110 010 110 000.011 110 110 808 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 110 010 110 000.011 110 110 808(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
110 010 110 000.011 110 110 808(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: 110 010 110 000.
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;
  • 110 010 110 000 ÷ 2 = 55 005 055 000 + 0;
  • 55 005 055 000 ÷ 2 = 27 502 527 500 + 0;
  • 27 502 527 500 ÷ 2 = 13 751 263 750 + 0;
  • 13 751 263 750 ÷ 2 = 6 875 631 875 + 0;
  • 6 875 631 875 ÷ 2 = 3 437 815 937 + 1;
  • 3 437 815 937 ÷ 2 = 1 718 907 968 + 1;
  • 1 718 907 968 ÷ 2 = 859 453 984 + 0;
  • 859 453 984 ÷ 2 = 429 726 992 + 0;
  • 429 726 992 ÷ 2 = 214 863 496 + 0;
  • 214 863 496 ÷ 2 = 107 431 748 + 0;
  • 107 431 748 ÷ 2 = 53 715 874 + 0;
  • 53 715 874 ÷ 2 = 26 857 937 + 0;
  • 26 857 937 ÷ 2 = 13 428 968 + 1;
  • 13 428 968 ÷ 2 = 6 714 484 + 0;
  • 6 714 484 ÷ 2 = 3 357 242 + 0;
  • 3 357 242 ÷ 2 = 1 678 621 + 0;
  • 1 678 621 ÷ 2 = 839 310 + 1;
  • 839 310 ÷ 2 = 419 655 + 0;
  • 419 655 ÷ 2 = 209 827 + 1;
  • 209 827 ÷ 2 = 104 913 + 1;
  • 104 913 ÷ 2 = 52 456 + 1;
  • 52 456 ÷ 2 = 26 228 + 0;
  • 26 228 ÷ 2 = 13 114 + 0;
  • 13 114 ÷ 2 = 6 557 + 0;
  • 6 557 ÷ 2 = 3 278 + 1;
  • 3 278 ÷ 2 = 1 639 + 0;
  • 1 639 ÷ 2 = 819 + 1;
  • 819 ÷ 2 = 409 + 1;
  • 409 ÷ 2 = 204 + 1;
  • 204 ÷ 2 = 102 + 0;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

110 010 110 000(10) =

1 1001 1001 1101 0001 1101 0001 0000 0011 0000(2)


3. Convert to binary (base 2) the fractional part: 0.011 110 110 808.

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) 0.011 110 110 808 × 2 = 0 + 0.022 220 221 616;
  • 2) 0.022 220 221 616 × 2 = 0 + 0.044 440 443 232;
  • 3) 0.044 440 443 232 × 2 = 0 + 0.088 880 886 464;
  • 4) 0.088 880 886 464 × 2 = 0 + 0.177 761 772 928;
  • 5) 0.177 761 772 928 × 2 = 0 + 0.355 523 545 856;
  • 6) 0.355 523 545 856 × 2 = 0 + 0.711 047 091 712;
  • 7) 0.711 047 091 712 × 2 = 1 + 0.422 094 183 424;
  • 8) 0.422 094 183 424 × 2 = 0 + 0.844 188 366 848;
  • 9) 0.844 188 366 848 × 2 = 1 + 0.688 376 733 696;
  • 10) 0.688 376 733 696 × 2 = 1 + 0.376 753 467 392;
  • 11) 0.376 753 467 392 × 2 = 0 + 0.753 506 934 784;
  • 12) 0.753 506 934 784 × 2 = 1 + 0.507 013 869 568;
  • 13) 0.507 013 869 568 × 2 = 1 + 0.014 027 739 136;
  • 14) 0.014 027 739 136 × 2 = 0 + 0.028 055 478 272;
  • 15) 0.028 055 478 272 × 2 = 0 + 0.056 110 956 544;
  • 16) 0.056 110 956 544 × 2 = 0 + 0.112 221 913 088;
  • 17) 0.112 221 913 088 × 2 = 0 + 0.224 443 826 176;
  • 18) 0.224 443 826 176 × 2 = 0 + 0.448 887 652 352;
  • 19) 0.448 887 652 352 × 2 = 0 + 0.897 775 304 704;
  • 20) 0.897 775 304 704 × 2 = 1 + 0.795 550 609 408;
  • 21) 0.795 550 609 408 × 2 = 1 + 0.591 101 218 816;
  • 22) 0.591 101 218 816 × 2 = 1 + 0.182 202 437 632;
  • 23) 0.182 202 437 632 × 2 = 0 + 0.364 404 875 264;
  • 24) 0.364 404 875 264 × 2 = 0 + 0.728 809 750 528;

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:


0.011 110 110 808(10) =

0.0000 0010 1101 1000 0001 1100(2)

5. Positive number before normalization:

110 010 110 000.011 110 110 808(10) =

1 1001 1001 1101 0001 1101 0001 0000 0011 0000.0000 0010 1101 1000 0001 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 36 positions to the left, so that only one non zero digit remains to the left of it:


110 010 110 000.011 110 110 808(10) =

1 1001 1001 1101 0001 1101 0001 0000 0011 0000.0000 0010 1101 1000 0001 1100(2) =

1 1001 1001 1101 0001 1101 0001 0000 0011 0000.0000 0010 1101 1000 0001 1100(2) × 20 =

1.1001 1001 1101 0001 1101 0001 0000 0011 0000 0000 0010 1101 1000 0001 1100(2) × 236


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): 36

Mantissa (not normalized):
1.1001 1001 1101 0001 1101 0001 0000 0011 0000 0000 0010 1101 1000 0001 1100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

36 + 2(8-1) - 1 =

(36 + 127)(10) =

163(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;
  • 163 ÷ 2 = 81 + 1;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

163(10) =

1010 0011(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 100 1100 1110 1000 1110 1000 1 0000 0011 0000 0000 0010 1101 1000 0001 1100 =

100 1100 1110 1000 1110 1000


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) =
1010 0011

Mantissa (23 bits) =
100 1100 1110 1000 1110 1000


Decimal number 110 010 110 000.011 110 110 808 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1010 0011 - 100 1100 1110 1000 1110 1000

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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