0.734 683 969 263 929 692 480 460 336 203 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.734 683 969 263 929 692 480 460 336 203(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
0.734 683 969 263 929 692 480 460 336 203(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: 0.734 683 969 263 929 692 480 460 336 203.

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) 0.734 683 969 263 929 692 480 460 336 203 × 2 = 1 + 0.469 367 938 527 859 384 960 920 672 406;
  • 2) 0.469 367 938 527 859 384 960 920 672 406 × 2 = 0 + 0.938 735 877 055 718 769 921 841 344 812;
  • 3) 0.938 735 877 055 718 769 921 841 344 812 × 2 = 1 + 0.877 471 754 111 437 539 843 682 689 624;
  • 4) 0.877 471 754 111 437 539 843 682 689 624 × 2 = 1 + 0.754 943 508 222 875 079 687 365 379 248;
  • 5) 0.754 943 508 222 875 079 687 365 379 248 × 2 = 1 + 0.509 887 016 445 750 159 374 730 758 496;
  • 6) 0.509 887 016 445 750 159 374 730 758 496 × 2 = 1 + 0.019 774 032 891 500 318 749 461 516 992;
  • 7) 0.019 774 032 891 500 318 749 461 516 992 × 2 = 0 + 0.039 548 065 783 000 637 498 923 033 984;
  • 8) 0.039 548 065 783 000 637 498 923 033 984 × 2 = 0 + 0.079 096 131 566 001 274 997 846 067 968;
  • 9) 0.079 096 131 566 001 274 997 846 067 968 × 2 = 0 + 0.158 192 263 132 002 549 995 692 135 936;
  • 10) 0.158 192 263 132 002 549 995 692 135 936 × 2 = 0 + 0.316 384 526 264 005 099 991 384 271 872;
  • 11) 0.316 384 526 264 005 099 991 384 271 872 × 2 = 0 + 0.632 769 052 528 010 199 982 768 543 744;
  • 12) 0.632 769 052 528 010 199 982 768 543 744 × 2 = 1 + 0.265 538 105 056 020 399 965 537 087 488;
  • 13) 0.265 538 105 056 020 399 965 537 087 488 × 2 = 0 + 0.531 076 210 112 040 799 931 074 174 976;
  • 14) 0.531 076 210 112 040 799 931 074 174 976 × 2 = 1 + 0.062 152 420 224 081 599 862 148 349 952;
  • 15) 0.062 152 420 224 081 599 862 148 349 952 × 2 = 0 + 0.124 304 840 448 163 199 724 296 699 904;
  • 16) 0.124 304 840 448 163 199 724 296 699 904 × 2 = 0 + 0.248 609 680 896 326 399 448 593 399 808;
  • 17) 0.248 609 680 896 326 399 448 593 399 808 × 2 = 0 + 0.497 219 361 792 652 798 897 186 799 616;
  • 18) 0.497 219 361 792 652 798 897 186 799 616 × 2 = 0 + 0.994 438 723 585 305 597 794 373 599 232;
  • 19) 0.994 438 723 585 305 597 794 373 599 232 × 2 = 1 + 0.988 877 447 170 611 195 588 747 198 464;
  • 20) 0.988 877 447 170 611 195 588 747 198 464 × 2 = 1 + 0.977 754 894 341 222 391 177 494 396 928;
  • 21) 0.977 754 894 341 222 391 177 494 396 928 × 2 = 1 + 0.955 509 788 682 444 782 354 988 793 856;
  • 22) 0.955 509 788 682 444 782 354 988 793 856 × 2 = 1 + 0.911 019 577 364 889 564 709 977 587 712;
  • 23) 0.911 019 577 364 889 564 709 977 587 712 × 2 = 1 + 0.822 039 154 729 779 129 419 955 175 424;
  • 24) 0.822 039 154 729 779 129 419 955 175 424 × 2 = 1 + 0.644 078 309 459 558 258 839 910 350 848;

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:


0.734 683 969 263 929 692 480 460 336 203(10) =

0.1011 1100 0001 0100 0011 1111(2)

5. Positive number before normalization:

0.734 683 969 263 929 692 480 460 336 203(10) =

0.1011 1100 0001 0100 0011 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.734 683 969 263 929 692 480 460 336 203(10) =

0.1011 1100 0001 0100 0011 1111(2) =

0.1011 1100 0001 0100 0011 1111(2) × 20 =

1.0111 1000 0010 1000 0111 111(2) × 2-1


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): -1

Mantissa (not normalized):
1.0111 1000 0010 1000 0111 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 127)(10) =

126(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;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

126(10) =

0111 1110(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. 011 1100 0001 0100 0011 1111 =

011 1100 0001 0100 0011 1111


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 1110

Mantissa (23 bits) =
011 1100 0001 0100 0011 1111


Decimal number 0.734 683 969 263 929 692 480 460 336 203 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1110 - 011 1100 0001 0100 0011 1111

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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