-0.000 000 000 742 147 676 69 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 69(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 69(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 69| = 0.000 000 000 742 147 676 69


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 69.

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 69 × 2 = 0 + 0.000 000 001 484 295 353 38;
  • 2) 0.000 000 001 484 295 353 38 × 2 = 0 + 0.000 000 002 968 590 706 76;
  • 3) 0.000 000 002 968 590 706 76 × 2 = 0 + 0.000 000 005 937 181 413 52;
  • 4) 0.000 000 005 937 181 413 52 × 2 = 0 + 0.000 000 011 874 362 827 04;
  • 5) 0.000 000 011 874 362 827 04 × 2 = 0 + 0.000 000 023 748 725 654 08;
  • 6) 0.000 000 023 748 725 654 08 × 2 = 0 + 0.000 000 047 497 451 308 16;
  • 7) 0.000 000 047 497 451 308 16 × 2 = 0 + 0.000 000 094 994 902 616 32;
  • 8) 0.000 000 094 994 902 616 32 × 2 = 0 + 0.000 000 189 989 805 232 64;
  • 9) 0.000 000 189 989 805 232 64 × 2 = 0 + 0.000 000 379 979 610 465 28;
  • 10) 0.000 000 379 979 610 465 28 × 2 = 0 + 0.000 000 759 959 220 930 56;
  • 11) 0.000 000 759 959 220 930 56 × 2 = 0 + 0.000 001 519 918 441 861 12;
  • 12) 0.000 001 519 918 441 861 12 × 2 = 0 + 0.000 003 039 836 883 722 24;
  • 13) 0.000 003 039 836 883 722 24 × 2 = 0 + 0.000 006 079 673 767 444 48;
  • 14) 0.000 006 079 673 767 444 48 × 2 = 0 + 0.000 012 159 347 534 888 96;
  • 15) 0.000 012 159 347 534 888 96 × 2 = 0 + 0.000 024 318 695 069 777 92;
  • 16) 0.000 024 318 695 069 777 92 × 2 = 0 + 0.000 048 637 390 139 555 84;
  • 17) 0.000 048 637 390 139 555 84 × 2 = 0 + 0.000 097 274 780 279 111 68;
  • 18) 0.000 097 274 780 279 111 68 × 2 = 0 + 0.000 194 549 560 558 223 36;
  • 19) 0.000 194 549 560 558 223 36 × 2 = 0 + 0.000 389 099 121 116 446 72;
  • 20) 0.000 389 099 121 116 446 72 × 2 = 0 + 0.000 778 198 242 232 893 44;
  • 21) 0.000 778 198 242 232 893 44 × 2 = 0 + 0.001 556 396 484 465 786 88;
  • 22) 0.001 556 396 484 465 786 88 × 2 = 0 + 0.003 112 792 968 931 573 76;
  • 23) 0.003 112 792 968 931 573 76 × 2 = 0 + 0.006 225 585 937 863 147 52;
  • 24) 0.006 225 585 937 863 147 52 × 2 = 0 + 0.012 451 171 875 726 295 04;
  • 25) 0.012 451 171 875 726 295 04 × 2 = 0 + 0.024 902 343 751 452 590 08;
  • 26) 0.024 902 343 751 452 590 08 × 2 = 0 + 0.049 804 687 502 905 180 16;
  • 27) 0.049 804 687 502 905 180 16 × 2 = 0 + 0.099 609 375 005 810 360 32;
  • 28) 0.099 609 375 005 810 360 32 × 2 = 0 + 0.199 218 750 011 620 720 64;
  • 29) 0.199 218 750 011 620 720 64 × 2 = 0 + 0.398 437 500 023 241 441 28;
  • 30) 0.398 437 500 023 241 441 28 × 2 = 0 + 0.796 875 000 046 482 882 56;
  • 31) 0.796 875 000 046 482 882 56 × 2 = 1 + 0.593 750 000 092 965 765 12;
  • 32) 0.593 750 000 092 965 765 12 × 2 = 1 + 0.187 500 000 185 931 530 24;
  • 33) 0.187 500 000 185 931 530 24 × 2 = 0 + 0.375 000 000 371 863 060 48;
  • 34) 0.375 000 000 371 863 060 48 × 2 = 0 + 0.750 000 000 743 726 120 96;
  • 35) 0.750 000 000 743 726 120 96 × 2 = 1 + 0.500 000 001 487 452 241 92;
  • 36) 0.500 000 001 487 452 241 92 × 2 = 1 + 0.000 000 002 974 904 483 84;
  • 37) 0.000 000 002 974 904 483 84 × 2 = 0 + 0.000 000 005 949 808 967 68;
  • 38) 0.000 000 005 949 808 967 68 × 2 = 0 + 0.000 000 011 899 617 935 36;
  • 39) 0.000 000 011 899 617 935 36 × 2 = 0 + 0.000 000 023 799 235 870 72;
  • 40) 0.000 000 023 799 235 870 72 × 2 = 0 + 0.000 000 047 598 471 741 44;
  • 41) 0.000 000 047 598 471 741 44 × 2 = 0 + 0.000 000 095 196 943 482 88;
  • 42) 0.000 000 095 196 943 482 88 × 2 = 0 + 0.000 000 190 393 886 965 76;
  • 43) 0.000 000 190 393 886 965 76 × 2 = 0 + 0.000 000 380 787 773 931 52;
  • 44) 0.000 000 380 787 773 931 52 × 2 = 0 + 0.000 000 761 575 547 863 04;
  • 45) 0.000 000 761 575 547 863 04 × 2 = 0 + 0.000 001 523 151 095 726 08;
  • 46) 0.000 001 523 151 095 726 08 × 2 = 0 + 0.000 003 046 302 191 452 16;
  • 47) 0.000 003 046 302 191 452 16 × 2 = 0 + 0.000 006 092 604 382 904 32;
  • 48) 0.000 006 092 604 382 904 32 × 2 = 0 + 0.000 012 185 208 765 808 64;
  • 49) 0.000 012 185 208 765 808 64 × 2 = 0 + 0.000 024 370 417 531 617 28;
  • 50) 0.000 024 370 417 531 617 28 × 2 = 0 + 0.000 048 740 835 063 234 56;
  • 51) 0.000 048 740 835 063 234 56 × 2 = 0 + 0.000 097 481 670 126 469 12;
  • 52) 0.000 097 481 670 126 469 12 × 2 = 0 + 0.000 194 963 340 252 938 24;
  • 53) 0.000 194 963 340 252 938 24 × 2 = 0 + 0.000 389 926 680 505 876 48;
  • 54) 0.000 389 926 680 505 876 48 × 2 = 0 + 0.000 779 853 361 011 752 96;

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 69(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 69(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 69(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 69 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