-0.000 000 000 742 147 675 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 9(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
-0.000 000 000 742 147 675 9(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:

|-0.000 000 000 742 147 675 9| = 0.000 000 000 742 147 675 9


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: 0.000 000 000 742 147 675 9.

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) 0.000 000 000 742 147 675 9 × 2 = 0 + 0.000 000 001 484 295 351 8;
  • 2) 0.000 000 001 484 295 351 8 × 2 = 0 + 0.000 000 002 968 590 703 6;
  • 3) 0.000 000 002 968 590 703 6 × 2 = 0 + 0.000 000 005 937 181 407 2;
  • 4) 0.000 000 005 937 181 407 2 × 2 = 0 + 0.000 000 011 874 362 814 4;
  • 5) 0.000 000 011 874 362 814 4 × 2 = 0 + 0.000 000 023 748 725 628 8;
  • 6) 0.000 000 023 748 725 628 8 × 2 = 0 + 0.000 000 047 497 451 257 6;
  • 7) 0.000 000 047 497 451 257 6 × 2 = 0 + 0.000 000 094 994 902 515 2;
  • 8) 0.000 000 094 994 902 515 2 × 2 = 0 + 0.000 000 189 989 805 030 4;
  • 9) 0.000 000 189 989 805 030 4 × 2 = 0 + 0.000 000 379 979 610 060 8;
  • 10) 0.000 000 379 979 610 060 8 × 2 = 0 + 0.000 000 759 959 220 121 6;
  • 11) 0.000 000 759 959 220 121 6 × 2 = 0 + 0.000 001 519 918 440 243 2;
  • 12) 0.000 001 519 918 440 243 2 × 2 = 0 + 0.000 003 039 836 880 486 4;
  • 13) 0.000 003 039 836 880 486 4 × 2 = 0 + 0.000 006 079 673 760 972 8;
  • 14) 0.000 006 079 673 760 972 8 × 2 = 0 + 0.000 012 159 347 521 945 6;
  • 15) 0.000 012 159 347 521 945 6 × 2 = 0 + 0.000 024 318 695 043 891 2;
  • 16) 0.000 024 318 695 043 891 2 × 2 = 0 + 0.000 048 637 390 087 782 4;
  • 17) 0.000 048 637 390 087 782 4 × 2 = 0 + 0.000 097 274 780 175 564 8;
  • 18) 0.000 097 274 780 175 564 8 × 2 = 0 + 0.000 194 549 560 351 129 6;
  • 19) 0.000 194 549 560 351 129 6 × 2 = 0 + 0.000 389 099 120 702 259 2;
  • 20) 0.000 389 099 120 702 259 2 × 2 = 0 + 0.000 778 198 241 404 518 4;
  • 21) 0.000 778 198 241 404 518 4 × 2 = 0 + 0.001 556 396 482 809 036 8;
  • 22) 0.001 556 396 482 809 036 8 × 2 = 0 + 0.003 112 792 965 618 073 6;
  • 23) 0.003 112 792 965 618 073 6 × 2 = 0 + 0.006 225 585 931 236 147 2;
  • 24) 0.006 225 585 931 236 147 2 × 2 = 0 + 0.012 451 171 862 472 294 4;
  • 25) 0.012 451 171 862 472 294 4 × 2 = 0 + 0.024 902 343 724 944 588 8;
  • 26) 0.024 902 343 724 944 588 8 × 2 = 0 + 0.049 804 687 449 889 177 6;
  • 27) 0.049 804 687 449 889 177 6 × 2 = 0 + 0.099 609 374 899 778 355 2;
  • 28) 0.099 609 374 899 778 355 2 × 2 = 0 + 0.199 218 749 799 556 710 4;
  • 29) 0.199 218 749 799 556 710 4 × 2 = 0 + 0.398 437 499 599 113 420 8;
  • 30) 0.398 437 499 599 113 420 8 × 2 = 0 + 0.796 874 999 198 226 841 6;
  • 31) 0.796 874 999 198 226 841 6 × 2 = 1 + 0.593 749 998 396 453 683 2;
  • 32) 0.593 749 998 396 453 683 2 × 2 = 1 + 0.187 499 996 792 907 366 4;
  • 33) 0.187 499 996 792 907 366 4 × 2 = 0 + 0.374 999 993 585 814 732 8;
  • 34) 0.374 999 993 585 814 732 8 × 2 = 0 + 0.749 999 987 171 629 465 6;
  • 35) 0.749 999 987 171 629 465 6 × 2 = 1 + 0.499 999 974 343 258 931 2;
  • 36) 0.499 999 974 343 258 931 2 × 2 = 0 + 0.999 999 948 686 517 862 4;
  • 37) 0.999 999 948 686 517 862 4 × 2 = 1 + 0.999 999 897 373 035 724 8;
  • 38) 0.999 999 897 373 035 724 8 × 2 = 1 + 0.999 999 794 746 071 449 6;
  • 39) 0.999 999 794 746 071 449 6 × 2 = 1 + 0.999 999 589 492 142 899 2;
  • 40) 0.999 999 589 492 142 899 2 × 2 = 1 + 0.999 999 178 984 285 798 4;
  • 41) 0.999 999 178 984 285 798 4 × 2 = 1 + 0.999 998 357 968 571 596 8;
  • 42) 0.999 998 357 968 571 596 8 × 2 = 1 + 0.999 996 715 937 143 193 6;
  • 43) 0.999 996 715 937 143 193 6 × 2 = 1 + 0.999 993 431 874 286 387 2;
  • 44) 0.999 993 431 874 286 387 2 × 2 = 1 + 0.999 986 863 748 572 774 4;
  • 45) 0.999 986 863 748 572 774 4 × 2 = 1 + 0.999 973 727 497 145 548 8;
  • 46) 0.999 973 727 497 145 548 8 × 2 = 1 + 0.999 947 454 994 291 097 6;
  • 47) 0.999 947 454 994 291 097 6 × 2 = 1 + 0.999 894 909 988 582 195 2;
  • 48) 0.999 894 909 988 582 195 2 × 2 = 1 + 0.999 789 819 977 164 390 4;
  • 49) 0.999 789 819 977 164 390 4 × 2 = 1 + 0.999 579 639 954 328 780 8;
  • 50) 0.999 579 639 954 328 780 8 × 2 = 1 + 0.999 159 279 908 657 561 6;
  • 51) 0.999 159 279 908 657 561 6 × 2 = 1 + 0.998 318 559 817 315 123 2;
  • 52) 0.998 318 559 817 315 123 2 × 2 = 1 + 0.996 637 119 634 630 246 4;
  • 53) 0.996 637 119 634 630 246 4 × 2 = 1 + 0.993 274 239 269 260 492 8;
  • 54) 0.993 274 239 269 260 492 8 × 2 = 1 + 0.986 548 478 538 520 985 6;

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:


0.000 000 000 742 147 675 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 9(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:


0.000 000 000 742 147 675 9(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 -0.000 000 000 742 147 675 9 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