0.000 000 110 008 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 008 54(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 54(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 54.

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 54 × 2 = 0 + 0.000 000 220 017 08;
  • 2) 0.000 000 220 017 08 × 2 = 0 + 0.000 000 440 034 16;
  • 3) 0.000 000 440 034 16 × 2 = 0 + 0.000 000 880 068 32;
  • 4) 0.000 000 880 068 32 × 2 = 0 + 0.000 001 760 136 64;
  • 5) 0.000 001 760 136 64 × 2 = 0 + 0.000 003 520 273 28;
  • 6) 0.000 003 520 273 28 × 2 = 0 + 0.000 007 040 546 56;
  • 7) 0.000 007 040 546 56 × 2 = 0 + 0.000 014 081 093 12;
  • 8) 0.000 014 081 093 12 × 2 = 0 + 0.000 028 162 186 24;
  • 9) 0.000 028 162 186 24 × 2 = 0 + 0.000 056 324 372 48;
  • 10) 0.000 056 324 372 48 × 2 = 0 + 0.000 112 648 744 96;
  • 11) 0.000 112 648 744 96 × 2 = 0 + 0.000 225 297 489 92;
  • 12) 0.000 225 297 489 92 × 2 = 0 + 0.000 450 594 979 84;
  • 13) 0.000 450 594 979 84 × 2 = 0 + 0.000 901 189 959 68;
  • 14) 0.000 901 189 959 68 × 2 = 0 + 0.001 802 379 919 36;
  • 15) 0.001 802 379 919 36 × 2 = 0 + 0.003 604 759 838 72;
  • 16) 0.003 604 759 838 72 × 2 = 0 + 0.007 209 519 677 44;
  • 17) 0.007 209 519 677 44 × 2 = 0 + 0.014 419 039 354 88;
  • 18) 0.014 419 039 354 88 × 2 = 0 + 0.028 838 078 709 76;
  • 19) 0.028 838 078 709 76 × 2 = 0 + 0.057 676 157 419 52;
  • 20) 0.057 676 157 419 52 × 2 = 0 + 0.115 352 314 839 04;
  • 21) 0.115 352 314 839 04 × 2 = 0 + 0.230 704 629 678 08;
  • 22) 0.230 704 629 678 08 × 2 = 0 + 0.461 409 259 356 16;
  • 23) 0.461 409 259 356 16 × 2 = 0 + 0.922 818 518 712 32;
  • 24) 0.922 818 518 712 32 × 2 = 1 + 0.845 637 037 424 64;
  • 25) 0.845 637 037 424 64 × 2 = 1 + 0.691 274 074 849 28;
  • 26) 0.691 274 074 849 28 × 2 = 1 + 0.382 548 149 698 56;
  • 27) 0.382 548 149 698 56 × 2 = 0 + 0.765 096 299 397 12;
  • 28) 0.765 096 299 397 12 × 2 = 1 + 0.530 192 598 794 24;
  • 29) 0.530 192 598 794 24 × 2 = 1 + 0.060 385 197 588 48;
  • 30) 0.060 385 197 588 48 × 2 = 0 + 0.120 770 395 176 96;
  • 31) 0.120 770 395 176 96 × 2 = 0 + 0.241 540 790 353 92;
  • 32) 0.241 540 790 353 92 × 2 = 0 + 0.483 081 580 707 84;
  • 33) 0.483 081 580 707 84 × 2 = 0 + 0.966 163 161 415 68;
  • 34) 0.966 163 161 415 68 × 2 = 1 + 0.932 326 322 831 36;
  • 35) 0.932 326 322 831 36 × 2 = 1 + 0.864 652 645 662 72;
  • 36) 0.864 652 645 662 72 × 2 = 1 + 0.729 305 291 325 44;
  • 37) 0.729 305 291 325 44 × 2 = 1 + 0.458 610 582 650 88;
  • 38) 0.458 610 582 650 88 × 2 = 0 + 0.917 221 165 301 76;
  • 39) 0.917 221 165 301 76 × 2 = 1 + 0.834 442 330 603 52;
  • 40) 0.834 442 330 603 52 × 2 = 1 + 0.668 884 661 207 04;
  • 41) 0.668 884 661 207 04 × 2 = 1 + 0.337 769 322 414 08;
  • 42) 0.337 769 322 414 08 × 2 = 0 + 0.675 538 644 828 16;
  • 43) 0.675 538 644 828 16 × 2 = 1 + 0.351 077 289 656 32;
  • 44) 0.351 077 289 656 32 × 2 = 0 + 0.702 154 579 312 64;
  • 45) 0.702 154 579 312 64 × 2 = 1 + 0.404 309 158 625 28;
  • 46) 0.404 309 158 625 28 × 2 = 0 + 0.808 618 317 250 56;
  • 47) 0.808 618 317 250 56 × 2 = 1 + 0.617 236 634 501 12;

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 54(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1010 101(2)

5. Positive number before normalization:

0.000 000 110 008 54(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1010 101(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 54(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1010 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1011 1010 101(2) × 20 =

1.1101 1000 0111 1011 1010 101(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 1011 1010 101


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 1101 1101 0101 =

110 1100 0011 1101 1101 0101


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 1101 1101 0101


Decimal number 0.000 000 110 008 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1101 1101 0101

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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