-0.000 000 000 13 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 13(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 13(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 13| = 0.000 000 000 13


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 13.

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 13 × 2 = 0 + 0.000 000 000 26;
  • 2) 0.000 000 000 26 × 2 = 0 + 0.000 000 000 52;
  • 3) 0.000 000 000 52 × 2 = 0 + 0.000 000 001 04;
  • 4) 0.000 000 001 04 × 2 = 0 + 0.000 000 002 08;
  • 5) 0.000 000 002 08 × 2 = 0 + 0.000 000 004 16;
  • 6) 0.000 000 004 16 × 2 = 0 + 0.000 000 008 32;
  • 7) 0.000 000 008 32 × 2 = 0 + 0.000 000 016 64;
  • 8) 0.000 000 016 64 × 2 = 0 + 0.000 000 033 28;
  • 9) 0.000 000 033 28 × 2 = 0 + 0.000 000 066 56;
  • 10) 0.000 000 066 56 × 2 = 0 + 0.000 000 133 12;
  • 11) 0.000 000 133 12 × 2 = 0 + 0.000 000 266 24;
  • 12) 0.000 000 266 24 × 2 = 0 + 0.000 000 532 48;
  • 13) 0.000 000 532 48 × 2 = 0 + 0.000 001 064 96;
  • 14) 0.000 001 064 96 × 2 = 0 + 0.000 002 129 92;
  • 15) 0.000 002 129 92 × 2 = 0 + 0.000 004 259 84;
  • 16) 0.000 004 259 84 × 2 = 0 + 0.000 008 519 68;
  • 17) 0.000 008 519 68 × 2 = 0 + 0.000 017 039 36;
  • 18) 0.000 017 039 36 × 2 = 0 + 0.000 034 078 72;
  • 19) 0.000 034 078 72 × 2 = 0 + 0.000 068 157 44;
  • 20) 0.000 068 157 44 × 2 = 0 + 0.000 136 314 88;
  • 21) 0.000 136 314 88 × 2 = 0 + 0.000 272 629 76;
  • 22) 0.000 272 629 76 × 2 = 0 + 0.000 545 259 52;
  • 23) 0.000 545 259 52 × 2 = 0 + 0.001 090 519 04;
  • 24) 0.001 090 519 04 × 2 = 0 + 0.002 181 038 08;
  • 25) 0.002 181 038 08 × 2 = 0 + 0.004 362 076 16;
  • 26) 0.004 362 076 16 × 2 = 0 + 0.008 724 152 32;
  • 27) 0.008 724 152 32 × 2 = 0 + 0.017 448 304 64;
  • 28) 0.017 448 304 64 × 2 = 0 + 0.034 896 609 28;
  • 29) 0.034 896 609 28 × 2 = 0 + 0.069 793 218 56;
  • 30) 0.069 793 218 56 × 2 = 0 + 0.139 586 437 12;
  • 31) 0.139 586 437 12 × 2 = 0 + 0.279 172 874 24;
  • 32) 0.279 172 874 24 × 2 = 0 + 0.558 345 748 48;
  • 33) 0.558 345 748 48 × 2 = 1 + 0.116 691 496 96;
  • 34) 0.116 691 496 96 × 2 = 0 + 0.233 382 993 92;
  • 35) 0.233 382 993 92 × 2 = 0 + 0.466 765 987 84;
  • 36) 0.466 765 987 84 × 2 = 0 + 0.933 531 975 68;
  • 37) 0.933 531 975 68 × 2 = 1 + 0.867 063 951 36;
  • 38) 0.867 063 951 36 × 2 = 1 + 0.734 127 902 72;
  • 39) 0.734 127 902 72 × 2 = 1 + 0.468 255 805 44;
  • 40) 0.468 255 805 44 × 2 = 0 + 0.936 511 610 88;
  • 41) 0.936 511 610 88 × 2 = 1 + 0.873 023 221 76;
  • 42) 0.873 023 221 76 × 2 = 1 + 0.746 046 443 52;
  • 43) 0.746 046 443 52 × 2 = 1 + 0.492 092 887 04;
  • 44) 0.492 092 887 04 × 2 = 0 + 0.984 185 774 08;
  • 45) 0.984 185 774 08 × 2 = 1 + 0.968 371 548 16;
  • 46) 0.968 371 548 16 × 2 = 1 + 0.936 743 096 32;
  • 47) 0.936 743 096 32 × 2 = 1 + 0.873 486 192 64;
  • 48) 0.873 486 192 64 × 2 = 1 + 0.746 972 385 28;
  • 49) 0.746 972 385 28 × 2 = 1 + 0.493 944 770 56;
  • 50) 0.493 944 770 56 × 2 = 0 + 0.987 889 541 12;
  • 51) 0.987 889 541 12 × 2 = 1 + 0.975 779 082 24;
  • 52) 0.975 779 082 24 × 2 = 1 + 0.951 558 164 48;
  • 53) 0.951 558 164 48 × 2 = 1 + 0.903 116 328 96;
  • 54) 0.903 116 328 96 × 2 = 1 + 0.806 232 657 92;
  • 55) 0.806 232 657 92 × 2 = 1 + 0.612 465 315 84;
  • 56) 0.612 465 315 84 × 2 = 1 + 0.224 930 631 68;

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 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1000 1110 1110 1111 1011 1111(2)

6. Positive number before normalization:

0.000 000 000 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1000 1110 1110 1111 1011 1111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 33 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1000 1110 1110 1111 1011 1111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1000 1110 1110 1111 1011 1111(2) × 20 =

1.0001 1101 1101 1111 0111 111(2) × 2-33


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): -33

Mantissa (not normalized):
1.0001 1101 1101 1111 0111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-33 + 2(8-1) - 1 =

(-33 + 127)(10) =

94(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;
  • 94 ÷ 2 = 47 + 0;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

94(10) =

0101 1110(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. 000 1110 1110 1111 1011 1111 =

000 1110 1110 1111 1011 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) =
0101 1110

Mantissa (23 bits) =
000 1110 1110 1111 1011 1111


Decimal number -0.000 000 000 13 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1110 - 000 1110 1110 1111 1011 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