-0.000 000 000 741 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 14(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
-0.000 000 000 741 14(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:

|-0.000 000 000 741 14| = 0.000 000 000 741 14


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: 0.000 000 000 741 14.

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) 0.000 000 000 741 14 × 2 = 0 + 0.000 000 001 482 28;
  • 2) 0.000 000 001 482 28 × 2 = 0 + 0.000 000 002 964 56;
  • 3) 0.000 000 002 964 56 × 2 = 0 + 0.000 000 005 929 12;
  • 4) 0.000 000 005 929 12 × 2 = 0 + 0.000 000 011 858 24;
  • 5) 0.000 000 011 858 24 × 2 = 0 + 0.000 000 023 716 48;
  • 6) 0.000 000 023 716 48 × 2 = 0 + 0.000 000 047 432 96;
  • 7) 0.000 000 047 432 96 × 2 = 0 + 0.000 000 094 865 92;
  • 8) 0.000 000 094 865 92 × 2 = 0 + 0.000 000 189 731 84;
  • 9) 0.000 000 189 731 84 × 2 = 0 + 0.000 000 379 463 68;
  • 10) 0.000 000 379 463 68 × 2 = 0 + 0.000 000 758 927 36;
  • 11) 0.000 000 758 927 36 × 2 = 0 + 0.000 001 517 854 72;
  • 12) 0.000 001 517 854 72 × 2 = 0 + 0.000 003 035 709 44;
  • 13) 0.000 003 035 709 44 × 2 = 0 + 0.000 006 071 418 88;
  • 14) 0.000 006 071 418 88 × 2 = 0 + 0.000 012 142 837 76;
  • 15) 0.000 012 142 837 76 × 2 = 0 + 0.000 024 285 675 52;
  • 16) 0.000 024 285 675 52 × 2 = 0 + 0.000 048 571 351 04;
  • 17) 0.000 048 571 351 04 × 2 = 0 + 0.000 097 142 702 08;
  • 18) 0.000 097 142 702 08 × 2 = 0 + 0.000 194 285 404 16;
  • 19) 0.000 194 285 404 16 × 2 = 0 + 0.000 388 570 808 32;
  • 20) 0.000 388 570 808 32 × 2 = 0 + 0.000 777 141 616 64;
  • 21) 0.000 777 141 616 64 × 2 = 0 + 0.001 554 283 233 28;
  • 22) 0.001 554 283 233 28 × 2 = 0 + 0.003 108 566 466 56;
  • 23) 0.003 108 566 466 56 × 2 = 0 + 0.006 217 132 933 12;
  • 24) 0.006 217 132 933 12 × 2 = 0 + 0.012 434 265 866 24;
  • 25) 0.012 434 265 866 24 × 2 = 0 + 0.024 868 531 732 48;
  • 26) 0.024 868 531 732 48 × 2 = 0 + 0.049 737 063 464 96;
  • 27) 0.049 737 063 464 96 × 2 = 0 + 0.099 474 126 929 92;
  • 28) 0.099 474 126 929 92 × 2 = 0 + 0.198 948 253 859 84;
  • 29) 0.198 948 253 859 84 × 2 = 0 + 0.397 896 507 719 68;
  • 30) 0.397 896 507 719 68 × 2 = 0 + 0.795 793 015 439 36;
  • 31) 0.795 793 015 439 36 × 2 = 1 + 0.591 586 030 878 72;
  • 32) 0.591 586 030 878 72 × 2 = 1 + 0.183 172 061 757 44;
  • 33) 0.183 172 061 757 44 × 2 = 0 + 0.366 344 123 514 88;
  • 34) 0.366 344 123 514 88 × 2 = 0 + 0.732 688 247 029 76;
  • 35) 0.732 688 247 029 76 × 2 = 1 + 0.465 376 494 059 52;
  • 36) 0.465 376 494 059 52 × 2 = 0 + 0.930 752 988 119 04;
  • 37) 0.930 752 988 119 04 × 2 = 1 + 0.861 505 976 238 08;
  • 38) 0.861 505 976 238 08 × 2 = 1 + 0.723 011 952 476 16;
  • 39) 0.723 011 952 476 16 × 2 = 1 + 0.446 023 904 952 32;
  • 40) 0.446 023 904 952 32 × 2 = 0 + 0.892 047 809 904 64;
  • 41) 0.892 047 809 904 64 × 2 = 1 + 0.784 095 619 809 28;
  • 42) 0.784 095 619 809 28 × 2 = 1 + 0.568 191 239 618 56;
  • 43) 0.568 191 239 618 56 × 2 = 1 + 0.136 382 479 237 12;
  • 44) 0.136 382 479 237 12 × 2 = 0 + 0.272 764 958 474 24;
  • 45) 0.272 764 958 474 24 × 2 = 0 + 0.545 529 916 948 48;
  • 46) 0.545 529 916 948 48 × 2 = 1 + 0.091 059 833 896 96;
  • 47) 0.091 059 833 896 96 × 2 = 0 + 0.182 119 667 793 92;
  • 48) 0.182 119 667 793 92 × 2 = 0 + 0.364 239 335 587 84;
  • 49) 0.364 239 335 587 84 × 2 = 0 + 0.728 478 671 175 68;
  • 50) 0.728 478 671 175 68 × 2 = 1 + 0.456 957 342 351 36;
  • 51) 0.456 957 342 351 36 × 2 = 0 + 0.913 914 684 702 72;
  • 52) 0.913 914 684 702 72 × 2 = 1 + 0.827 829 369 405 44;
  • 53) 0.827 829 369 405 44 × 2 = 1 + 0.655 658 738 810 88;
  • 54) 0.655 658 738 810 88 × 2 = 1 + 0.311 317 477 621 76;

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:


0.000 000 000 741 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 0100 0101 11(2)

6. Positive number before normalization:

0.000 000 000 741 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 0100 0101 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:


0.000 000 000 741 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 0100 0101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 0100 0101 11(2) × 20 =

1.1001 0111 0111 0010 0010 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 0111 0010 0010 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 1011 1001 0001 0111 =

100 1011 1011 1001 0001 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 1011 1001 0001 0111


Decimal number -0.000 000 000 741 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1011 1001 0001 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