0.000 000 110 008 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 008 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
0.000 000 110 008 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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 008 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) 0.000 000 110 008 93 × 2 = 0 + 0.000 000 220 017 86;
  • 2) 0.000 000 220 017 86 × 2 = 0 + 0.000 000 440 035 72;
  • 3) 0.000 000 440 035 72 × 2 = 0 + 0.000 000 880 071 44;
  • 4) 0.000 000 880 071 44 × 2 = 0 + 0.000 001 760 142 88;
  • 5) 0.000 001 760 142 88 × 2 = 0 + 0.000 003 520 285 76;
  • 6) 0.000 003 520 285 76 × 2 = 0 + 0.000 007 040 571 52;
  • 7) 0.000 007 040 571 52 × 2 = 0 + 0.000 014 081 143 04;
  • 8) 0.000 014 081 143 04 × 2 = 0 + 0.000 028 162 286 08;
  • 9) 0.000 028 162 286 08 × 2 = 0 + 0.000 056 324 572 16;
  • 10) 0.000 056 324 572 16 × 2 = 0 + 0.000 112 649 144 32;
  • 11) 0.000 112 649 144 32 × 2 = 0 + 0.000 225 298 288 64;
  • 12) 0.000 225 298 288 64 × 2 = 0 + 0.000 450 596 577 28;
  • 13) 0.000 450 596 577 28 × 2 = 0 + 0.000 901 193 154 56;
  • 14) 0.000 901 193 154 56 × 2 = 0 + 0.001 802 386 309 12;
  • 15) 0.001 802 386 309 12 × 2 = 0 + 0.003 604 772 618 24;
  • 16) 0.003 604 772 618 24 × 2 = 0 + 0.007 209 545 236 48;
  • 17) 0.007 209 545 236 48 × 2 = 0 + 0.014 419 090 472 96;
  • 18) 0.014 419 090 472 96 × 2 = 0 + 0.028 838 180 945 92;
  • 19) 0.028 838 180 945 92 × 2 = 0 + 0.057 676 361 891 84;
  • 20) 0.057 676 361 891 84 × 2 = 0 + 0.115 352 723 783 68;
  • 21) 0.115 352 723 783 68 × 2 = 0 + 0.230 705 447 567 36;
  • 22) 0.230 705 447 567 36 × 2 = 0 + 0.461 410 895 134 72;
  • 23) 0.461 410 895 134 72 × 2 = 0 + 0.922 821 790 269 44;
  • 24) 0.922 821 790 269 44 × 2 = 1 + 0.845 643 580 538 88;
  • 25) 0.845 643 580 538 88 × 2 = 1 + 0.691 287 161 077 76;
  • 26) 0.691 287 161 077 76 × 2 = 1 + 0.382 574 322 155 52;
  • 27) 0.382 574 322 155 52 × 2 = 0 + 0.765 148 644 311 04;
  • 28) 0.765 148 644 311 04 × 2 = 1 + 0.530 297 288 622 08;
  • 29) 0.530 297 288 622 08 × 2 = 1 + 0.060 594 577 244 16;
  • 30) 0.060 594 577 244 16 × 2 = 0 + 0.121 189 154 488 32;
  • 31) 0.121 189 154 488 32 × 2 = 0 + 0.242 378 308 976 64;
  • 32) 0.242 378 308 976 64 × 2 = 0 + 0.484 756 617 953 28;
  • 33) 0.484 756 617 953 28 × 2 = 0 + 0.969 513 235 906 56;
  • 34) 0.969 513 235 906 56 × 2 = 1 + 0.939 026 471 813 12;
  • 35) 0.939 026 471 813 12 × 2 = 1 + 0.878 052 943 626 24;
  • 36) 0.878 052 943 626 24 × 2 = 1 + 0.756 105 887 252 48;
  • 37) 0.756 105 887 252 48 × 2 = 1 + 0.512 211 774 504 96;
  • 38) 0.512 211 774 504 96 × 2 = 1 + 0.024 423 549 009 92;
  • 39) 0.024 423 549 009 92 × 2 = 0 + 0.048 847 098 019 84;
  • 40) 0.048 847 098 019 84 × 2 = 0 + 0.097 694 196 039 68;
  • 41) 0.097 694 196 039 68 × 2 = 0 + 0.195 388 392 079 36;
  • 42) 0.195 388 392 079 36 × 2 = 0 + 0.390 776 784 158 72;
  • 43) 0.390 776 784 158 72 × 2 = 0 + 0.781 553 568 317 44;
  • 44) 0.781 553 568 317 44 × 2 = 1 + 0.563 107 136 634 88;
  • 45) 0.563 107 136 634 88 × 2 = 1 + 0.126 214 273 269 76;
  • 46) 0.126 214 273 269 76 × 2 = 0 + 0.252 428 546 539 52;
  • 47) 0.252 428 546 539 52 × 2 = 0 + 0.504 857 093 079 04;

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).


4. 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:


0.000 000 110 008 93(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 100(2)

5. Positive number before normalization:

0.000 000 110 008 93(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 100(2)

6. 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:


0.000 000 110 008 93(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 100(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 100(2) × 20 =

1.1101 1000 0111 1100 0001 100(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1000 0111 1100 0001 100


8. 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)


9. 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;

10. 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)


11. 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 1100 0011 1110 0000 1100 =

110 1100 0011 1110 0000 1100


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 1110 0000 1100


Decimal number 0.000 000 110 008 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0000 1100

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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