-0.000 000 000 741 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 93(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 93(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 93| = 0.000 000 000 741 93


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 93.

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 93 × 2 = 0 + 0.000 000 001 483 86;
  • 2) 0.000 000 001 483 86 × 2 = 0 + 0.000 000 002 967 72;
  • 3) 0.000 000 002 967 72 × 2 = 0 + 0.000 000 005 935 44;
  • 4) 0.000 000 005 935 44 × 2 = 0 + 0.000 000 011 870 88;
  • 5) 0.000 000 011 870 88 × 2 = 0 + 0.000 000 023 741 76;
  • 6) 0.000 000 023 741 76 × 2 = 0 + 0.000 000 047 483 52;
  • 7) 0.000 000 047 483 52 × 2 = 0 + 0.000 000 094 967 04;
  • 8) 0.000 000 094 967 04 × 2 = 0 + 0.000 000 189 934 08;
  • 9) 0.000 000 189 934 08 × 2 = 0 + 0.000 000 379 868 16;
  • 10) 0.000 000 379 868 16 × 2 = 0 + 0.000 000 759 736 32;
  • 11) 0.000 000 759 736 32 × 2 = 0 + 0.000 001 519 472 64;
  • 12) 0.000 001 519 472 64 × 2 = 0 + 0.000 003 038 945 28;
  • 13) 0.000 003 038 945 28 × 2 = 0 + 0.000 006 077 890 56;
  • 14) 0.000 006 077 890 56 × 2 = 0 + 0.000 012 155 781 12;
  • 15) 0.000 012 155 781 12 × 2 = 0 + 0.000 024 311 562 24;
  • 16) 0.000 024 311 562 24 × 2 = 0 + 0.000 048 623 124 48;
  • 17) 0.000 048 623 124 48 × 2 = 0 + 0.000 097 246 248 96;
  • 18) 0.000 097 246 248 96 × 2 = 0 + 0.000 194 492 497 92;
  • 19) 0.000 194 492 497 92 × 2 = 0 + 0.000 388 984 995 84;
  • 20) 0.000 388 984 995 84 × 2 = 0 + 0.000 777 969 991 68;
  • 21) 0.000 777 969 991 68 × 2 = 0 + 0.001 555 939 983 36;
  • 22) 0.001 555 939 983 36 × 2 = 0 + 0.003 111 879 966 72;
  • 23) 0.003 111 879 966 72 × 2 = 0 + 0.006 223 759 933 44;
  • 24) 0.006 223 759 933 44 × 2 = 0 + 0.012 447 519 866 88;
  • 25) 0.012 447 519 866 88 × 2 = 0 + 0.024 895 039 733 76;
  • 26) 0.024 895 039 733 76 × 2 = 0 + 0.049 790 079 467 52;
  • 27) 0.049 790 079 467 52 × 2 = 0 + 0.099 580 158 935 04;
  • 28) 0.099 580 158 935 04 × 2 = 0 + 0.199 160 317 870 08;
  • 29) 0.199 160 317 870 08 × 2 = 0 + 0.398 320 635 740 16;
  • 30) 0.398 320 635 740 16 × 2 = 0 + 0.796 641 271 480 32;
  • 31) 0.796 641 271 480 32 × 2 = 1 + 0.593 282 542 960 64;
  • 32) 0.593 282 542 960 64 × 2 = 1 + 0.186 565 085 921 28;
  • 33) 0.186 565 085 921 28 × 2 = 0 + 0.373 130 171 842 56;
  • 34) 0.373 130 171 842 56 × 2 = 0 + 0.746 260 343 685 12;
  • 35) 0.746 260 343 685 12 × 2 = 1 + 0.492 520 687 370 24;
  • 36) 0.492 520 687 370 24 × 2 = 0 + 0.985 041 374 740 48;
  • 37) 0.985 041 374 740 48 × 2 = 1 + 0.970 082 749 480 96;
  • 38) 0.970 082 749 480 96 × 2 = 1 + 0.940 165 498 961 92;
  • 39) 0.940 165 498 961 92 × 2 = 1 + 0.880 330 997 923 84;
  • 40) 0.880 330 997 923 84 × 2 = 1 + 0.760 661 995 847 68;
  • 41) 0.760 661 995 847 68 × 2 = 1 + 0.521 323 991 695 36;
  • 42) 0.521 323 991 695 36 × 2 = 1 + 0.042 647 983 390 72;
  • 43) 0.042 647 983 390 72 × 2 = 0 + 0.085 295 966 781 44;
  • 44) 0.085 295 966 781 44 × 2 = 0 + 0.170 591 933 562 88;
  • 45) 0.170 591 933 562 88 × 2 = 0 + 0.341 183 867 125 76;
  • 46) 0.341 183 867 125 76 × 2 = 0 + 0.682 367 734 251 52;
  • 47) 0.682 367 734 251 52 × 2 = 1 + 0.364 735 468 503 04;
  • 48) 0.364 735 468 503 04 × 2 = 0 + 0.729 470 937 006 08;
  • 49) 0.729 470 937 006 08 × 2 = 1 + 0.458 941 874 012 16;
  • 50) 0.458 941 874 012 16 × 2 = 0 + 0.917 883 748 024 32;
  • 51) 0.917 883 748 024 32 × 2 = 1 + 0.835 767 496 048 64;
  • 52) 0.835 767 496 048 64 × 2 = 1 + 0.671 534 992 097 28;
  • 53) 0.671 534 992 097 28 × 2 = 1 + 0.343 069 984 194 56;
  • 54) 0.343 069 984 194 56 × 2 = 0 + 0.686 139 968 389 12;

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 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0010 1011 10(2)

6. Positive number before normalization:

0.000 000 000 741 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0010 1011 10(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 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0010 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0010 1011 10(2) × 20 =

1.1001 0111 1110 0001 0101 110(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 1110 0001 0101 110


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 0000 1010 1110 =

100 1011 1111 0000 1010 1110


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 0000 1010 1110


Decimal number -0.000 000 000 741 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0000 1010 1110

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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