29.275 010 000 011 110 101 000 110 009 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 29.275 010 000 011 110 101 000 110 009 63(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
29.275 010 000 011 110 101 000 110 009 63(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: 29.
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;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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.

29(10) =

1 1101(2)


3. Convert to binary (base 2) the fractional part: 0.275 010 000 011 110 101 000 110 009 63.

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) 0.275 010 000 011 110 101 000 110 009 63 × 2 = 0 + 0.550 020 000 022 220 202 000 220 019 26;
  • 2) 0.550 020 000 022 220 202 000 220 019 26 × 2 = 1 + 0.100 040 000 044 440 404 000 440 038 52;
  • 3) 0.100 040 000 044 440 404 000 440 038 52 × 2 = 0 + 0.200 080 000 088 880 808 000 880 077 04;
  • 4) 0.200 080 000 088 880 808 000 880 077 04 × 2 = 0 + 0.400 160 000 177 761 616 001 760 154 08;
  • 5) 0.400 160 000 177 761 616 001 760 154 08 × 2 = 0 + 0.800 320 000 355 523 232 003 520 308 16;
  • 6) 0.800 320 000 355 523 232 003 520 308 16 × 2 = 1 + 0.600 640 000 711 046 464 007 040 616 32;
  • 7) 0.600 640 000 711 046 464 007 040 616 32 × 2 = 1 + 0.201 280 001 422 092 928 014 081 232 64;
  • 8) 0.201 280 001 422 092 928 014 081 232 64 × 2 = 0 + 0.402 560 002 844 185 856 028 162 465 28;
  • 9) 0.402 560 002 844 185 856 028 162 465 28 × 2 = 0 + 0.805 120 005 688 371 712 056 324 930 56;
  • 10) 0.805 120 005 688 371 712 056 324 930 56 × 2 = 1 + 0.610 240 011 376 743 424 112 649 861 12;
  • 11) 0.610 240 011 376 743 424 112 649 861 12 × 2 = 1 + 0.220 480 022 753 486 848 225 299 722 24;
  • 12) 0.220 480 022 753 486 848 225 299 722 24 × 2 = 0 + 0.440 960 045 506 973 696 450 599 444 48;
  • 13) 0.440 960 045 506 973 696 450 599 444 48 × 2 = 0 + 0.881 920 091 013 947 392 901 198 888 96;
  • 14) 0.881 920 091 013 947 392 901 198 888 96 × 2 = 1 + 0.763 840 182 027 894 785 802 397 777 92;
  • 15) 0.763 840 182 027 894 785 802 397 777 92 × 2 = 1 + 0.527 680 364 055 789 571 604 795 555 84;
  • 16) 0.527 680 364 055 789 571 604 795 555 84 × 2 = 1 + 0.055 360 728 111 579 143 209 591 111 68;
  • 17) 0.055 360 728 111 579 143 209 591 111 68 × 2 = 0 + 0.110 721 456 223 158 286 419 182 223 36;
  • 18) 0.110 721 456 223 158 286 419 182 223 36 × 2 = 0 + 0.221 442 912 446 316 572 838 364 446 72;
  • 19) 0.221 442 912 446 316 572 838 364 446 72 × 2 = 0 + 0.442 885 824 892 633 145 676 728 893 44;
  • 20) 0.442 885 824 892 633 145 676 728 893 44 × 2 = 0 + 0.885 771 649 785 266 291 353 457 786 88;
  • 21) 0.885 771 649 785 266 291 353 457 786 88 × 2 = 1 + 0.771 543 299 570 532 582 706 915 573 76;
  • 22) 0.771 543 299 570 532 582 706 915 573 76 × 2 = 1 + 0.543 086 599 141 065 165 413 831 147 52;
  • 23) 0.543 086 599 141 065 165 413 831 147 52 × 2 = 1 + 0.086 173 198 282 130 330 827 662 295 04;
  • 24) 0.086 173 198 282 130 330 827 662 295 04 × 2 = 0 + 0.172 346 396 564 260 661 655 324 590 08;

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:


0.275 010 000 011 110 101 000 110 009 63(10) =

0.0100 0110 0110 0111 0000 1110(2)

5. Positive number before normalization:

29.275 010 000 011 110 101 000 110 009 63(10) =

1 1101.0100 0110 0110 0111 0000 1110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


29.275 010 000 011 110 101 000 110 009 63(10) =

1 1101.0100 0110 0110 0111 0000 1110(2) =

1 1101.0100 0110 0110 0111 0000 1110(2) × 20 =

1.1101 0100 0110 0110 0111 0000 1110(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): 4

Mantissa (not normalized):
1.1101 0100 0110 0110 0111 0000 1110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(8-1) - 1 =

(4 + 127)(10) =

131(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;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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) =

131(10) =

1000 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. 110 1010 0011 0011 0011 1000 0 1110 =

110 1010 0011 0011 0011 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) =
1000 0011

Mantissa (23 bits) =
110 1010 0011 0011 0011 1000


Decimal number 29.275 010 000 011 110 101 000 110 009 63 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0011 - 110 1010 0011 0011 0011 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