32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 11 000 001 110 101 109 999 999 999 999 978 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 11 000 001 110 101 109 999 999 999 999 978(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;
  • 11 000 001 110 101 109 999 999 999 999 978 ÷ 2 = 5 500 000 555 050 554 999 999 999 999 989 + 0;
  • 5 500 000 555 050 554 999 999 999 999 989 ÷ 2 = 2 750 000 277 525 277 499 999 999 999 994 + 1;
  • 2 750 000 277 525 277 499 999 999 999 994 ÷ 2 = 1 375 000 138 762 638 749 999 999 999 997 + 0;
  • 1 375 000 138 762 638 749 999 999 999 997 ÷ 2 = 687 500 069 381 319 374 999 999 999 998 + 1;
  • 687 500 069 381 319 374 999 999 999 998 ÷ 2 = 343 750 034 690 659 687 499 999 999 999 + 0;
  • 343 750 034 690 659 687 499 999 999 999 ÷ 2 = 171 875 017 345 329 843 749 999 999 999 + 1;
  • 171 875 017 345 329 843 749 999 999 999 ÷ 2 = 85 937 508 672 664 921 874 999 999 999 + 1;
  • 85 937 508 672 664 921 874 999 999 999 ÷ 2 = 42 968 754 336 332 460 937 499 999 999 + 1;
  • 42 968 754 336 332 460 937 499 999 999 ÷ 2 = 21 484 377 168 166 230 468 749 999 999 + 1;
  • 21 484 377 168 166 230 468 749 999 999 ÷ 2 = 10 742 188 584 083 115 234 374 999 999 + 1;
  • 10 742 188 584 083 115 234 374 999 999 ÷ 2 = 5 371 094 292 041 557 617 187 499 999 + 1;
  • 5 371 094 292 041 557 617 187 499 999 ÷ 2 = 2 685 547 146 020 778 808 593 749 999 + 1;
  • 2 685 547 146 020 778 808 593 749 999 ÷ 2 = 1 342 773 573 010 389 404 296 874 999 + 1;
  • 1 342 773 573 010 389 404 296 874 999 ÷ 2 = 671 386 786 505 194 702 148 437 499 + 1;
  • 671 386 786 505 194 702 148 437 499 ÷ 2 = 335 693 393 252 597 351 074 218 749 + 1;
  • 335 693 393 252 597 351 074 218 749 ÷ 2 = 167 846 696 626 298 675 537 109 374 + 1;
  • 167 846 696 626 298 675 537 109 374 ÷ 2 = 83 923 348 313 149 337 768 554 687 + 0;
  • 83 923 348 313 149 337 768 554 687 ÷ 2 = 41 961 674 156 574 668 884 277 343 + 1;
  • 41 961 674 156 574 668 884 277 343 ÷ 2 = 20 980 837 078 287 334 442 138 671 + 1;
  • 20 980 837 078 287 334 442 138 671 ÷ 2 = 10 490 418 539 143 667 221 069 335 + 1;
  • 10 490 418 539 143 667 221 069 335 ÷ 2 = 5 245 209 269 571 833 610 534 667 + 1;
  • 5 245 209 269 571 833 610 534 667 ÷ 2 = 2 622 604 634 785 916 805 267 333 + 1;
  • 2 622 604 634 785 916 805 267 333 ÷ 2 = 1 311 302 317 392 958 402 633 666 + 1;
  • 1 311 302 317 392 958 402 633 666 ÷ 2 = 655 651 158 696 479 201 316 833 + 0;
  • 655 651 158 696 479 201 316 833 ÷ 2 = 327 825 579 348 239 600 658 416 + 1;
  • 327 825 579 348 239 600 658 416 ÷ 2 = 163 912 789 674 119 800 329 208 + 0;
  • 163 912 789 674 119 800 329 208 ÷ 2 = 81 956 394 837 059 900 164 604 + 0;
  • 81 956 394 837 059 900 164 604 ÷ 2 = 40 978 197 418 529 950 082 302 + 0;
  • 40 978 197 418 529 950 082 302 ÷ 2 = 20 489 098 709 264 975 041 151 + 0;
  • 20 489 098 709 264 975 041 151 ÷ 2 = 10 244 549 354 632 487 520 575 + 1;
  • 10 244 549 354 632 487 520 575 ÷ 2 = 5 122 274 677 316 243 760 287 + 1;
  • 5 122 274 677 316 243 760 287 ÷ 2 = 2 561 137 338 658 121 880 143 + 1;
  • 2 561 137 338 658 121 880 143 ÷ 2 = 1 280 568 669 329 060 940 071 + 1;
  • 1 280 568 669 329 060 940 071 ÷ 2 = 640 284 334 664 530 470 035 + 1;
  • 640 284 334 664 530 470 035 ÷ 2 = 320 142 167 332 265 235 017 + 1;
  • 320 142 167 332 265 235 017 ÷ 2 = 160 071 083 666 132 617 508 + 1;
  • 160 071 083 666 132 617 508 ÷ 2 = 80 035 541 833 066 308 754 + 0;
  • 80 035 541 833 066 308 754 ÷ 2 = 40 017 770 916 533 154 377 + 0;
  • 40 017 770 916 533 154 377 ÷ 2 = 20 008 885 458 266 577 188 + 1;
  • 20 008 885 458 266 577 188 ÷ 2 = 10 004 442 729 133 288 594 + 0;
  • 10 004 442 729 133 288 594 ÷ 2 = 5 002 221 364 566 644 297 + 0;
  • 5 002 221 364 566 644 297 ÷ 2 = 2 501 110 682 283 322 148 + 1;
  • 2 501 110 682 283 322 148 ÷ 2 = 1 250 555 341 141 661 074 + 0;
  • 1 250 555 341 141 661 074 ÷ 2 = 625 277 670 570 830 537 + 0;
  • 625 277 670 570 830 537 ÷ 2 = 312 638 835 285 415 268 + 1;
  • 312 638 835 285 415 268 ÷ 2 = 156 319 417 642 707 634 + 0;
  • 156 319 417 642 707 634 ÷ 2 = 78 159 708 821 353 817 + 0;
  • 78 159 708 821 353 817 ÷ 2 = 39 079 854 410 676 908 + 1;
  • 39 079 854 410 676 908 ÷ 2 = 19 539 927 205 338 454 + 0;
  • 19 539 927 205 338 454 ÷ 2 = 9 769 963 602 669 227 + 0;
  • 9 769 963 602 669 227 ÷ 2 = 4 884 981 801 334 613 + 1;
  • 4 884 981 801 334 613 ÷ 2 = 2 442 490 900 667 306 + 1;
  • 2 442 490 900 667 306 ÷ 2 = 1 221 245 450 333 653 + 0;
  • 1 221 245 450 333 653 ÷ 2 = 610 622 725 166 826 + 1;
  • 610 622 725 166 826 ÷ 2 = 305 311 362 583 413 + 0;
  • 305 311 362 583 413 ÷ 2 = 152 655 681 291 706 + 1;
  • 152 655 681 291 706 ÷ 2 = 76 327 840 645 853 + 0;
  • 76 327 840 645 853 ÷ 2 = 38 163 920 322 926 + 1;
  • 38 163 920 322 926 ÷ 2 = 19 081 960 161 463 + 0;
  • 19 081 960 161 463 ÷ 2 = 9 540 980 080 731 + 1;
  • 9 540 980 080 731 ÷ 2 = 4 770 490 040 365 + 1;
  • 4 770 490 040 365 ÷ 2 = 2 385 245 020 182 + 1;
  • 2 385 245 020 182 ÷ 2 = 1 192 622 510 091 + 0;
  • 1 192 622 510 091 ÷ 2 = 596 311 255 045 + 1;
  • 596 311 255 045 ÷ 2 = 298 155 627 522 + 1;
  • 298 155 627 522 ÷ 2 = 149 077 813 761 + 0;
  • 149 077 813 761 ÷ 2 = 74 538 906 880 + 1;
  • 74 538 906 880 ÷ 2 = 37 269 453 440 + 0;
  • 37 269 453 440 ÷ 2 = 18 634 726 720 + 0;
  • 18 634 726 720 ÷ 2 = 9 317 363 360 + 0;
  • 9 317 363 360 ÷ 2 = 4 658 681 680 + 0;
  • 4 658 681 680 ÷ 2 = 2 329 340 840 + 0;
  • 2 329 340 840 ÷ 2 = 1 164 670 420 + 0;
  • 1 164 670 420 ÷ 2 = 582 335 210 + 0;
  • 582 335 210 ÷ 2 = 291 167 605 + 0;
  • 291 167 605 ÷ 2 = 145 583 802 + 1;
  • 145 583 802 ÷ 2 = 72 791 901 + 0;
  • 72 791 901 ÷ 2 = 36 395 950 + 1;
  • 36 395 950 ÷ 2 = 18 197 975 + 0;
  • 18 197 975 ÷ 2 = 9 098 987 + 1;
  • 9 098 987 ÷ 2 = 4 549 493 + 1;
  • 4 549 493 ÷ 2 = 2 274 746 + 1;
  • 2 274 746 ÷ 2 = 1 137 373 + 0;
  • 1 137 373 ÷ 2 = 568 686 + 1;
  • 568 686 ÷ 2 = 284 343 + 0;
  • 284 343 ÷ 2 = 142 171 + 1;
  • 142 171 ÷ 2 = 71 085 + 1;
  • 71 085 ÷ 2 = 35 542 + 1;
  • 35 542 ÷ 2 = 17 771 + 0;
  • 17 771 ÷ 2 = 8 885 + 1;
  • 8 885 ÷ 2 = 4 442 + 1;
  • 4 442 ÷ 2 = 2 221 + 0;
  • 2 221 ÷ 2 = 1 110 + 1;
  • 1 110 ÷ 2 = 555 + 0;
  • 555 ÷ 2 = 277 + 1;
  • 277 ÷ 2 = 138 + 1;
  • 138 ÷ 2 = 69 + 0;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 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.


11 000 001 110 101 109 999 999 999 999 978(10) =


1000 1010 1101 0110 1110 1011 1010 1000 0000 0101 1011 1010 1010 1100 1001 0010 0100 1111 1110 0001 0111 1110 1111 1111 1110 1010(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 103 positions to the left, so that only one non zero digit remains to the left of it:


11 000 001 110 101 109 999 999 999 999 978(10) =


1000 1010 1101 0110 1110 1011 1010 1000 0000 0101 1011 1010 1010 1100 1001 0010 0100 1111 1110 0001 0111 1110 1111 1111 1110 1010(2) =


1000 1010 1101 0110 1110 1011 1010 1000 0000 0101 1011 1010 1010 1100 1001 0010 0100 1111 1110 0001 0111 1110 1111 1111 1110 1010(2) × 20 =


1.0001 0101 1010 1101 1101 0111 0101 0000 0000 1011 0111 0101 0101 1001 0010 0100 1001 1111 1100 0010 1111 1101 1111 1111 1101 010(2) × 2103


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): 103


Mantissa (not normalized):
1.0001 0101 1010 1101 1101 0111 0101 0000 0000 1011 0111 0101 0101 1001 0010 0100 1001 1111 1100 0010 1111 1101 1111 1111 1101 010


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


103 + 2(8-1) - 1 =


(103 + 127)(10) =


230(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;
  • 230 ÷ 2 = 115 + 0;
  • 115 ÷ 2 = 57 + 1;
  • 57 ÷ 2 = 28 + 1;
  • 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) =


230(10) =


1110 0110(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. 000 1010 1101 0110 1110 1011 1010 1000 0000 0101 1011 1010 1010 1100 1001 0010 0100 1111 1110 0001 0111 1110 1111 1111 1110 1010 =


000 1010 1101 0110 1110 1011


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 0110


Mantissa (23 bits) =
000 1010 1101 0110 1110 1011


The base ten decimal number 11 000 001 110 101 109 999 999 999 999 978 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1110 0110 - 000 1010 1101 0110 1110 1011

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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