1.175 494 350 822 287 507 968 740 21 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.175 494 350 822 287 507 968 740 21(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
1.175 494 350 822 287 507 968 740 21(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.175 494 350 822 287 507 968 740 21.

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) 0.175 494 350 822 287 507 968 740 21 × 2 = 0 + 0.350 988 701 644 575 015 937 480 42;
  • 2) 0.350 988 701 644 575 015 937 480 42 × 2 = 0 + 0.701 977 403 289 150 031 874 960 84;
  • 3) 0.701 977 403 289 150 031 874 960 84 × 2 = 1 + 0.403 954 806 578 300 063 749 921 68;
  • 4) 0.403 954 806 578 300 063 749 921 68 × 2 = 0 + 0.807 909 613 156 600 127 499 843 36;
  • 5) 0.807 909 613 156 600 127 499 843 36 × 2 = 1 + 0.615 819 226 313 200 254 999 686 72;
  • 6) 0.615 819 226 313 200 254 999 686 72 × 2 = 1 + 0.231 638 452 626 400 509 999 373 44;
  • 7) 0.231 638 452 626 400 509 999 373 44 × 2 = 0 + 0.463 276 905 252 801 019 998 746 88;
  • 8) 0.463 276 905 252 801 019 998 746 88 × 2 = 0 + 0.926 553 810 505 602 039 997 493 76;
  • 9) 0.926 553 810 505 602 039 997 493 76 × 2 = 1 + 0.853 107 621 011 204 079 994 987 52;
  • 10) 0.853 107 621 011 204 079 994 987 52 × 2 = 1 + 0.706 215 242 022 408 159 989 975 04;
  • 11) 0.706 215 242 022 408 159 989 975 04 × 2 = 1 + 0.412 430 484 044 816 319 979 950 08;
  • 12) 0.412 430 484 044 816 319 979 950 08 × 2 = 0 + 0.824 860 968 089 632 639 959 900 16;
  • 13) 0.824 860 968 089 632 639 959 900 16 × 2 = 1 + 0.649 721 936 179 265 279 919 800 32;
  • 14) 0.649 721 936 179 265 279 919 800 32 × 2 = 1 + 0.299 443 872 358 530 559 839 600 64;
  • 15) 0.299 443 872 358 530 559 839 600 64 × 2 = 0 + 0.598 887 744 717 061 119 679 201 28;
  • 16) 0.598 887 744 717 061 119 679 201 28 × 2 = 1 + 0.197 775 489 434 122 239 358 402 56;
  • 17) 0.197 775 489 434 122 239 358 402 56 × 2 = 0 + 0.395 550 978 868 244 478 716 805 12;
  • 18) 0.395 550 978 868 244 478 716 805 12 × 2 = 0 + 0.791 101 957 736 488 957 433 610 24;
  • 19) 0.791 101 957 736 488 957 433 610 24 × 2 = 1 + 0.582 203 915 472 977 914 867 220 48;
  • 20) 0.582 203 915 472 977 914 867 220 48 × 2 = 1 + 0.164 407 830 945 955 829 734 440 96;
  • 21) 0.164 407 830 945 955 829 734 440 96 × 2 = 0 + 0.328 815 661 891 911 659 468 881 92;
  • 22) 0.328 815 661 891 911 659 468 881 92 × 2 = 0 + 0.657 631 323 783 823 318 937 763 84;
  • 23) 0.657 631 323 783 823 318 937 763 84 × 2 = 1 + 0.315 262 647 567 646 637 875 527 68;
  • 24) 0.315 262 647 567 646 637 875 527 68 × 2 = 0 + 0.630 525 295 135 293 275 751 055 36;

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:


0.175 494 350 822 287 507 968 740 21(10) =

0.0010 1100 1110 1101 0011 0010(2)

5. Positive number before normalization:

1.175 494 350 822 287 507 968 740 21(10) =

1.0010 1100 1110 1101 0011 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.175 494 350 822 287 507 968 740 21(10) =

1.0010 1100 1110 1101 0011 0010(2) =

1.0010 1100 1110 1101 0011 0010(2) × 20


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): 0

Mantissa (not normalized):
1.0010 1100 1110 1101 0011 0010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(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;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

127(10) =

0111 1111(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. 001 0110 0111 0110 1001 1001 0 =

001 0110 0111 0110 1001 1001


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) =
0111 1111

Mantissa (23 bits) =
001 0110 0111 0110 1001 1001


Decimal number 1.175 494 350 822 287 507 968 740 21 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 001 0110 0111 0110 1001 1001

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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