-0.000 000 000 742 147 676 76 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 76(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 76(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 76| = 0.000 000 000 742 147 676 76


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 76.

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 76 × 2 = 0 + 0.000 000 001 484 295 353 52;
  • 2) 0.000 000 001 484 295 353 52 × 2 = 0 + 0.000 000 002 968 590 707 04;
  • 3) 0.000 000 002 968 590 707 04 × 2 = 0 + 0.000 000 005 937 181 414 08;
  • 4) 0.000 000 005 937 181 414 08 × 2 = 0 + 0.000 000 011 874 362 828 16;
  • 5) 0.000 000 011 874 362 828 16 × 2 = 0 + 0.000 000 023 748 725 656 32;
  • 6) 0.000 000 023 748 725 656 32 × 2 = 0 + 0.000 000 047 497 451 312 64;
  • 7) 0.000 000 047 497 451 312 64 × 2 = 0 + 0.000 000 094 994 902 625 28;
  • 8) 0.000 000 094 994 902 625 28 × 2 = 0 + 0.000 000 189 989 805 250 56;
  • 9) 0.000 000 189 989 805 250 56 × 2 = 0 + 0.000 000 379 979 610 501 12;
  • 10) 0.000 000 379 979 610 501 12 × 2 = 0 + 0.000 000 759 959 221 002 24;
  • 11) 0.000 000 759 959 221 002 24 × 2 = 0 + 0.000 001 519 918 442 004 48;
  • 12) 0.000 001 519 918 442 004 48 × 2 = 0 + 0.000 003 039 836 884 008 96;
  • 13) 0.000 003 039 836 884 008 96 × 2 = 0 + 0.000 006 079 673 768 017 92;
  • 14) 0.000 006 079 673 768 017 92 × 2 = 0 + 0.000 012 159 347 536 035 84;
  • 15) 0.000 012 159 347 536 035 84 × 2 = 0 + 0.000 024 318 695 072 071 68;
  • 16) 0.000 024 318 695 072 071 68 × 2 = 0 + 0.000 048 637 390 144 143 36;
  • 17) 0.000 048 637 390 144 143 36 × 2 = 0 + 0.000 097 274 780 288 286 72;
  • 18) 0.000 097 274 780 288 286 72 × 2 = 0 + 0.000 194 549 560 576 573 44;
  • 19) 0.000 194 549 560 576 573 44 × 2 = 0 + 0.000 389 099 121 153 146 88;
  • 20) 0.000 389 099 121 153 146 88 × 2 = 0 + 0.000 778 198 242 306 293 76;
  • 21) 0.000 778 198 242 306 293 76 × 2 = 0 + 0.001 556 396 484 612 587 52;
  • 22) 0.001 556 396 484 612 587 52 × 2 = 0 + 0.003 112 792 969 225 175 04;
  • 23) 0.003 112 792 969 225 175 04 × 2 = 0 + 0.006 225 585 938 450 350 08;
  • 24) 0.006 225 585 938 450 350 08 × 2 = 0 + 0.012 451 171 876 900 700 16;
  • 25) 0.012 451 171 876 900 700 16 × 2 = 0 + 0.024 902 343 753 801 400 32;
  • 26) 0.024 902 343 753 801 400 32 × 2 = 0 + 0.049 804 687 507 602 800 64;
  • 27) 0.049 804 687 507 602 800 64 × 2 = 0 + 0.099 609 375 015 205 601 28;
  • 28) 0.099 609 375 015 205 601 28 × 2 = 0 + 0.199 218 750 030 411 202 56;
  • 29) 0.199 218 750 030 411 202 56 × 2 = 0 + 0.398 437 500 060 822 405 12;
  • 30) 0.398 437 500 060 822 405 12 × 2 = 0 + 0.796 875 000 121 644 810 24;
  • 31) 0.796 875 000 121 644 810 24 × 2 = 1 + 0.593 750 000 243 289 620 48;
  • 32) 0.593 750 000 243 289 620 48 × 2 = 1 + 0.187 500 000 486 579 240 96;
  • 33) 0.187 500 000 486 579 240 96 × 2 = 0 + 0.375 000 000 973 158 481 92;
  • 34) 0.375 000 000 973 158 481 92 × 2 = 0 + 0.750 000 001 946 316 963 84;
  • 35) 0.750 000 001 946 316 963 84 × 2 = 1 + 0.500 000 003 892 633 927 68;
  • 36) 0.500 000 003 892 633 927 68 × 2 = 1 + 0.000 000 007 785 267 855 36;
  • 37) 0.000 000 007 785 267 855 36 × 2 = 0 + 0.000 000 015 570 535 710 72;
  • 38) 0.000 000 015 570 535 710 72 × 2 = 0 + 0.000 000 031 141 071 421 44;
  • 39) 0.000 000 031 141 071 421 44 × 2 = 0 + 0.000 000 062 282 142 842 88;
  • 40) 0.000 000 062 282 142 842 88 × 2 = 0 + 0.000 000 124 564 285 685 76;
  • 41) 0.000 000 124 564 285 685 76 × 2 = 0 + 0.000 000 249 128 571 371 52;
  • 42) 0.000 000 249 128 571 371 52 × 2 = 0 + 0.000 000 498 257 142 743 04;
  • 43) 0.000 000 498 257 142 743 04 × 2 = 0 + 0.000 000 996 514 285 486 08;
  • 44) 0.000 000 996 514 285 486 08 × 2 = 0 + 0.000 001 993 028 570 972 16;
  • 45) 0.000 001 993 028 570 972 16 × 2 = 0 + 0.000 003 986 057 141 944 32;
  • 46) 0.000 003 986 057 141 944 32 × 2 = 0 + 0.000 007 972 114 283 888 64;
  • 47) 0.000 007 972 114 283 888 64 × 2 = 0 + 0.000 015 944 228 567 777 28;
  • 48) 0.000 015 944 228 567 777 28 × 2 = 0 + 0.000 031 888 457 135 554 56;
  • 49) 0.000 031 888 457 135 554 56 × 2 = 0 + 0.000 063 776 914 271 109 12;
  • 50) 0.000 063 776 914 271 109 12 × 2 = 0 + 0.000 127 553 828 542 218 24;
  • 51) 0.000 127 553 828 542 218 24 × 2 = 0 + 0.000 255 107 657 084 436 48;
  • 52) 0.000 255 107 657 084 436 48 × 2 = 0 + 0.000 510 215 314 168 872 96;
  • 53) 0.000 510 215 314 168 872 96 × 2 = 0 + 0.001 020 430 628 337 745 92;
  • 54) 0.001 020 430 628 337 745 92 × 2 = 0 + 0.002 040 861 256 675 491 84;

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 76(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 76(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 76(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 76 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