-0.000 000 000 742 147 676 648 86 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 86(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
-0.000 000 000 742 147 676 648 86(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:

|-0.000 000 000 742 147 676 648 86| = 0.000 000 000 742 147 676 648 86


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: 0.000 000 000 742 147 676 648 86.

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) 0.000 000 000 742 147 676 648 86 × 2 = 0 + 0.000 000 001 484 295 353 297 72;
  • 2) 0.000 000 001 484 295 353 297 72 × 2 = 0 + 0.000 000 002 968 590 706 595 44;
  • 3) 0.000 000 002 968 590 706 595 44 × 2 = 0 + 0.000 000 005 937 181 413 190 88;
  • 4) 0.000 000 005 937 181 413 190 88 × 2 = 0 + 0.000 000 011 874 362 826 381 76;
  • 5) 0.000 000 011 874 362 826 381 76 × 2 = 0 + 0.000 000 023 748 725 652 763 52;
  • 6) 0.000 000 023 748 725 652 763 52 × 2 = 0 + 0.000 000 047 497 451 305 527 04;
  • 7) 0.000 000 047 497 451 305 527 04 × 2 = 0 + 0.000 000 094 994 902 611 054 08;
  • 8) 0.000 000 094 994 902 611 054 08 × 2 = 0 + 0.000 000 189 989 805 222 108 16;
  • 9) 0.000 000 189 989 805 222 108 16 × 2 = 0 + 0.000 000 379 979 610 444 216 32;
  • 10) 0.000 000 379 979 610 444 216 32 × 2 = 0 + 0.000 000 759 959 220 888 432 64;
  • 11) 0.000 000 759 959 220 888 432 64 × 2 = 0 + 0.000 001 519 918 441 776 865 28;
  • 12) 0.000 001 519 918 441 776 865 28 × 2 = 0 + 0.000 003 039 836 883 553 730 56;
  • 13) 0.000 003 039 836 883 553 730 56 × 2 = 0 + 0.000 006 079 673 767 107 461 12;
  • 14) 0.000 006 079 673 767 107 461 12 × 2 = 0 + 0.000 012 159 347 534 214 922 24;
  • 15) 0.000 012 159 347 534 214 922 24 × 2 = 0 + 0.000 024 318 695 068 429 844 48;
  • 16) 0.000 024 318 695 068 429 844 48 × 2 = 0 + 0.000 048 637 390 136 859 688 96;
  • 17) 0.000 048 637 390 136 859 688 96 × 2 = 0 + 0.000 097 274 780 273 719 377 92;
  • 18) 0.000 097 274 780 273 719 377 92 × 2 = 0 + 0.000 194 549 560 547 438 755 84;
  • 19) 0.000 194 549 560 547 438 755 84 × 2 = 0 + 0.000 389 099 121 094 877 511 68;
  • 20) 0.000 389 099 121 094 877 511 68 × 2 = 0 + 0.000 778 198 242 189 755 023 36;
  • 21) 0.000 778 198 242 189 755 023 36 × 2 = 0 + 0.001 556 396 484 379 510 046 72;
  • 22) 0.001 556 396 484 379 510 046 72 × 2 = 0 + 0.003 112 792 968 759 020 093 44;
  • 23) 0.003 112 792 968 759 020 093 44 × 2 = 0 + 0.006 225 585 937 518 040 186 88;
  • 24) 0.006 225 585 937 518 040 186 88 × 2 = 0 + 0.012 451 171 875 036 080 373 76;
  • 25) 0.012 451 171 875 036 080 373 76 × 2 = 0 + 0.024 902 343 750 072 160 747 52;
  • 26) 0.024 902 343 750 072 160 747 52 × 2 = 0 + 0.049 804 687 500 144 321 495 04;
  • 27) 0.049 804 687 500 144 321 495 04 × 2 = 0 + 0.099 609 375 000 288 642 990 08;
  • 28) 0.099 609 375 000 288 642 990 08 × 2 = 0 + 0.199 218 750 000 577 285 980 16;
  • 29) 0.199 218 750 000 577 285 980 16 × 2 = 0 + 0.398 437 500 001 154 571 960 32;
  • 30) 0.398 437 500 001 154 571 960 32 × 2 = 0 + 0.796 875 000 002 309 143 920 64;
  • 31) 0.796 875 000 002 309 143 920 64 × 2 = 1 + 0.593 750 000 004 618 287 841 28;
  • 32) 0.593 750 000 004 618 287 841 28 × 2 = 1 + 0.187 500 000 009 236 575 682 56;
  • 33) 0.187 500 000 009 236 575 682 56 × 2 = 0 + 0.375 000 000 018 473 151 365 12;
  • 34) 0.375 000 000 018 473 151 365 12 × 2 = 0 + 0.750 000 000 036 946 302 730 24;
  • 35) 0.750 000 000 036 946 302 730 24 × 2 = 1 + 0.500 000 000 073 892 605 460 48;
  • 36) 0.500 000 000 073 892 605 460 48 × 2 = 1 + 0.000 000 000 147 785 210 920 96;
  • 37) 0.000 000 000 147 785 210 920 96 × 2 = 0 + 0.000 000 000 295 570 421 841 92;
  • 38) 0.000 000 000 295 570 421 841 92 × 2 = 0 + 0.000 000 000 591 140 843 683 84;
  • 39) 0.000 000 000 591 140 843 683 84 × 2 = 0 + 0.000 000 001 182 281 687 367 68;
  • 40) 0.000 000 001 182 281 687 367 68 × 2 = 0 + 0.000 000 002 364 563 374 735 36;
  • 41) 0.000 000 002 364 563 374 735 36 × 2 = 0 + 0.000 000 004 729 126 749 470 72;
  • 42) 0.000 000 004 729 126 749 470 72 × 2 = 0 + 0.000 000 009 458 253 498 941 44;
  • 43) 0.000 000 009 458 253 498 941 44 × 2 = 0 + 0.000 000 018 916 506 997 882 88;
  • 44) 0.000 000 018 916 506 997 882 88 × 2 = 0 + 0.000 000 037 833 013 995 765 76;
  • 45) 0.000 000 037 833 013 995 765 76 × 2 = 0 + 0.000 000 075 666 027 991 531 52;
  • 46) 0.000 000 075 666 027 991 531 52 × 2 = 0 + 0.000 000 151 332 055 983 063 04;
  • 47) 0.000 000 151 332 055 983 063 04 × 2 = 0 + 0.000 000 302 664 111 966 126 08;
  • 48) 0.000 000 302 664 111 966 126 08 × 2 = 0 + 0.000 000 605 328 223 932 252 16;
  • 49) 0.000 000 605 328 223 932 252 16 × 2 = 0 + 0.000 001 210 656 447 864 504 32;
  • 50) 0.000 001 210 656 447 864 504 32 × 2 = 0 + 0.000 002 421 312 895 729 008 64;
  • 51) 0.000 002 421 312 895 729 008 64 × 2 = 0 + 0.000 004 842 625 791 458 017 28;
  • 52) 0.000 004 842 625 791 458 017 28 × 2 = 0 + 0.000 009 685 251 582 916 034 56;
  • 53) 0.000 009 685 251 582 916 034 56 × 2 = 0 + 0.000 019 370 503 165 832 069 12;
  • 54) 0.000 019 370 503 165 832 069 12 × 2 = 0 + 0.000 038 741 006 331 664 138 24;

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:


0.000 000 000 742 147 676 648 86(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648 86(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:


0.000 000 000 742 147 676 648 86(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 -0.000 000 000 742 147 676 648 86 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