-0.000 013 504 595 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 013 504 595(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
-0.000 013 504 595(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:

|-0.000 013 504 595| = 0.000 013 504 595


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: 0.000 013 504 595.

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) 0.000 013 504 595 × 2 = 0 + 0.000 027 009 19;
  • 2) 0.000 027 009 19 × 2 = 0 + 0.000 054 018 38;
  • 3) 0.000 054 018 38 × 2 = 0 + 0.000 108 036 76;
  • 4) 0.000 108 036 76 × 2 = 0 + 0.000 216 073 52;
  • 5) 0.000 216 073 52 × 2 = 0 + 0.000 432 147 04;
  • 6) 0.000 432 147 04 × 2 = 0 + 0.000 864 294 08;
  • 7) 0.000 864 294 08 × 2 = 0 + 0.001 728 588 16;
  • 8) 0.001 728 588 16 × 2 = 0 + 0.003 457 176 32;
  • 9) 0.003 457 176 32 × 2 = 0 + 0.006 914 352 64;
  • 10) 0.006 914 352 64 × 2 = 0 + 0.013 828 705 28;
  • 11) 0.013 828 705 28 × 2 = 0 + 0.027 657 410 56;
  • 12) 0.027 657 410 56 × 2 = 0 + 0.055 314 821 12;
  • 13) 0.055 314 821 12 × 2 = 0 + 0.110 629 642 24;
  • 14) 0.110 629 642 24 × 2 = 0 + 0.221 259 284 48;
  • 15) 0.221 259 284 48 × 2 = 0 + 0.442 518 568 96;
  • 16) 0.442 518 568 96 × 2 = 0 + 0.885 037 137 92;
  • 17) 0.885 037 137 92 × 2 = 1 + 0.770 074 275 84;
  • 18) 0.770 074 275 84 × 2 = 1 + 0.540 148 551 68;
  • 19) 0.540 148 551 68 × 2 = 1 + 0.080 297 103 36;
  • 20) 0.080 297 103 36 × 2 = 0 + 0.160 594 206 72;
  • 21) 0.160 594 206 72 × 2 = 0 + 0.321 188 413 44;
  • 22) 0.321 188 413 44 × 2 = 0 + 0.642 376 826 88;
  • 23) 0.642 376 826 88 × 2 = 1 + 0.284 753 653 76;
  • 24) 0.284 753 653 76 × 2 = 0 + 0.569 507 307 52;
  • 25) 0.569 507 307 52 × 2 = 1 + 0.139 014 615 04;
  • 26) 0.139 014 615 04 × 2 = 0 + 0.278 029 230 08;
  • 27) 0.278 029 230 08 × 2 = 0 + 0.556 058 460 16;
  • 28) 0.556 058 460 16 × 2 = 1 + 0.112 116 920 32;
  • 29) 0.112 116 920 32 × 2 = 0 + 0.224 233 840 64;
  • 30) 0.224 233 840 64 × 2 = 0 + 0.448 467 681 28;
  • 31) 0.448 467 681 28 × 2 = 0 + 0.896 935 362 56;
  • 32) 0.896 935 362 56 × 2 = 1 + 0.793 870 725 12;
  • 33) 0.793 870 725 12 × 2 = 1 + 0.587 741 450 24;
  • 34) 0.587 741 450 24 × 2 = 1 + 0.175 482 900 48;
  • 35) 0.175 482 900 48 × 2 = 0 + 0.350 965 800 96;
  • 36) 0.350 965 800 96 × 2 = 0 + 0.701 931 601 92;
  • 37) 0.701 931 601 92 × 2 = 1 + 0.403 863 203 84;
  • 38) 0.403 863 203 84 × 2 = 0 + 0.807 726 407 68;
  • 39) 0.807 726 407 68 × 2 = 1 + 0.615 452 815 36;
  • 40) 0.615 452 815 36 × 2 = 1 + 0.230 905 630 72;

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:


0.000 013 504 595(10) =

0.0000 0000 0000 0000 1110 0010 1001 0001 1100 1011(2)

6. Positive number before normalization:

0.000 013 504 595(10) =

0.0000 0000 0000 0000 1110 0010 1001 0001 1100 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 013 504 595(10) =

0.0000 0000 0000 0000 1110 0010 1001 0001 1100 1011(2) =

0.0000 0000 0000 0000 1110 0010 1001 0001 1100 1011(2) × 20 =

1.1100 0101 0010 0011 1001 011(2) × 2-17


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): -17

Mantissa (not normalized):
1.1100 0101 0010 0011 1001 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(8-1) - 1 =

(-17 + 127)(10) =

110(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;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

110(10) =

0110 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. 110 0010 1001 0001 1100 1011 =

110 0010 1001 0001 1100 1011


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 1110

Mantissa (23 bits) =
110 0010 1001 0001 1100 1011


Decimal number -0.000 013 504 595 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1110 - 110 0010 1001 0001 1100 1011

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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