-0.000 000 100 100 092 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 100 100 092 5(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
-0.000 000 100 100 092 5(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:

|-0.000 000 100 100 092 5| = 0.000 000 100 100 092 5


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: 0.000 000 100 100 092 5.

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) 0.000 000 100 100 092 5 × 2 = 0 + 0.000 000 200 200 185;
  • 2) 0.000 000 200 200 185 × 2 = 0 + 0.000 000 400 400 37;
  • 3) 0.000 000 400 400 37 × 2 = 0 + 0.000 000 800 800 74;
  • 4) 0.000 000 800 800 74 × 2 = 0 + 0.000 001 601 601 48;
  • 5) 0.000 001 601 601 48 × 2 = 0 + 0.000 003 203 202 96;
  • 6) 0.000 003 203 202 96 × 2 = 0 + 0.000 006 406 405 92;
  • 7) 0.000 006 406 405 92 × 2 = 0 + 0.000 012 812 811 84;
  • 8) 0.000 012 812 811 84 × 2 = 0 + 0.000 025 625 623 68;
  • 9) 0.000 025 625 623 68 × 2 = 0 + 0.000 051 251 247 36;
  • 10) 0.000 051 251 247 36 × 2 = 0 + 0.000 102 502 494 72;
  • 11) 0.000 102 502 494 72 × 2 = 0 + 0.000 205 004 989 44;
  • 12) 0.000 205 004 989 44 × 2 = 0 + 0.000 410 009 978 88;
  • 13) 0.000 410 009 978 88 × 2 = 0 + 0.000 820 019 957 76;
  • 14) 0.000 820 019 957 76 × 2 = 0 + 0.001 640 039 915 52;
  • 15) 0.001 640 039 915 52 × 2 = 0 + 0.003 280 079 831 04;
  • 16) 0.003 280 079 831 04 × 2 = 0 + 0.006 560 159 662 08;
  • 17) 0.006 560 159 662 08 × 2 = 0 + 0.013 120 319 324 16;
  • 18) 0.013 120 319 324 16 × 2 = 0 + 0.026 240 638 648 32;
  • 19) 0.026 240 638 648 32 × 2 = 0 + 0.052 481 277 296 64;
  • 20) 0.052 481 277 296 64 × 2 = 0 + 0.104 962 554 593 28;
  • 21) 0.104 962 554 593 28 × 2 = 0 + 0.209 925 109 186 56;
  • 22) 0.209 925 109 186 56 × 2 = 0 + 0.419 850 218 373 12;
  • 23) 0.419 850 218 373 12 × 2 = 0 + 0.839 700 436 746 24;
  • 24) 0.839 700 436 746 24 × 2 = 1 + 0.679 400 873 492 48;
  • 25) 0.679 400 873 492 48 × 2 = 1 + 0.358 801 746 984 96;
  • 26) 0.358 801 746 984 96 × 2 = 0 + 0.717 603 493 969 92;
  • 27) 0.717 603 493 969 92 × 2 = 1 + 0.435 206 987 939 84;
  • 28) 0.435 206 987 939 84 × 2 = 0 + 0.870 413 975 879 68;
  • 29) 0.870 413 975 879 68 × 2 = 1 + 0.740 827 951 759 36;
  • 30) 0.740 827 951 759 36 × 2 = 1 + 0.481 655 903 518 72;
  • 31) 0.481 655 903 518 72 × 2 = 0 + 0.963 311 807 037 44;
  • 32) 0.963 311 807 037 44 × 2 = 1 + 0.926 623 614 074 88;
  • 33) 0.926 623 614 074 88 × 2 = 1 + 0.853 247 228 149 76;
  • 34) 0.853 247 228 149 76 × 2 = 1 + 0.706 494 456 299 52;
  • 35) 0.706 494 456 299 52 × 2 = 1 + 0.412 988 912 599 04;
  • 36) 0.412 988 912 599 04 × 2 = 0 + 0.825 977 825 198 08;
  • 37) 0.825 977 825 198 08 × 2 = 1 + 0.651 955 650 396 16;
  • 38) 0.651 955 650 396 16 × 2 = 1 + 0.303 911 300 792 32;
  • 39) 0.303 911 300 792 32 × 2 = 0 + 0.607 822 601 584 64;
  • 40) 0.607 822 601 584 64 × 2 = 1 + 0.215 645 203 169 28;
  • 41) 0.215 645 203 169 28 × 2 = 0 + 0.431 290 406 338 56;
  • 42) 0.431 290 406 338 56 × 2 = 0 + 0.862 580 812 677 12;
  • 43) 0.862 580 812 677 12 × 2 = 1 + 0.725 161 625 354 24;
  • 44) 0.725 161 625 354 24 × 2 = 1 + 0.450 323 250 708 48;
  • 45) 0.450 323 250 708 48 × 2 = 0 + 0.900 646 501 416 96;
  • 46) 0.900 646 501 416 96 × 2 = 1 + 0.801 293 002 833 92;
  • 47) 0.801 293 002 833 92 × 2 = 1 + 0.602 586 005 667 84;

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:


0.000 000 100 100 092 5(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0011 011(2)

6. Positive number before normalization:

0.000 000 100 100 092 5(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0011 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 100 100 092 5(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0011 011(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1110 1101 0011 011(2) × 20 =

1.1010 1101 1110 1101 0011 011(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1101 1110 1101 0011 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 101 0110 1111 0110 1001 1011 =

101 0110 1111 0110 1001 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 0111

Mantissa (23 bits) =
101 0110 1111 0110 1001 1011


Decimal number -0.000 000 100 100 092 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1111 0110 1001 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