-0.000 000 003 64 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 003 64(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
-0.000 000 003 64(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:

|-0.000 000 003 64| = 0.000 000 003 64


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: 0.000 000 003 64.

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) 0.000 000 003 64 × 2 = 0 + 0.000 000 007 28;
  • 2) 0.000 000 007 28 × 2 = 0 + 0.000 000 014 56;
  • 3) 0.000 000 014 56 × 2 = 0 + 0.000 000 029 12;
  • 4) 0.000 000 029 12 × 2 = 0 + 0.000 000 058 24;
  • 5) 0.000 000 058 24 × 2 = 0 + 0.000 000 116 48;
  • 6) 0.000 000 116 48 × 2 = 0 + 0.000 000 232 96;
  • 7) 0.000 000 232 96 × 2 = 0 + 0.000 000 465 92;
  • 8) 0.000 000 465 92 × 2 = 0 + 0.000 000 931 84;
  • 9) 0.000 000 931 84 × 2 = 0 + 0.000 001 863 68;
  • 10) 0.000 001 863 68 × 2 = 0 + 0.000 003 727 36;
  • 11) 0.000 003 727 36 × 2 = 0 + 0.000 007 454 72;
  • 12) 0.000 007 454 72 × 2 = 0 + 0.000 014 909 44;
  • 13) 0.000 014 909 44 × 2 = 0 + 0.000 029 818 88;
  • 14) 0.000 029 818 88 × 2 = 0 + 0.000 059 637 76;
  • 15) 0.000 059 637 76 × 2 = 0 + 0.000 119 275 52;
  • 16) 0.000 119 275 52 × 2 = 0 + 0.000 238 551 04;
  • 17) 0.000 238 551 04 × 2 = 0 + 0.000 477 102 08;
  • 18) 0.000 477 102 08 × 2 = 0 + 0.000 954 204 16;
  • 19) 0.000 954 204 16 × 2 = 0 + 0.001 908 408 32;
  • 20) 0.001 908 408 32 × 2 = 0 + 0.003 816 816 64;
  • 21) 0.003 816 816 64 × 2 = 0 + 0.007 633 633 28;
  • 22) 0.007 633 633 28 × 2 = 0 + 0.015 267 266 56;
  • 23) 0.015 267 266 56 × 2 = 0 + 0.030 534 533 12;
  • 24) 0.030 534 533 12 × 2 = 0 + 0.061 069 066 24;
  • 25) 0.061 069 066 24 × 2 = 0 + 0.122 138 132 48;
  • 26) 0.122 138 132 48 × 2 = 0 + 0.244 276 264 96;
  • 27) 0.244 276 264 96 × 2 = 0 + 0.488 552 529 92;
  • 28) 0.488 552 529 92 × 2 = 0 + 0.977 105 059 84;
  • 29) 0.977 105 059 84 × 2 = 1 + 0.954 210 119 68;
  • 30) 0.954 210 119 68 × 2 = 1 + 0.908 420 239 36;
  • 31) 0.908 420 239 36 × 2 = 1 + 0.816 840 478 72;
  • 32) 0.816 840 478 72 × 2 = 1 + 0.633 680 957 44;
  • 33) 0.633 680 957 44 × 2 = 1 + 0.267 361 914 88;
  • 34) 0.267 361 914 88 × 2 = 0 + 0.534 723 829 76;
  • 35) 0.534 723 829 76 × 2 = 1 + 0.069 447 659 52;
  • 36) 0.069 447 659 52 × 2 = 0 + 0.138 895 319 04;
  • 37) 0.138 895 319 04 × 2 = 0 + 0.277 790 638 08;
  • 38) 0.277 790 638 08 × 2 = 0 + 0.555 581 276 16;
  • 39) 0.555 581 276 16 × 2 = 1 + 0.111 162 552 32;
  • 40) 0.111 162 552 32 × 2 = 0 + 0.222 325 104 64;
  • 41) 0.222 325 104 64 × 2 = 0 + 0.444 650 209 28;
  • 42) 0.444 650 209 28 × 2 = 0 + 0.889 300 418 56;
  • 43) 0.889 300 418 56 × 2 = 1 + 0.778 600 837 12;
  • 44) 0.778 600 837 12 × 2 = 1 + 0.557 201 674 24;
  • 45) 0.557 201 674 24 × 2 = 1 + 0.114 403 348 48;
  • 46) 0.114 403 348 48 × 2 = 0 + 0.228 806 696 96;
  • 47) 0.228 806 696 96 × 2 = 0 + 0.457 613 393 92;
  • 48) 0.457 613 393 92 × 2 = 0 + 0.915 226 787 84;
  • 49) 0.915 226 787 84 × 2 = 1 + 0.830 453 575 68;
  • 50) 0.830 453 575 68 × 2 = 1 + 0.660 907 151 36;
  • 51) 0.660 907 151 36 × 2 = 1 + 0.321 814 302 72;
  • 52) 0.321 814 302 72 × 2 = 0 + 0.643 628 605 44;

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:


0.000 000 003 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 1111 1010 0010 0011 1000 1110(2)

6. Positive number before normalization:

0.000 000 003 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 1111 1010 0010 0011 1000 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 003 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 1111 1010 0010 0011 1000 1110(2) =

0.0000 0000 0000 0000 0000 0000 0000 1111 1010 0010 0011 1000 1110(2) × 20 =

1.1111 0100 0100 0111 0001 110(2) × 2-29


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): -29

Mantissa (not normalized):
1.1111 0100 0100 0111 0001 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 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) =

98(10) =

0110 0010(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. 111 1010 0010 0011 1000 1110 =

111 1010 0010 0011 1000 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 0010

Mantissa (23 bits) =
111 1010 0010 0011 1000 1110


Decimal number -0.000 000 003 64 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 111 1010 0010 0011 1000 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