0.734 683 969 263 929 692 480 460 335 84 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.734 683 969 263 929 692 480 460 335 84(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
0.734 683 969 263 929 692 480 460 335 84(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: 0.734 683 969 263 929 692 480 460 335 84.

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) 0.734 683 969 263 929 692 480 460 335 84 × 2 = 1 + 0.469 367 938 527 859 384 960 920 671 68;
  • 2) 0.469 367 938 527 859 384 960 920 671 68 × 2 = 0 + 0.938 735 877 055 718 769 921 841 343 36;
  • 3) 0.938 735 877 055 718 769 921 841 343 36 × 2 = 1 + 0.877 471 754 111 437 539 843 682 686 72;
  • 4) 0.877 471 754 111 437 539 843 682 686 72 × 2 = 1 + 0.754 943 508 222 875 079 687 365 373 44;
  • 5) 0.754 943 508 222 875 079 687 365 373 44 × 2 = 1 + 0.509 887 016 445 750 159 374 730 746 88;
  • 6) 0.509 887 016 445 750 159 374 730 746 88 × 2 = 1 + 0.019 774 032 891 500 318 749 461 493 76;
  • 7) 0.019 774 032 891 500 318 749 461 493 76 × 2 = 0 + 0.039 548 065 783 000 637 498 922 987 52;
  • 8) 0.039 548 065 783 000 637 498 922 987 52 × 2 = 0 + 0.079 096 131 566 001 274 997 845 975 04;
  • 9) 0.079 096 131 566 001 274 997 845 975 04 × 2 = 0 + 0.158 192 263 132 002 549 995 691 950 08;
  • 10) 0.158 192 263 132 002 549 995 691 950 08 × 2 = 0 + 0.316 384 526 264 005 099 991 383 900 16;
  • 11) 0.316 384 526 264 005 099 991 383 900 16 × 2 = 0 + 0.632 769 052 528 010 199 982 767 800 32;
  • 12) 0.632 769 052 528 010 199 982 767 800 32 × 2 = 1 + 0.265 538 105 056 020 399 965 535 600 64;
  • 13) 0.265 538 105 056 020 399 965 535 600 64 × 2 = 0 + 0.531 076 210 112 040 799 931 071 201 28;
  • 14) 0.531 076 210 112 040 799 931 071 201 28 × 2 = 1 + 0.062 152 420 224 081 599 862 142 402 56;
  • 15) 0.062 152 420 224 081 599 862 142 402 56 × 2 = 0 + 0.124 304 840 448 163 199 724 284 805 12;
  • 16) 0.124 304 840 448 163 199 724 284 805 12 × 2 = 0 + 0.248 609 680 896 326 399 448 569 610 24;
  • 17) 0.248 609 680 896 326 399 448 569 610 24 × 2 = 0 + 0.497 219 361 792 652 798 897 139 220 48;
  • 18) 0.497 219 361 792 652 798 897 139 220 48 × 2 = 0 + 0.994 438 723 585 305 597 794 278 440 96;
  • 19) 0.994 438 723 585 305 597 794 278 440 96 × 2 = 1 + 0.988 877 447 170 611 195 588 556 881 92;
  • 20) 0.988 877 447 170 611 195 588 556 881 92 × 2 = 1 + 0.977 754 894 341 222 391 177 113 763 84;
  • 21) 0.977 754 894 341 222 391 177 113 763 84 × 2 = 1 + 0.955 509 788 682 444 782 354 227 527 68;
  • 22) 0.955 509 788 682 444 782 354 227 527 68 × 2 = 1 + 0.911 019 577 364 889 564 708 455 055 36;
  • 23) 0.911 019 577 364 889 564 708 455 055 36 × 2 = 1 + 0.822 039 154 729 779 129 416 910 110 72;
  • 24) 0.822 039 154 729 779 129 416 910 110 72 × 2 = 1 + 0.644 078 309 459 558 258 833 820 221 44;

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:


0.734 683 969 263 929 692 480 460 335 84(10) =

0.1011 1100 0001 0100 0011 1111(2)

5. Positive number before normalization:

0.734 683 969 263 929 692 480 460 335 84(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:


0.734 683 969 263 929 692 480 460 335 84(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 0.734 683 969 263 929 692 480 460 335 84 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