-0.000 000 000 742 147 49 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 49(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 49(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 49| = 0.000 000 000 742 147 49


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 49.

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 49 × 2 = 0 + 0.000 000 001 484 294 98;
  • 2) 0.000 000 001 484 294 98 × 2 = 0 + 0.000 000 002 968 589 96;
  • 3) 0.000 000 002 968 589 96 × 2 = 0 + 0.000 000 005 937 179 92;
  • 4) 0.000 000 005 937 179 92 × 2 = 0 + 0.000 000 011 874 359 84;
  • 5) 0.000 000 011 874 359 84 × 2 = 0 + 0.000 000 023 748 719 68;
  • 6) 0.000 000 023 748 719 68 × 2 = 0 + 0.000 000 047 497 439 36;
  • 7) 0.000 000 047 497 439 36 × 2 = 0 + 0.000 000 094 994 878 72;
  • 8) 0.000 000 094 994 878 72 × 2 = 0 + 0.000 000 189 989 757 44;
  • 9) 0.000 000 189 989 757 44 × 2 = 0 + 0.000 000 379 979 514 88;
  • 10) 0.000 000 379 979 514 88 × 2 = 0 + 0.000 000 759 959 029 76;
  • 11) 0.000 000 759 959 029 76 × 2 = 0 + 0.000 001 519 918 059 52;
  • 12) 0.000 001 519 918 059 52 × 2 = 0 + 0.000 003 039 836 119 04;
  • 13) 0.000 003 039 836 119 04 × 2 = 0 + 0.000 006 079 672 238 08;
  • 14) 0.000 006 079 672 238 08 × 2 = 0 + 0.000 012 159 344 476 16;
  • 15) 0.000 012 159 344 476 16 × 2 = 0 + 0.000 024 318 688 952 32;
  • 16) 0.000 024 318 688 952 32 × 2 = 0 + 0.000 048 637 377 904 64;
  • 17) 0.000 048 637 377 904 64 × 2 = 0 + 0.000 097 274 755 809 28;
  • 18) 0.000 097 274 755 809 28 × 2 = 0 + 0.000 194 549 511 618 56;
  • 19) 0.000 194 549 511 618 56 × 2 = 0 + 0.000 389 099 023 237 12;
  • 20) 0.000 389 099 023 237 12 × 2 = 0 + 0.000 778 198 046 474 24;
  • 21) 0.000 778 198 046 474 24 × 2 = 0 + 0.001 556 396 092 948 48;
  • 22) 0.001 556 396 092 948 48 × 2 = 0 + 0.003 112 792 185 896 96;
  • 23) 0.003 112 792 185 896 96 × 2 = 0 + 0.006 225 584 371 793 92;
  • 24) 0.006 225 584 371 793 92 × 2 = 0 + 0.012 451 168 743 587 84;
  • 25) 0.012 451 168 743 587 84 × 2 = 0 + 0.024 902 337 487 175 68;
  • 26) 0.024 902 337 487 175 68 × 2 = 0 + 0.049 804 674 974 351 36;
  • 27) 0.049 804 674 974 351 36 × 2 = 0 + 0.099 609 349 948 702 72;
  • 28) 0.099 609 349 948 702 72 × 2 = 0 + 0.199 218 699 897 405 44;
  • 29) 0.199 218 699 897 405 44 × 2 = 0 + 0.398 437 399 794 810 88;
  • 30) 0.398 437 399 794 810 88 × 2 = 0 + 0.796 874 799 589 621 76;
  • 31) 0.796 874 799 589 621 76 × 2 = 1 + 0.593 749 599 179 243 52;
  • 32) 0.593 749 599 179 243 52 × 2 = 1 + 0.187 499 198 358 487 04;
  • 33) 0.187 499 198 358 487 04 × 2 = 0 + 0.374 998 396 716 974 08;
  • 34) 0.374 998 396 716 974 08 × 2 = 0 + 0.749 996 793 433 948 16;
  • 35) 0.749 996 793 433 948 16 × 2 = 1 + 0.499 993 586 867 896 32;
  • 36) 0.499 993 586 867 896 32 × 2 = 0 + 0.999 987 173 735 792 64;
  • 37) 0.999 987 173 735 792 64 × 2 = 1 + 0.999 974 347 471 585 28;
  • 38) 0.999 974 347 471 585 28 × 2 = 1 + 0.999 948 694 943 170 56;
  • 39) 0.999 948 694 943 170 56 × 2 = 1 + 0.999 897 389 886 341 12;
  • 40) 0.999 897 389 886 341 12 × 2 = 1 + 0.999 794 779 772 682 24;
  • 41) 0.999 794 779 772 682 24 × 2 = 1 + 0.999 589 559 545 364 48;
  • 42) 0.999 589 559 545 364 48 × 2 = 1 + 0.999 179 119 090 728 96;
  • 43) 0.999 179 119 090 728 96 × 2 = 1 + 0.998 358 238 181 457 92;
  • 44) 0.998 358 238 181 457 92 × 2 = 1 + 0.996 716 476 362 915 84;
  • 45) 0.996 716 476 362 915 84 × 2 = 1 + 0.993 432 952 725 831 68;
  • 46) 0.993 432 952 725 831 68 × 2 = 1 + 0.986 865 905 451 663 36;
  • 47) 0.986 865 905 451 663 36 × 2 = 1 + 0.973 731 810 903 326 72;
  • 48) 0.973 731 810 903 326 72 × 2 = 1 + 0.947 463 621 806 653 44;
  • 49) 0.947 463 621 806 653 44 × 2 = 1 + 0.894 927 243 613 306 88;
  • 50) 0.894 927 243 613 306 88 × 2 = 1 + 0.789 854 487 226 613 76;
  • 51) 0.789 854 487 226 613 76 × 2 = 1 + 0.579 708 974 453 227 52;
  • 52) 0.579 708 974 453 227 52 × 2 = 1 + 0.159 417 948 906 455 04;
  • 53) 0.159 417 948 906 455 04 × 2 = 0 + 0.318 835 897 812 910 08;
  • 54) 0.318 835 897 812 910 08 × 2 = 0 + 0.637 671 795 625 820 16;

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 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 49 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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