-0.000 000 000 742 147 22 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 22(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 22(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 22| = 0.000 000 000 742 147 22


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 22.

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 22 × 2 = 0 + 0.000 000 001 484 294 44;
  • 2) 0.000 000 001 484 294 44 × 2 = 0 + 0.000 000 002 968 588 88;
  • 3) 0.000 000 002 968 588 88 × 2 = 0 + 0.000 000 005 937 177 76;
  • 4) 0.000 000 005 937 177 76 × 2 = 0 + 0.000 000 011 874 355 52;
  • 5) 0.000 000 011 874 355 52 × 2 = 0 + 0.000 000 023 748 711 04;
  • 6) 0.000 000 023 748 711 04 × 2 = 0 + 0.000 000 047 497 422 08;
  • 7) 0.000 000 047 497 422 08 × 2 = 0 + 0.000 000 094 994 844 16;
  • 8) 0.000 000 094 994 844 16 × 2 = 0 + 0.000 000 189 989 688 32;
  • 9) 0.000 000 189 989 688 32 × 2 = 0 + 0.000 000 379 979 376 64;
  • 10) 0.000 000 379 979 376 64 × 2 = 0 + 0.000 000 759 958 753 28;
  • 11) 0.000 000 759 958 753 28 × 2 = 0 + 0.000 001 519 917 506 56;
  • 12) 0.000 001 519 917 506 56 × 2 = 0 + 0.000 003 039 835 013 12;
  • 13) 0.000 003 039 835 013 12 × 2 = 0 + 0.000 006 079 670 026 24;
  • 14) 0.000 006 079 670 026 24 × 2 = 0 + 0.000 012 159 340 052 48;
  • 15) 0.000 012 159 340 052 48 × 2 = 0 + 0.000 024 318 680 104 96;
  • 16) 0.000 024 318 680 104 96 × 2 = 0 + 0.000 048 637 360 209 92;
  • 17) 0.000 048 637 360 209 92 × 2 = 0 + 0.000 097 274 720 419 84;
  • 18) 0.000 097 274 720 419 84 × 2 = 0 + 0.000 194 549 440 839 68;
  • 19) 0.000 194 549 440 839 68 × 2 = 0 + 0.000 389 098 881 679 36;
  • 20) 0.000 389 098 881 679 36 × 2 = 0 + 0.000 778 197 763 358 72;
  • 21) 0.000 778 197 763 358 72 × 2 = 0 + 0.001 556 395 526 717 44;
  • 22) 0.001 556 395 526 717 44 × 2 = 0 + 0.003 112 791 053 434 88;
  • 23) 0.003 112 791 053 434 88 × 2 = 0 + 0.006 225 582 106 869 76;
  • 24) 0.006 225 582 106 869 76 × 2 = 0 + 0.012 451 164 213 739 52;
  • 25) 0.012 451 164 213 739 52 × 2 = 0 + 0.024 902 328 427 479 04;
  • 26) 0.024 902 328 427 479 04 × 2 = 0 + 0.049 804 656 854 958 08;
  • 27) 0.049 804 656 854 958 08 × 2 = 0 + 0.099 609 313 709 916 16;
  • 28) 0.099 609 313 709 916 16 × 2 = 0 + 0.199 218 627 419 832 32;
  • 29) 0.199 218 627 419 832 32 × 2 = 0 + 0.398 437 254 839 664 64;
  • 30) 0.398 437 254 839 664 64 × 2 = 0 + 0.796 874 509 679 329 28;
  • 31) 0.796 874 509 679 329 28 × 2 = 1 + 0.593 749 019 358 658 56;
  • 32) 0.593 749 019 358 658 56 × 2 = 1 + 0.187 498 038 717 317 12;
  • 33) 0.187 498 038 717 317 12 × 2 = 0 + 0.374 996 077 434 634 24;
  • 34) 0.374 996 077 434 634 24 × 2 = 0 + 0.749 992 154 869 268 48;
  • 35) 0.749 992 154 869 268 48 × 2 = 1 + 0.499 984 309 738 536 96;
  • 36) 0.499 984 309 738 536 96 × 2 = 0 + 0.999 968 619 477 073 92;
  • 37) 0.999 968 619 477 073 92 × 2 = 1 + 0.999 937 238 954 147 84;
  • 38) 0.999 937 238 954 147 84 × 2 = 1 + 0.999 874 477 908 295 68;
  • 39) 0.999 874 477 908 295 68 × 2 = 1 + 0.999 748 955 816 591 36;
  • 40) 0.999 748 955 816 591 36 × 2 = 1 + 0.999 497 911 633 182 72;
  • 41) 0.999 497 911 633 182 72 × 2 = 1 + 0.998 995 823 266 365 44;
  • 42) 0.998 995 823 266 365 44 × 2 = 1 + 0.997 991 646 532 730 88;
  • 43) 0.997 991 646 532 730 88 × 2 = 1 + 0.995 983 293 065 461 76;
  • 44) 0.995 983 293 065 461 76 × 2 = 1 + 0.991 966 586 130 923 52;
  • 45) 0.991 966 586 130 923 52 × 2 = 1 + 0.983 933 172 261 847 04;
  • 46) 0.983 933 172 261 847 04 × 2 = 1 + 0.967 866 344 523 694 08;
  • 47) 0.967 866 344 523 694 08 × 2 = 1 + 0.935 732 689 047 388 16;
  • 48) 0.935 732 689 047 388 16 × 2 = 1 + 0.871 465 378 094 776 32;
  • 49) 0.871 465 378 094 776 32 × 2 = 1 + 0.742 930 756 189 552 64;
  • 50) 0.742 930 756 189 552 64 × 2 = 1 + 0.485 861 512 379 105 28;
  • 51) 0.485 861 512 379 105 28 × 2 = 0 + 0.971 723 024 758 210 56;
  • 52) 0.971 723 024 758 210 56 × 2 = 1 + 0.943 446 049 516 421 12;
  • 53) 0.943 446 049 516 421 12 × 2 = 1 + 0.886 892 099 032 842 24;
  • 54) 0.886 892 099 032 842 24 × 2 = 1 + 0.773 784 198 065 684 48;

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 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 11(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 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 11(2) × 20 =

1.1001 0111 1111 1111 1110 111(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 0111 1111 1111 1110 111


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 1011 1111 1111 1111 0111 =

100 1011 1111 1111 1111 0111


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 1011 1111 1111 1111 0111


Decimal number -0.000 000 000 742 147 22 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 0111

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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