4 722 365 919 919 691 792 383.999 511 707 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 4 722 365 919 919 691 792 383.999 511 707 6(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
4 722 365 919 919 691 792 383.999 511 707 6(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: 4 722 365 919 919 691 792 383.
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;
  • 4 722 365 919 919 691 792 383 ÷ 2 = 2 361 182 959 959 845 896 191 + 1;
  • 2 361 182 959 959 845 896 191 ÷ 2 = 1 180 591 479 979 922 948 095 + 1;
  • 1 180 591 479 979 922 948 095 ÷ 2 = 590 295 739 989 961 474 047 + 1;
  • 590 295 739 989 961 474 047 ÷ 2 = 295 147 869 994 980 737 023 + 1;
  • 295 147 869 994 980 737 023 ÷ 2 = 147 573 934 997 490 368 511 + 1;
  • 147 573 934 997 490 368 511 ÷ 2 = 73 786 967 498 745 184 255 + 1;
  • 73 786 967 498 745 184 255 ÷ 2 = 36 893 483 749 372 592 127 + 1;
  • 36 893 483 749 372 592 127 ÷ 2 = 18 446 741 874 686 296 063 + 1;
  • 18 446 741 874 686 296 063 ÷ 2 = 9 223 370 937 343 148 031 + 1;
  • 9 223 370 937 343 148 031 ÷ 2 = 4 611 685 468 671 574 015 + 1;
  • 4 611 685 468 671 574 015 ÷ 2 = 2 305 842 734 335 787 007 + 1;
  • 2 305 842 734 335 787 007 ÷ 2 = 1 152 921 367 167 893 503 + 1;
  • 1 152 921 367 167 893 503 ÷ 2 = 576 460 683 583 946 751 + 1;
  • 576 460 683 583 946 751 ÷ 2 = 288 230 341 791 973 375 + 1;
  • 288 230 341 791 973 375 ÷ 2 = 144 115 170 895 986 687 + 1;
  • 144 115 170 895 986 687 ÷ 2 = 72 057 585 447 993 343 + 1;
  • 72 057 585 447 993 343 ÷ 2 = 36 028 792 723 996 671 + 1;
  • 36 028 792 723 996 671 ÷ 2 = 18 014 396 361 998 335 + 1;
  • 18 014 396 361 998 335 ÷ 2 = 9 007 198 180 999 167 + 1;
  • 9 007 198 180 999 167 ÷ 2 = 4 503 599 090 499 583 + 1;
  • 4 503 599 090 499 583 ÷ 2 = 2 251 799 545 249 791 + 1;
  • 2 251 799 545 249 791 ÷ 2 = 1 125 899 772 624 895 + 1;
  • 1 125 899 772 624 895 ÷ 2 = 562 949 886 312 447 + 1;
  • 562 949 886 312 447 ÷ 2 = 281 474 943 156 223 + 1;
  • 281 474 943 156 223 ÷ 2 = 140 737 471 578 111 + 1;
  • 140 737 471 578 111 ÷ 2 = 70 368 735 789 055 + 1;
  • 70 368 735 789 055 ÷ 2 = 35 184 367 894 527 + 1;
  • 35 184 367 894 527 ÷ 2 = 17 592 183 947 263 + 1;
  • 17 592 183 947 263 ÷ 2 = 8 796 091 973 631 + 1;
  • 8 796 091 973 631 ÷ 2 = 4 398 045 986 815 + 1;
  • 4 398 045 986 815 ÷ 2 = 2 199 022 993 407 + 1;
  • 2 199 022 993 407 ÷ 2 = 1 099 511 496 703 + 1;
  • 1 099 511 496 703 ÷ 2 = 549 755 748 351 + 1;
  • 549 755 748 351 ÷ 2 = 274 877 874 175 + 1;
  • 274 877 874 175 ÷ 2 = 137 438 937 087 + 1;
  • 137 438 937 087 ÷ 2 = 68 719 468 543 + 1;
  • 68 719 468 543 ÷ 2 = 34 359 734 271 + 1;
  • 34 359 734 271 ÷ 2 = 17 179 867 135 + 1;
  • 17 179 867 135 ÷ 2 = 8 589 933 567 + 1;
  • 8 589 933 567 ÷ 2 = 4 294 966 783 + 1;
  • 4 294 966 783 ÷ 2 = 2 147 483 391 + 1;
  • 2 147 483 391 ÷ 2 = 1 073 741 695 + 1;
  • 1 073 741 695 ÷ 2 = 536 870 847 + 1;
  • 536 870 847 ÷ 2 = 268 435 423 + 1;
  • 268 435 423 ÷ 2 = 134 217 711 + 1;
  • 134 217 711 ÷ 2 = 67 108 855 + 1;
  • 67 108 855 ÷ 2 = 33 554 427 + 1;
  • 33 554 427 ÷ 2 = 16 777 213 + 1;
  • 16 777 213 ÷ 2 = 8 388 606 + 1;
  • 8 388 606 ÷ 2 = 4 194 303 + 0;
  • 4 194 303 ÷ 2 = 2 097 151 + 1;
  • 2 097 151 ÷ 2 = 1 048 575 + 1;
  • 1 048 575 ÷ 2 = 524 287 + 1;
  • 524 287 ÷ 2 = 262 143 + 1;
  • 262 143 ÷ 2 = 131 071 + 1;
  • 131 071 ÷ 2 = 65 535 + 1;
  • 65 535 ÷ 2 = 32 767 + 1;
  • 32 767 ÷ 2 = 16 383 + 1;
  • 16 383 ÷ 2 = 8 191 + 1;
  • 8 191 ÷ 2 = 4 095 + 1;
  • 4 095 ÷ 2 = 2 047 + 1;
  • 2 047 ÷ 2 = 1 023 + 1;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

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.

4 722 365 919 919 691 792 383(10) =

1111 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2)


3. Convert to binary (base 2) the fractional part: 0.999 511 707 6.

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) 0.999 511 707 6 × 2 = 1 + 0.999 023 415 2;
  • 2) 0.999 023 415 2 × 2 = 1 + 0.998 046 830 4;
  • 3) 0.998 046 830 4 × 2 = 1 + 0.996 093 660 8;
  • 4) 0.996 093 660 8 × 2 = 1 + 0.992 187 321 6;
  • 5) 0.992 187 321 6 × 2 = 1 + 0.984 374 643 2;
  • 6) 0.984 374 643 2 × 2 = 1 + 0.968 749 286 4;
  • 7) 0.968 749 286 4 × 2 = 1 + 0.937 498 572 8;
  • 8) 0.937 498 572 8 × 2 = 1 + 0.874 997 145 6;
  • 9) 0.874 997 145 6 × 2 = 1 + 0.749 994 291 2;
  • 10) 0.749 994 291 2 × 2 = 1 + 0.499 988 582 4;
  • 11) 0.499 988 582 4 × 2 = 0 + 0.999 977 164 8;
  • 12) 0.999 977 164 8 × 2 = 1 + 0.999 954 329 6;
  • 13) 0.999 954 329 6 × 2 = 1 + 0.999 908 659 2;
  • 14) 0.999 908 659 2 × 2 = 1 + 0.999 817 318 4;
  • 15) 0.999 817 318 4 × 2 = 1 + 0.999 634 636 8;
  • 16) 0.999 634 636 8 × 2 = 1 + 0.999 269 273 6;
  • 17) 0.999 269 273 6 × 2 = 1 + 0.998 538 547 2;
  • 18) 0.998 538 547 2 × 2 = 1 + 0.997 077 094 4;
  • 19) 0.997 077 094 4 × 2 = 1 + 0.994 154 188 8;
  • 20) 0.994 154 188 8 × 2 = 1 + 0.988 308 377 6;
  • 21) 0.988 308 377 6 × 2 = 1 + 0.976 616 755 2;
  • 22) 0.976 616 755 2 × 2 = 1 + 0.953 233 510 4;
  • 23) 0.953 233 510 4 × 2 = 1 + 0.906 467 020 8;
  • 24) 0.906 467 020 8 × 2 = 1 + 0.812 934 041 6;

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:


0.999 511 707 6(10) =

0.1111 1111 1101 1111 1111 1111(2)

5. Positive number before normalization:

4 722 365 919 919 691 792 383.999 511 707 6(10) =

1111 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111.1111 1111 1101 1111 1111 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 71 positions to the left, so that only one non zero digit remains to the left of it:


4 722 365 919 919 691 792 383.999 511 707 6(10) =

1111 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111.1111 1111 1101 1111 1111 1111(2) =

1111 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111.1111 1111 1101 1111 1111 1111(2) × 20 =

1.1111 1111 1111 1111 1111 1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1011 1111 1111 111(2) × 271


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): 71

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1011 1111 1111 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

71 + 2(8-1) - 1 =

(71 + 127)(10) =

198(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;
  • 198 ÷ 2 = 99 + 0;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

198(10) =

1100 0110(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. 111 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1111 1111 =

111 1111 1111 1111 1111 1101


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) =
1100 0110

Mantissa (23 bits) =
111 1111 1111 1111 1111 1101


Decimal number 4 722 365 919 919 691 792 383.999 511 707 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1100 0110 - 111 1111 1111 1111 1111 1101

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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