Decimal to 32 Bit IEEE 754 Binary: Convert Number 5 972 200 000 000 000 000 000 005 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 5 972 200 000 000 000 000 000 005(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

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;
  • 5 972 200 000 000 000 000 000 005 ÷ 2 = 2 986 100 000 000 000 000 000 002 + 1;
  • 2 986 100 000 000 000 000 000 002 ÷ 2 = 1 493 050 000 000 000 000 000 001 + 0;
  • 1 493 050 000 000 000 000 000 001 ÷ 2 = 746 525 000 000 000 000 000 000 + 1;
  • 746 525 000 000 000 000 000 000 ÷ 2 = 373 262 500 000 000 000 000 000 + 0;
  • 373 262 500 000 000 000 000 000 ÷ 2 = 186 631 250 000 000 000 000 000 + 0;
  • 186 631 250 000 000 000 000 000 ÷ 2 = 93 315 625 000 000 000 000 000 + 0;
  • 93 315 625 000 000 000 000 000 ÷ 2 = 46 657 812 500 000 000 000 000 + 0;
  • 46 657 812 500 000 000 000 000 ÷ 2 = 23 328 906 250 000 000 000 000 + 0;
  • 23 328 906 250 000 000 000 000 ÷ 2 = 11 664 453 125 000 000 000 000 + 0;
  • 11 664 453 125 000 000 000 000 ÷ 2 = 5 832 226 562 500 000 000 000 + 0;
  • 5 832 226 562 500 000 000 000 ÷ 2 = 2 916 113 281 250 000 000 000 + 0;
  • 2 916 113 281 250 000 000 000 ÷ 2 = 1 458 056 640 625 000 000 000 + 0;
  • 1 458 056 640 625 000 000 000 ÷ 2 = 729 028 320 312 500 000 000 + 0;
  • 729 028 320 312 500 000 000 ÷ 2 = 364 514 160 156 250 000 000 + 0;
  • 364 514 160 156 250 000 000 ÷ 2 = 182 257 080 078 125 000 000 + 0;
  • 182 257 080 078 125 000 000 ÷ 2 = 91 128 540 039 062 500 000 + 0;
  • 91 128 540 039 062 500 000 ÷ 2 = 45 564 270 019 531 250 000 + 0;
  • 45 564 270 019 531 250 000 ÷ 2 = 22 782 135 009 765 625 000 + 0;
  • 22 782 135 009 765 625 000 ÷ 2 = 11 391 067 504 882 812 500 + 0;
  • 11 391 067 504 882 812 500 ÷ 2 = 5 695 533 752 441 406 250 + 0;
  • 5 695 533 752 441 406 250 ÷ 2 = 2 847 766 876 220 703 125 + 0;
  • 2 847 766 876 220 703 125 ÷ 2 = 1 423 883 438 110 351 562 + 1;
  • 1 423 883 438 110 351 562 ÷ 2 = 711 941 719 055 175 781 + 0;
  • 711 941 719 055 175 781 ÷ 2 = 355 970 859 527 587 890 + 1;
  • 355 970 859 527 587 890 ÷ 2 = 177 985 429 763 793 945 + 0;
  • 177 985 429 763 793 945 ÷ 2 = 88 992 714 881 896 972 + 1;
  • 88 992 714 881 896 972 ÷ 2 = 44 496 357 440 948 486 + 0;
  • 44 496 357 440 948 486 ÷ 2 = 22 248 178 720 474 243 + 0;
  • 22 248 178 720 474 243 ÷ 2 = 11 124 089 360 237 121 + 1;
  • 11 124 089 360 237 121 ÷ 2 = 5 562 044 680 118 560 + 1;
  • 5 562 044 680 118 560 ÷ 2 = 2 781 022 340 059 280 + 0;
  • 2 781 022 340 059 280 ÷ 2 = 1 390 511 170 029 640 + 0;
  • 1 390 511 170 029 640 ÷ 2 = 695 255 585 014 820 + 0;
  • 695 255 585 014 820 ÷ 2 = 347 627 792 507 410 + 0;
  • 347 627 792 507 410 ÷ 2 = 173 813 896 253 705 + 0;
  • 173 813 896 253 705 ÷ 2 = 86 906 948 126 852 + 1;
  • 86 906 948 126 852 ÷ 2 = 43 453 474 063 426 + 0;
  • 43 453 474 063 426 ÷ 2 = 21 726 737 031 713 + 0;
  • 21 726 737 031 713 ÷ 2 = 10 863 368 515 856 + 1;
  • 10 863 368 515 856 ÷ 2 = 5 431 684 257 928 + 0;
  • 5 431 684 257 928 ÷ 2 = 2 715 842 128 964 + 0;
  • 2 715 842 128 964 ÷ 2 = 1 357 921 064 482 + 0;
  • 1 357 921 064 482 ÷ 2 = 678 960 532 241 + 0;
  • 678 960 532 241 ÷ 2 = 339 480 266 120 + 1;
  • 339 480 266 120 ÷ 2 = 169 740 133 060 + 0;
  • 169 740 133 060 ÷ 2 = 84 870 066 530 + 0;
  • 84 870 066 530 ÷ 2 = 42 435 033 265 + 0;
  • 42 435 033 265 ÷ 2 = 21 217 516 632 + 1;
  • 21 217 516 632 ÷ 2 = 10 608 758 316 + 0;
  • 10 608 758 316 ÷ 2 = 5 304 379 158 + 0;
  • 5 304 379 158 ÷ 2 = 2 652 189 579 + 0;
  • 2 652 189 579 ÷ 2 = 1 326 094 789 + 1;
  • 1 326 094 789 ÷ 2 = 663 047 394 + 1;
  • 663 047 394 ÷ 2 = 331 523 697 + 0;
  • 331 523 697 ÷ 2 = 165 761 848 + 1;
  • 165 761 848 ÷ 2 = 82 880 924 + 0;
  • 82 880 924 ÷ 2 = 41 440 462 + 0;
  • 41 440 462 ÷ 2 = 20 720 231 + 0;
  • 20 720 231 ÷ 2 = 10 360 115 + 1;
  • 10 360 115 ÷ 2 = 5 180 057 + 1;
  • 5 180 057 ÷ 2 = 2 590 028 + 1;
  • 2 590 028 ÷ 2 = 1 295 014 + 0;
  • 1 295 014 ÷ 2 = 647 507 + 0;
  • 647 507 ÷ 2 = 323 753 + 1;
  • 323 753 ÷ 2 = 161 876 + 1;
  • 161 876 ÷ 2 = 80 938 + 0;
  • 80 938 ÷ 2 = 40 469 + 0;
  • 40 469 ÷ 2 = 20 234 + 1;
  • 20 234 ÷ 2 = 10 117 + 0;
  • 10 117 ÷ 2 = 5 058 + 1;
  • 5 058 ÷ 2 = 2 529 + 0;
  • 2 529 ÷ 2 = 1 264 + 1;
  • 1 264 ÷ 2 = 632 + 0;
  • 632 ÷ 2 = 316 + 0;
  • 316 ÷ 2 = 158 + 0;
  • 158 ÷ 2 = 79 + 0;
  • 79 ÷ 2 = 39 + 1;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.

5 972 200 000 000 000 000 000 005(10) =


100 1111 0000 1010 1001 1001 1100 0101 1000 1000 1000 0100 1000 0011 0010 1010 0000 0000 0000 0000 0101(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 82 positions to the left, so that only one non zero digit remains to the left of it:


5 972 200 000 000 000 000 000 005(10) =


100 1111 0000 1010 1001 1001 1100 0101 1000 1000 1000 0100 1000 0011 0010 1010 0000 0000 0000 0000 0101(2) =


100 1111 0000 1010 1001 1001 1100 0101 1000 1000 1000 0100 1000 0011 0010 1010 0000 0000 0000 0000 0101(2) × 20 =


1.0011 1100 0010 1010 0110 0111 0001 0110 0010 0010 0001 0010 0000 1100 1010 1000 0000 0000 0000 0001 01(2) × 282


4. 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): 82


Mantissa (not normalized):
1.0011 1100 0010 1010 0110 0111 0001 0110 0010 0010 0001 0010 0000 1100 1010 1000 0000 0000 0000 0001 01


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


82 + 2(8-1) - 1 =


(82 + 127)(10) =


209(10)


6. 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;
  • 209 ÷ 2 = 104 + 1;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

7. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


209(10) =


1101 0001(2)


8. 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 1110 0001 0101 0011 0011 100 0101 1000 1000 1000 0100 1000 0011 0010 1010 0000 0000 0000 0000 0101 =


001 1110 0001 0101 0011 0011


9. 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) =
1101 0001


Mantissa (23 bits) =
001 1110 0001 0101 0011 0011


The base ten decimal number 5 972 200 000 000 000 000 000 005 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

0 - 1101 0001 - 001 1110 0001 0101 0011 0011

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