 124 424 704 × 2 = 0 + 0.000 778 198 248 849 408;
  • 21) 0.000 778 198 248 849 408 × 2 = 0 + 0.001 556 396 497 698 816;
  • 22) 0.001 556 396 497 698 816 × 2 = 0 + 0.003 112 792 995 397 632;
  • 23) 0.003 112 792 995 397 632 × 2 = 0 + 0.006 225 585 990 795 264;
  • 24) 0.006 225 585 990 795 264 × 2 = 0 + 0.012 451 171 981 590 528;
  • 25) 0.012 451 171 981 590 528 × 2 = 0 + 0.024 902 343 963 181 056;
  • 26) 0.024 902 343 963 181 056 × 2 = 0 + 0.049 804 687 926 362 112;
  • 27) 0.049 804 687 926 362 112 × 2 = 0 + 0.099 609 375 852 724 224;
  • 28) 0.099 609 375 852 724 224 × 2 = 0 + 0.199 218 751 705 448 448;
  • 29) 0.199 218 751 705 448 448 × 2 = 0 + 0.398 437 503 410 896 896;
  • 30) 0.398 437 503 410 896 896 × 2 = 0 + 0.796 875 006 821 793 792;
  • 31) 0.796 875 006 821 793 792 × 2 = 1 + 0.593 750 013 643 587 584;
  • 32) 0.593 750 013 643 587 584 × 2 = 1 + 0.187 500 027 287 175 168;
  • 33) 0.187 500 027 287 175 168 × 2 = 0 + 0.375 000 054 574 350 336;
  • 34) 0.375 000 054 574 350 336 × 2 = 0 + 0.750 000 109 148 700 672;
  • 35) 0.750 000 109 148 700 672 × 2 = 1 + 0.500 000 218 297 401 344;
  • 36) 0.500 000 218 297 401 344 × 2 = 1 + 0.000 000 436 594 802 688;
  • 37) 0.000 000 436 594 802 688 × 2 = 0 + 0.000 000 873 189 605 376;
  • 38) 0.000 000 873 189 605 376 × 2 = 0 + 0.000 001 746 379 210 752;
  • 39) 0.000 001 746 379 210 752 × 2 = 0 + 0.000 003 492 758 421 504;
  • 40) 0.000 003 492 758 421 504 × 2 = 0 + 0.000 006 985 516 843 008;
  • 41) 0.000 006 985 516 843 008 × 2 = 0 + 0.000 013 971 033 686 016;
  • 42) 0.000 013 971 033 686 016 × 2 = 0 + 0.000 027 942 067 372 032;
  • 43) 0.000 027 942 067 372 032 × 2 = 0 + 0.000 055 884 134 744 064;
  • 44) 0.000 055 884 134 744 064 × 2 = 0 + 0.000 111 768 269 488 128;
  • 45) 0.000 111 768 269 488 128 × 2 = 0 + 0.000 223 536 538 976 256;
  • 46) 0.000 223 536 538 976 256 × 2 = 0 + 0.000 447 073 077 952 512;
  • 47) 0.000 447 073 077 952 512 × 2 = 0 + 0.000 894 146 155 905 024;
  • 48) 0.000 894 146 155 905 024 × 2 = 0 + 0.001 788 292 311 810 048;
  • 49) 0.001 788 292 311 810 048 × 2 = 0 + 0.003 576 584 623 620 096;
  • 50) 0.003 576 584 623 620 096 × 2 = 0 + 0.007 153 169 247 240 192;
  • 51) 0.007 153 169 247 240 192 × 2 = 0 + 0.014 306 338 494 480 384;
  • 52) 0.014 306 338 494 480 384 × 2 = 0 + 0.028 612 676 988 960 768;
  • 53) 0.028 612 676 988 960 768 × 2 = 0 + 0.057 225 353 977 921 536;
  • 54) 0.057 225 353 977 921 536 × 2 = 0 + 0.114 450 707 955 843 072;

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:


0.000 000 000 742 147 683(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 683(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:


0.000 000 000 742 147 683(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 -0.000 000 000 742 147 683 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