-0.000 000 000 742 147 676 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 3(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 3(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 3| = 0.000 000 000 742 147 676 3


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 3.

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 3 × 2 = 0 + 0.000 000 001 484 295 352 6;
  • 2) 0.000 000 001 484 295 352 6 × 2 = 0 + 0.000 000 002 968 590 705 2;
  • 3) 0.000 000 002 968 590 705 2 × 2 = 0 + 0.000 000 005 937 181 410 4;
  • 4) 0.000 000 005 937 181 410 4 × 2 = 0 + 0.000 000 011 874 362 820 8;
  • 5) 0.000 000 011 874 362 820 8 × 2 = 0 + 0.000 000 023 748 725 641 6;
  • 6) 0.000 000 023 748 725 641 6 × 2 = 0 + 0.000 000 047 497 451 283 2;
  • 7) 0.000 000 047 497 451 283 2 × 2 = 0 + 0.000 000 094 994 902 566 4;
  • 8) 0.000 000 094 994 902 566 4 × 2 = 0 + 0.000 000 189 989 805 132 8;
  • 9) 0.000 000 189 989 805 132 8 × 2 = 0 + 0.000 000 379 979 610 265 6;
  • 10) 0.000 000 379 979 610 265 6 × 2 = 0 + 0.000 000 759 959 220 531 2;
  • 11) 0.000 000 759 959 220 531 2 × 2 = 0 + 0.000 001 519 918 441 062 4;
  • 12) 0.000 001 519 918 441 062 4 × 2 = 0 + 0.000 003 039 836 882 124 8;
  • 13) 0.000 003 039 836 882 124 8 × 2 = 0 + 0.000 006 079 673 764 249 6;
  • 14) 0.000 006 079 673 764 249 6 × 2 = 0 + 0.000 012 159 347 528 499 2;
  • 15) 0.000 012 159 347 528 499 2 × 2 = 0 + 0.000 024 318 695 056 998 4;
  • 16) 0.000 024 318 695 056 998 4 × 2 = 0 + 0.000 048 637 390 113 996 8;
  • 17) 0.000 048 637 390 113 996 8 × 2 = 0 + 0.000 097 274 780 227 993 6;
  • 18) 0.000 097 274 780 227 993 6 × 2 = 0 + 0.000 194 549 560 455 987 2;
  • 19) 0.000 194 549 560 455 987 2 × 2 = 0 + 0.000 389 099 120 911 974 4;
  • 20) 0.000 389 099 120 911 974 4 × 2 = 0 + 0.000 778 198 241 823 948 8;
  • 21) 0.000 778 198 241 823 948 8 × 2 = 0 + 0.001 556 396 483 647 897 6;
  • 22) 0.001 556 396 483 647 897 6 × 2 = 0 + 0.003 112 792 967 295 795 2;
  • 23) 0.003 112 792 967 295 795 2 × 2 = 0 + 0.006 225 585 934 591 590 4;
  • 24) 0.006 225 585 934 591 590 4 × 2 = 0 + 0.012 451 171 869 183 180 8;
  • 25) 0.012 451 171 869 183 180 8 × 2 = 0 + 0.024 902 343 738 366 361 6;
  • 26) 0.024 902 343 738 366 361 6 × 2 = 0 + 0.049 804 687 476 732 723 2;
  • 27) 0.049 804 687 476 732 723 2 × 2 = 0 + 0.099 609 374 953 465 446 4;
  • 28) 0.099 609 374 953 465 446 4 × 2 = 0 + 0.199 218 749 906 930 892 8;
  • 29) 0.199 218 749 906 930 892 8 × 2 = 0 + 0.398 437 499 813 861 785 6;
  • 30) 0.398 437 499 813 861 785 6 × 2 = 0 + 0.796 874 999 627 723 571 2;
  • 31) 0.796 874 999 627 723 571 2 × 2 = 1 + 0.593 749 999 255 447 142 4;
  • 32) 0.593 749 999 255 447 142 4 × 2 = 1 + 0.187 499 998 510 894 284 8;
  • 33) 0.187 499 998 510 894 284 8 × 2 = 0 + 0.374 999 997 021 788 569 6;
  • 34) 0.374 999 997 021 788 569 6 × 2 = 0 + 0.749 999 994 043 577 139 2;
  • 35) 0.749 999 994 043 577 139 2 × 2 = 1 + 0.499 999 988 087 154 278 4;
  • 36) 0.499 999 988 087 154 278 4 × 2 = 0 + 0.999 999 976 174 308 556 8;
  • 37) 0.999 999 976 174 308 556 8 × 2 = 1 + 0.999 999 952 348 617 113 6;
  • 38) 0.999 999 952 348 617 113 6 × 2 = 1 + 0.999 999 904 697 234 227 2;
  • 39) 0.999 999 904 697 234 227 2 × 2 = 1 + 0.999 999 809 394 468 454 4;
  • 40) 0.999 999 809 394 468 454 4 × 2 = 1 + 0.999 999 618 788 936 908 8;
  • 41) 0.999 999 618 788 936 908 8 × 2 = 1 + 0.999 999 237 577 873 817 6;
  • 42) 0.999 999 237 577 873 817 6 × 2 = 1 + 0.999 998 475 155 747 635 2;
  • 43) 0.999 998 475 155 747 635 2 × 2 = 1 + 0.999 996 950 311 495 270 4;
  • 44) 0.999 996 950 311 495 270 4 × 2 = 1 + 0.999 993 900 622 990 540 8;
  • 45) 0.999 993 900 622 990 540 8 × 2 = 1 + 0.999 987 801 245 981 081 6;
  • 46) 0.999 987 801 245 981 081 6 × 2 = 1 + 0.999 975 602 491 962 163 2;
  • 47) 0.999 975 602 491 962 163 2 × 2 = 1 + 0.999 951 204 983 924 326 4;
  • 48) 0.999 951 204 983 924 326 4 × 2 = 1 + 0.999 902 409 967 848 652 8;
  • 49) 0.999 902 409 967 848 652 8 × 2 = 1 + 0.999 804 819 935 697 305 6;
  • 50) 0.999 804 819 935 697 305 6 × 2 = 1 + 0.999 609 639 871 394 611 2;
  • 51) 0.999 609 639 871 394 611 2 × 2 = 1 + 0.999 219 279 742 789 222 4;
  • 52) 0.999 219 279 742 789 222 4 × 2 = 1 + 0.998 438 559 485 578 444 8;
  • 53) 0.998 438 559 485 578 444 8 × 2 = 1 + 0.996 877 118 971 156 889 6;
  • 54) 0.996 877 118 971 156 889 6 × 2 = 1 + 0.993 754 237 942 313 779 2;

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 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 676 3(10) =

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

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

1.1001 0111 1111 1111 1111 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 1111 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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 676 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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