32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 1 000 100 010 010 000 000 000 000 000 006 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 1 000 100 010 010 000 000 000 000 000 006(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;
  • 1 000 100 010 010 000 000 000 000 000 006 ÷ 2 = 500 050 005 005 000 000 000 000 000 003 + 0;
  • 500 050 005 005 000 000 000 000 000 003 ÷ 2 = 250 025 002 502 500 000 000 000 000 001 + 1;
  • 250 025 002 502 500 000 000 000 000 001 ÷ 2 = 125 012 501 251 250 000 000 000 000 000 + 1;
  • 125 012 501 251 250 000 000 000 000 000 ÷ 2 = 62 506 250 625 625 000 000 000 000 000 + 0;
  • 62 506 250 625 625 000 000 000 000 000 ÷ 2 = 31 253 125 312 812 500 000 000 000 000 + 0;
  • 31 253 125 312 812 500 000 000 000 000 ÷ 2 = 15 626 562 656 406 250 000 000 000 000 + 0;
  • 15 626 562 656 406 250 000 000 000 000 ÷ 2 = 7 813 281 328 203 125 000 000 000 000 + 0;
  • 7 813 281 328 203 125 000 000 000 000 ÷ 2 = 3 906 640 664 101 562 500 000 000 000 + 0;
  • 3 906 640 664 101 562 500 000 000 000 ÷ 2 = 1 953 320 332 050 781 250 000 000 000 + 0;
  • 1 953 320 332 050 781 250 000 000 000 ÷ 2 = 976 660 166 025 390 625 000 000 000 + 0;
  • 976 660 166 025 390 625 000 000 000 ÷ 2 = 488 330 083 012 695 312 500 000 000 + 0;
  • 488 330 083 012 695 312 500 000 000 ÷ 2 = 244 165 041 506 347 656 250 000 000 + 0;
  • 244 165 041 506 347 656 250 000 000 ÷ 2 = 122 082 520 753 173 828 125 000 000 + 0;
  • 122 082 520 753 173 828 125 000 000 ÷ 2 = 61 041 260 376 586 914 062 500 000 + 0;
  • 61 041 260 376 586 914 062 500 000 ÷ 2 = 30 520 630 188 293 457 031 250 000 + 0;
  • 30 520 630 188 293 457 031 250 000 ÷ 2 = 15 260 315 094 146 728 515 625 000 + 0;
  • 15 260 315 094 146 728 515 625 000 ÷ 2 = 7 630 157 547 073 364 257 812 500 + 0;
  • 7 630 157 547 073 364 257 812 500 ÷ 2 = 3 815 078 773 536 682 128 906 250 + 0;
  • 3 815 078 773 536 682 128 906 250 ÷ 2 = 1 907 539 386 768 341 064 453 125 + 0;
  • 1 907 539 386 768 341 064 453 125 ÷ 2 = 953 769 693 384 170 532 226 562 + 1;
  • 953 769 693 384 170 532 226 562 ÷ 2 = 476 884 846 692 085 266 113 281 + 0;
  • 476 884 846 692 085 266 113 281 ÷ 2 = 238 442 423 346 042 633 056 640 + 1;
  • 238 442 423 346 042 633 056 640 ÷ 2 = 119 221 211 673 021 316 528 320 + 0;
  • 119 221 211 673 021 316 528 320 ÷ 2 = 59 610 605 836 510 658 264 160 + 0;
  • 59 610 605 836 510 658 264 160 ÷ 2 = 29 805 302 918 255 329 132 080 + 0;
  • 29 805 302 918 255 329 132 080 ÷ 2 = 14 902 651 459 127 664 566 040 + 0;
  • 14 902 651 459 127 664 566 040 ÷ 2 = 7 451 325 729 563 832 283 020 + 0;
  • 7 451 325 729 563 832 283 020 ÷ 2 = 3 725 662 864 781 916 141 510 + 0;
  • 3 725 662 864 781 916 141 510 ÷ 2 = 1 862 831 432 390 958 070 755 + 0;
  • 1 862 831 432 390 958 070 755 ÷ 2 = 931 415 716 195 479 035 377 + 1;
  • 931 415 716 195 479 035 377 ÷ 2 = 465 707 858 097 739 517 688 + 1;
  • 465 707 858 097 739 517 688 ÷ 2 = 232 853 929 048 869 758 844 + 0;
  • 232 853 929 048 869 758 844 ÷ 2 = 116 426 964 524 434 879 422 + 0;
  • 116 426 964 524 434 879 422 ÷ 2 = 58 213 482 262 217 439 711 + 0;
  • 58 213 482 262 217 439 711 ÷ 2 = 29 106 741 131 108 719 855 + 1;
  • 29 106 741 131 108 719 855 ÷ 2 = 14 553 370 565 554 359 927 + 1;
  • 14 553 370 565 554 359 927 ÷ 2 = 7 276 685 282 777 179 963 + 1;
  • 7 276 685 282 777 179 963 ÷ 2 = 3 638 342 641 388 589 981 + 1;
  • 3 638 342 641 388 589 981 ÷ 2 = 1 819 171 320 694 294 990 + 1;
  • 1 819 171 320 694 294 990 ÷ 2 = 909 585 660 347 147 495 + 0;
  • 909 585 660 347 147 495 ÷ 2 = 454 792 830 173 573 747 + 1;
  • 454 792 830 173 573 747 ÷ 2 = 227 396 415 086 786 873 + 1;
  • 227 396 415 086 786 873 ÷ 2 = 113 698 207 543 393 436 + 1;
  • 113 698 207 543 393 436 ÷ 2 = 56 849 103 771 696 718 + 0;
  • 56 849 103 771 696 718 ÷ 2 = 28 424 551 885 848 359 + 0;
  • 28 424 551 885 848 359 ÷ 2 = 14 212 275 942 924 179 + 1;
  • 14 212 275 942 924 179 ÷ 2 = 7 106 137 971 462 089 + 1;
  • 7 106 137 971 462 089 ÷ 2 = 3 553 068 985 731 044 + 1;
  • 3 553 068 985 731 044 ÷ 2 = 1 776 534 492 865 522 + 0;
  • 1 776 534 492 865 522 ÷ 2 = 888 267 246 432 761 + 0;
  • 888 267 246 432 761 ÷ 2 = 444 133 623 216 380 + 1;
  • 444 133 623 216 380 ÷ 2 = 222 066 811 608 190 + 0;
  • 222 066 811 608 190 ÷ 2 = 111 033 405 804 095 + 0;
  • 111 033 405 804 095 ÷ 2 = 55 516 702 902 047 + 1;
  • 55 516 702 902 047 ÷ 2 = 27 758 351 451 023 + 1;
  • 27 758 351 451 023 ÷ 2 = 13 879 175 725 511 + 1;
  • 13 879 175 725 511 ÷ 2 = 6 939 587 862 755 + 1;
  • 6 939 587 862 755 ÷ 2 = 3 469 793 931 377 + 1;
  • 3 469 793 931 377 ÷ 2 = 1 734 896 965 688 + 1;
  • 1 734 896 965 688 ÷ 2 = 867 448 482 844 + 0;
  • 867 448 482 844 ÷ 2 = 433 724 241 422 + 0;
  • 433 724 241 422 ÷ 2 = 216 862 120 711 + 0;
  • 216 862 120 711 ÷ 2 = 108 431 060 355 + 1;
  • 108 431 060 355 ÷ 2 = 54 215 530 177 + 1;
  • 54 215 530 177 ÷ 2 = 27 107 765 088 + 1;
  • 27 107 765 088 ÷ 2 = 13 553 882 544 + 0;
  • 13 553 882 544 ÷ 2 = 6 776 941 272 + 0;
  • 6 776 941 272 ÷ 2 = 3 388 470 636 + 0;
  • 3 388 470 636 ÷ 2 = 1 694 235 318 + 0;
  • 1 694 235 318 ÷ 2 = 847 117 659 + 0;
  • 847 117 659 ÷ 2 = 423 558 829 + 1;
  • 423 558 829 ÷ 2 = 211 779 414 + 1;
  • 211 779 414 ÷ 2 = 105 889 707 + 0;
  • 105 889 707 ÷ 2 = 52 944 853 + 1;
  • 52 944 853 ÷ 2 = 26 472 426 + 1;
  • 26 472 426 ÷ 2 = 13 236 213 + 0;
  • 13 236 213 ÷ 2 = 6 618 106 + 1;
  • 6 618 106 ÷ 2 = 3 309 053 + 0;
  • 3 309 053 ÷ 2 = 1 654 526 + 1;
  • 1 654 526 ÷ 2 = 827 263 + 0;
  • 827 263 ÷ 2 = 413 631 + 1;
  • 413 631 ÷ 2 = 206 815 + 1;
  • 206 815 ÷ 2 = 103 407 + 1;
  • 103 407 ÷ 2 = 51 703 + 1;
  • 51 703 ÷ 2 = 25 851 + 1;
  • 25 851 ÷ 2 = 12 925 + 1;
  • 12 925 ÷ 2 = 6 462 + 1;
  • 6 462 ÷ 2 = 3 231 + 0;
  • 3 231 ÷ 2 = 1 615 + 1;
  • 1 615 ÷ 2 = 807 + 1;
  • 807 ÷ 2 = 403 + 1;
  • 403 ÷ 2 = 201 + 1;
  • 201 ÷ 2 = 100 + 1;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.


1 000 100 010 010 000 000 000 000 000 006(10) =


1100 1001 1111 0111 1111 0101 0110 1100 0001 1100 0111 1110 0100 1110 0111 0111 1100 0110 0000 0010 1000 0000 0000 0000 0110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 99 positions to the left, so that only one non zero digit remains to the left of it:


1 000 100 010 010 000 000 000 000 000 006(10) =


1100 1001 1111 0111 1111 0101 0110 1100 0001 1100 0111 1110 0100 1110 0111 0111 1100 0110 0000 0010 1000 0000 0000 0000 0110(2) =


1100 1001 1111 0111 1111 0101 0110 1100 0001 1100 0111 1110 0100 1110 0111 0111 1100 0110 0000 0010 1000 0000 0000 0000 0110(2) × 20 =


1.1001 0011 1110 1111 1110 1010 1101 1000 0011 1000 1111 1100 1001 1100 1110 1111 1000 1100 0000 0101 0000 0000 0000 0000 110(2) × 299


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): 99


Mantissa (not normalized):
1.1001 0011 1110 1111 1110 1010 1101 1000 0011 1000 1111 1100 1001 1100 1110 1111 1000 1100 0000 0101 0000 0000 0000 0000 110


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


99 + 2(8-1) - 1 =


(99 + 127)(10) =


226(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;
  • 226 ÷ 2 = 113 + 0;
  • 113 ÷ 2 = 56 + 1;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =


226(10) =


1110 0010(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. 100 1001 1111 0111 1111 0101 0110 1100 0001 1100 0111 1110 0100 1110 0111 0111 1100 0110 0000 0010 1000 0000 0000 0000 0110 =


100 1001 1111 0111 1111 0101


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) =
1110 0010


Mantissa (23 bits) =
100 1001 1111 0111 1111 0101


The base ten decimal number 1 000 100 010 010 000 000 000 000 000 006 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1110 0010 - 100 1001 1111 0111 1111 0101

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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