0.000 000 110 011 04 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 011 04(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
0.000 000 110 011 04(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: 0.000 000 110 011 04.

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) 0.000 000 110 011 04 × 2 = 0 + 0.000 000 220 022 08;
  • 2) 0.000 000 220 022 08 × 2 = 0 + 0.000 000 440 044 16;
  • 3) 0.000 000 440 044 16 × 2 = 0 + 0.000 000 880 088 32;
  • 4) 0.000 000 880 088 32 × 2 = 0 + 0.000 001 760 176 64;
  • 5) 0.000 001 760 176 64 × 2 = 0 + 0.000 003 520 353 28;
  • 6) 0.000 003 520 353 28 × 2 = 0 + 0.000 007 040 706 56;
  • 7) 0.000 007 040 706 56 × 2 = 0 + 0.000 014 081 413 12;
  • 8) 0.000 014 081 413 12 × 2 = 0 + 0.000 028 162 826 24;
  • 9) 0.000 028 162 826 24 × 2 = 0 + 0.000 056 325 652 48;
  • 10) 0.000 056 325 652 48 × 2 = 0 + 0.000 112 651 304 96;
  • 11) 0.000 112 651 304 96 × 2 = 0 + 0.000 225 302 609 92;
  • 12) 0.000 225 302 609 92 × 2 = 0 + 0.000 450 605 219 84;
  • 13) 0.000 450 605 219 84 × 2 = 0 + 0.000 901 210 439 68;
  • 14) 0.000 901 210 439 68 × 2 = 0 + 0.001 802 420 879 36;
  • 15) 0.001 802 420 879 36 × 2 = 0 + 0.003 604 841 758 72;
  • 16) 0.003 604 841 758 72 × 2 = 0 + 0.007 209 683 517 44;
  • 17) 0.007 209 683 517 44 × 2 = 0 + 0.014 419 367 034 88;
  • 18) 0.014 419 367 034 88 × 2 = 0 + 0.028 838 734 069 76;
  • 19) 0.028 838 734 069 76 × 2 = 0 + 0.057 677 468 139 52;
  • 20) 0.057 677 468 139 52 × 2 = 0 + 0.115 354 936 279 04;
  • 21) 0.115 354 936 279 04 × 2 = 0 + 0.230 709 872 558 08;
  • 22) 0.230 709 872 558 08 × 2 = 0 + 0.461 419 745 116 16;
  • 23) 0.461 419 745 116 16 × 2 = 0 + 0.922 839 490 232 32;
  • 24) 0.922 839 490 232 32 × 2 = 1 + 0.845 678 980 464 64;
  • 25) 0.845 678 980 464 64 × 2 = 1 + 0.691 357 960 929 28;
  • 26) 0.691 357 960 929 28 × 2 = 1 + 0.382 715 921 858 56;
  • 27) 0.382 715 921 858 56 × 2 = 0 + 0.765 431 843 717 12;
  • 28) 0.765 431 843 717 12 × 2 = 1 + 0.530 863 687 434 24;
  • 29) 0.530 863 687 434 24 × 2 = 1 + 0.061 727 374 868 48;
  • 30) 0.061 727 374 868 48 × 2 = 0 + 0.123 454 749 736 96;
  • 31) 0.123 454 749 736 96 × 2 = 0 + 0.246 909 499 473 92;
  • 32) 0.246 909 499 473 92 × 2 = 0 + 0.493 818 998 947 84;
  • 33) 0.493 818 998 947 84 × 2 = 0 + 0.987 637 997 895 68;
  • 34) 0.987 637 997 895 68 × 2 = 1 + 0.975 275 995 791 36;
  • 35) 0.975 275 995 791 36 × 2 = 1 + 0.950 551 991 582 72;
  • 36) 0.950 551 991 582 72 × 2 = 1 + 0.901 103 983 165 44;
  • 37) 0.901 103 983 165 44 × 2 = 1 + 0.802 207 966 330 88;
  • 38) 0.802 207 966 330 88 × 2 = 1 + 0.604 415 932 661 76;
  • 39) 0.604 415 932 661 76 × 2 = 1 + 0.208 831 865 323 52;
  • 40) 0.208 831 865 323 52 × 2 = 0 + 0.417 663 730 647 04;
  • 41) 0.417 663 730 647 04 × 2 = 0 + 0.835 327 461 294 08;
  • 42) 0.835 327 461 294 08 × 2 = 1 + 0.670 654 922 588 16;
  • 43) 0.670 654 922 588 16 × 2 = 1 + 0.341 309 845 176 32;
  • 44) 0.341 309 845 176 32 × 2 = 0 + 0.682 619 690 352 64;
  • 45) 0.682 619 690 352 64 × 2 = 1 + 0.365 239 380 705 28;
  • 46) 0.365 239 380 705 28 × 2 = 0 + 0.730 478 761 410 56;
  • 47) 0.730 478 761 410 56 × 2 = 1 + 0.460 957 522 821 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:


0.000 000 110 011 04(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1110 0110 101(2)

5. Positive number before normalization:

0.000 000 110 011 04(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1110 0110 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:


0.000 000 110 011 04(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1110 0110 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1110 0110 101(2) × 20 =

1.1101 1000 0111 1110 0110 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 1110 0110 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 1111 0011 0101 =

110 1100 0011 1111 0011 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 1111 0011 0101


Decimal number 0.000 000 110 011 04 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1111 0011 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