32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 111 110 100 000 000 000 000 000 000 000 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 111 110 100 000 000 000 000 000 000 000(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (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;
  • 111 110 100 000 000 000 000 000 000 000 ÷ 2 = 55 555 050 000 000 000 000 000 000 000 + 0;
  • 55 555 050 000 000 000 000 000 000 000 ÷ 2 = 27 777 525 000 000 000 000 000 000 000 + 0;
  • 27 777 525 000 000 000 000 000 000 000 ÷ 2 = 13 888 762 500 000 000 000 000 000 000 + 0;
  • 13 888 762 500 000 000 000 000 000 000 ÷ 2 = 6 944 381 250 000 000 000 000 000 000 + 0;
  • 6 944 381 250 000 000 000 000 000 000 ÷ 2 = 3 472 190 625 000 000 000 000 000 000 + 0;
  • 3 472 190 625 000 000 000 000 000 000 ÷ 2 = 1 736 095 312 500 000 000 000 000 000 + 0;
  • 1 736 095 312 500 000 000 000 000 000 ÷ 2 = 868 047 656 250 000 000 000 000 000 + 0;
  • 868 047 656 250 000 000 000 000 000 ÷ 2 = 434 023 828 125 000 000 000 000 000 + 0;
  • 434 023 828 125 000 000 000 000 000 ÷ 2 = 217 011 914 062 500 000 000 000 000 + 0;
  • 217 011 914 062 500 000 000 000 000 ÷ 2 = 108 505 957 031 250 000 000 000 000 + 0;
  • 108 505 957 031 250 000 000 000 000 ÷ 2 = 54 252 978 515 625 000 000 000 000 + 0;
  • 54 252 978 515 625 000 000 000 000 ÷ 2 = 27 126 489 257 812 500 000 000 000 + 0;
  • 27 126 489 257 812 500 000 000 000 ÷ 2 = 13 563 244 628 906 250 000 000 000 + 0;
  • 13 563 244 628 906 250 000 000 000 ÷ 2 = 6 781 622 314 453 125 000 000 000 + 0;
  • 6 781 622 314 453 125 000 000 000 ÷ 2 = 3 390 811 157 226 562 500 000 000 + 0;
  • 3 390 811 157 226 562 500 000 000 ÷ 2 = 1 695 405 578 613 281 250 000 000 + 0;
  • 1 695 405 578 613 281 250 000 000 ÷ 2 = 847 702 789 306 640 625 000 000 + 0;
  • 847 702 789 306 640 625 000 000 ÷ 2 = 423 851 394 653 320 312 500 000 + 0;
  • 423 851 394 653 320 312 500 000 ÷ 2 = 211 925 697 326 660 156 250 000 + 0;
  • 211 925 697 326 660 156 250 000 ÷ 2 = 105 962 848 663 330 078 125 000 + 0;
  • 105 962 848 663 330 078 125 000 ÷ 2 = 52 981 424 331 665 039 062 500 + 0;
  • 52 981 424 331 665 039 062 500 ÷ 2 = 26 490 712 165 832 519 531 250 + 0;
  • 26 490 712 165 832 519 531 250 ÷ 2 = 13 245 356 082 916 259 765 625 + 0;
  • 13 245 356 082 916 259 765 625 ÷ 2 = 6 622 678 041 458 129 882 812 + 1;
  • 6 622 678 041 458 129 882 812 ÷ 2 = 3 311 339 020 729 064 941 406 + 0;
  • 3 311 339 020 729 064 941 406 ÷ 2 = 1 655 669 510 364 532 470 703 + 0;
  • 1 655 669 510 364 532 470 703 ÷ 2 = 827 834 755 182 266 235 351 + 1;
  • 827 834 755 182 266 235 351 ÷ 2 = 413 917 377 591 133 117 675 + 1;
  • 413 917 377 591 133 117 675 ÷ 2 = 206 958 688 795 566 558 837 + 1;
  • 206 958 688 795 566 558 837 ÷ 2 = 103 479 344 397 783 279 418 + 1;
  • 103 479 344 397 783 279 418 ÷ 2 = 51 739 672 198 891 639 709 + 0;
  • 51 739 672 198 891 639 709 ÷ 2 = 25 869 836 099 445 819 854 + 1;
  • 25 869 836 099 445 819 854 ÷ 2 = 12 934 918 049 722 909 927 + 0;
  • 12 934 918 049 722 909 927 ÷ 2 = 6 467 459 024 861 454 963 + 1;
  • 6 467 459 024 861 454 963 ÷ 2 = 3 233 729 512 430 727 481 + 1;
  • 3 233 729 512 430 727 481 ÷ 2 = 1 616 864 756 215 363 740 + 1;
  • 1 616 864 756 215 363 740 ÷ 2 = 808 432 378 107 681 870 + 0;
  • 808 432 378 107 681 870 ÷ 2 = 404 216 189 053 840 935 + 0;
  • 404 216 189 053 840 935 ÷ 2 = 202 108 094 526 920 467 + 1;
  • 202 108 094 526 920 467 ÷ 2 = 101 054 047 263 460 233 + 1;
  • 101 054 047 263 460 233 ÷ 2 = 50 527 023 631 730 116 + 1;
  • 50 527 023 631 730 116 ÷ 2 = 25 263 511 815 865 058 + 0;
  • 25 263 511 815 865 058 ÷ 2 = 12 631 755 907 932 529 + 0;
  • 12 631 755 907 932 529 ÷ 2 = 6 315 877 953 966 264 + 1;
  • 6 315 877 953 966 264 ÷ 2 = 3 157 938 976 983 132 + 0;
  • 3 157 938 976 983 132 ÷ 2 = 1 578 969 488 491 566 + 0;
  • 1 578 969 488 491 566 ÷ 2 = 789 484 744 245 783 + 0;
  • 789 484 744 245 783 ÷ 2 = 394 742 372 122 891 + 1;
  • 394 742 372 122 891 ÷ 2 = 197 371 186 061 445 + 1;
  • 197 371 186 061 445 ÷ 2 = 98 685 593 030 722 + 1;
  • 98 685 593 030 722 ÷ 2 = 49 342 796 515 361 + 0;
  • 49 342 796 515 361 ÷ 2 = 24 671 398 257 680 + 1;
  • 24 671 398 257 680 ÷ 2 = 12 335 699 128 840 + 0;
  • 12 335 699 128 840 ÷ 2 = 6 167 849 564 420 + 0;
  • 6 167 849 564 420 ÷ 2 = 3 083 924 782 210 + 0;
  • 3 083 924 782 210 ÷ 2 = 1 541 962 391 105 + 0;
  • 1 541 962 391 105 ÷ 2 = 770 981 195 552 + 1;
  • 770 981 195 552 ÷ 2 = 385 490 597 776 + 0;
  • 385 490 597 776 ÷ 2 = 192 745 298 888 + 0;
  • 192 745 298 888 ÷ 2 = 96 372 649 444 + 0;
  • 96 372 649 444 ÷ 2 = 48 186 324 722 + 0;
  • 48 186 324 722 ÷ 2 = 24 093 162 361 + 0;
  • 24 093 162 361 ÷ 2 = 12 046 581 180 + 1;
  • 12 046 581 180 ÷ 2 = 6 023 290 590 + 0;
  • 6 023 290 590 ÷ 2 = 3 011 645 295 + 0;
  • 3 011 645 295 ÷ 2 = 1 505 822 647 + 1;
  • 1 505 822 647 ÷ 2 = 752 911 323 + 1;
  • 752 911 323 ÷ 2 = 376 455 661 + 1;
  • 376 455 661 ÷ 2 = 188 227 830 + 1;
  • 188 227 830 ÷ 2 = 94 113 915 + 0;
  • 94 113 915 ÷ 2 = 47 056 957 + 1;
  • 47 056 957 ÷ 2 = 23 528 478 + 1;
  • 23 528 478 ÷ 2 = 11 764 239 + 0;
  • 11 764 239 ÷ 2 = 5 882 119 + 1;
  • 5 882 119 ÷ 2 = 2 941 059 + 1;
  • 2 941 059 ÷ 2 = 1 470 529 + 1;
  • 1 470 529 ÷ 2 = 735 264 + 1;
  • 735 264 ÷ 2 = 367 632 + 0;
  • 367 632 ÷ 2 = 183 816 + 0;
  • 183 816 ÷ 2 = 91 908 + 0;
  • 91 908 ÷ 2 = 45 954 + 0;
  • 45 954 ÷ 2 = 22 977 + 0;
  • 22 977 ÷ 2 = 11 488 + 1;
  • 11 488 ÷ 2 = 5 744 + 0;
  • 5 744 ÷ 2 = 2 872 + 0;
  • 2 872 ÷ 2 = 1 436 + 0;
  • 1 436 ÷ 2 = 718 + 0;
  • 718 ÷ 2 = 359 + 0;
  • 359 ÷ 2 = 179 + 1;
  • 179 ÷ 2 = 89 + 1;
  • 89 ÷ 2 = 44 + 1;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 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.


111 110 100 000 000 000 000 000 000 000(10) =


1 0110 0111 0000 0100 0001 1110 1101 1110 0100 0001 0000 1011 1000 1001 1100 1110 1011 1100 1000 0000 0000 0000 0000 0000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 96 positions to the left, so that only one non zero digit remains to the left of it:


111 110 100 000 000 000 000 000 000 000(10) =


1 0110 0111 0000 0100 0001 1110 1101 1110 0100 0001 0000 1011 1000 1001 1100 1110 1011 1100 1000 0000 0000 0000 0000 0000(2) =


1 0110 0111 0000 0100 0001 1110 1101 1110 0100 0001 0000 1011 1000 1001 1100 1110 1011 1100 1000 0000 0000 0000 0000 0000(2) × 20 =


1.0110 0111 0000 0100 0001 1110 1101 1110 0100 0001 0000 1011 1000 1001 1100 1110 1011 1100 1000 0000 0000 0000 0000 0000(2) × 296


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): 96


Mantissa (not normalized):
1.0110 0111 0000 0100 0001 1110 1101 1110 0100 0001 0000 1011 1000 1001 1100 1110 1011 1100 1000 0000 0000 0000 0000 0000


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


96 + 2(8-1) - 1 =


(96 + 127)(10) =


223(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;
  • 223 ÷ 2 = 111 + 1;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 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) =


223(10) =


1101 1111(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. 011 0011 1000 0010 0000 1111 0 1101 1110 0100 0001 0000 1011 1000 1001 1100 1110 1011 1100 1000 0000 0000 0000 0000 0000 =


011 0011 1000 0010 0000 1111


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 1111


Mantissa (23 bits) =
011 0011 1000 0010 0000 1111


The base ten decimal number 111 110 100 000 000 000 000 000 000 000 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1101 1111 - 011 0011 1000 0010 0000 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