Convert 1 000 111 011 011 009 999 999 999 999 994 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

1 000 111 011 011 009 999 999 999 999 994(10) to 32 bit single precision IEEE 754 binary floating point (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;
  • 1 000 111 011 011 009 999 999 999 999 994 ÷ 2 = 500 055 505 505 504 999 999 999 999 997 + 0;
  • 500 055 505 505 504 999 999 999 999 997 ÷ 2 = 250 027 752 752 752 499 999 999 999 998 + 1;
  • 250 027 752 752 752 499 999 999 999 998 ÷ 2 = 125 013 876 376 376 249 999 999 999 999 + 0;
  • 125 013 876 376 376 249 999 999 999 999 ÷ 2 = 62 506 938 188 188 124 999 999 999 999 + 1;
  • 62 506 938 188 188 124 999 999 999 999 ÷ 2 = 31 253 469 094 094 062 499 999 999 999 + 1;
  • 31 253 469 094 094 062 499 999 999 999 ÷ 2 = 15 626 734 547 047 031 249 999 999 999 + 1;
  • 15 626 734 547 047 031 249 999 999 999 ÷ 2 = 7 813 367 273 523 515 624 999 999 999 + 1;
  • 7 813 367 273 523 515 624 999 999 999 ÷ 2 = 3 906 683 636 761 757 812 499 999 999 + 1;
  • 3 906 683 636 761 757 812 499 999 999 ÷ 2 = 1 953 341 818 380 878 906 249 999 999 + 1;
  • 1 953 341 818 380 878 906 249 999 999 ÷ 2 = 976 670 909 190 439 453 124 999 999 + 1;
  • 976 670 909 190 439 453 124 999 999 ÷ 2 = 488 335 454 595 219 726 562 499 999 + 1;
  • 488 335 454 595 219 726 562 499 999 ÷ 2 = 244 167 727 297 609 863 281 249 999 + 1;
  • 244 167 727 297 609 863 281 249 999 ÷ 2 = 122 083 863 648 804 931 640 624 999 + 1;
  • 122 083 863 648 804 931 640 624 999 ÷ 2 = 61 041 931 824 402 465 820 312 499 + 1;
  • 61 041 931 824 402 465 820 312 499 ÷ 2 = 30 520 965 912 201 232 910 156 249 + 1;
  • 30 520 965 912 201 232 910 156 249 ÷ 2 = 15 260 482 956 100 616 455 078 124 + 1;
  • 15 260 482 956 100 616 455 078 124 ÷ 2 = 7 630 241 478 050 308 227 539 062 + 0;
  • 7 630 241 478 050 308 227 539 062 ÷ 2 = 3 815 120 739 025 154 113 769 531 + 0;
  • 3 815 120 739 025 154 113 769 531 ÷ 2 = 1 907 560 369 512 577 056 884 765 + 1;
  • 1 907 560 369 512 577 056 884 765 ÷ 2 = 953 780 184 756 288 528 442 382 + 1;
  • 953 780 184 756 288 528 442 382 ÷ 2 = 476 890 092 378 144 264 221 191 + 0;
  • 476 890 092 378 144 264 221 191 ÷ 2 = 238 445 046 189 072 132 110 595 + 1;
  • 238 445 046 189 072 132 110 595 ÷ 2 = 119 222 523 094 536 066 055 297 + 1;
  • 119 222 523 094 536 066 055 297 ÷ 2 = 59 611 261 547 268 033 027 648 + 1;
  • 59 611 261 547 268 033 027 648 ÷ 2 = 29 805 630 773 634 016 513 824 + 0;
  • 29 805 630 773 634 016 513 824 ÷ 2 = 14 902 815 386 817 008 256 912 + 0;
  • 14 902 815 386 817 008 256 912 ÷ 2 = 7 451 407 693 408 504 128 456 + 0;
  • 7 451 407 693 408 504 128 456 ÷ 2 = 3 725 703 846 704 252 064 228 + 0;
  • 3 725 703 846 704 252 064 228 ÷ 2 = 1 862 851 923 352 126 032 114 + 0;
  • 1 862 851 923 352 126 032 114 ÷ 2 = 931 425 961 676 063 016 057 + 0;
  • 931 425 961 676 063 016 057 ÷ 2 = 465 712 980 838 031 508 028 + 1;
  • 465 712 980 838 031 508 028 ÷ 2 = 232 856 490 419 015 754 014 + 0;
  • 232 856 490 419 015 754 014 ÷ 2 = 116 428 245 209 507 877 007 + 0;
  • 116 428 245 209 507 877 007 ÷ 2 = 58 214 122 604 753 938 503 + 1;
  • 58 214 122 604 753 938 503 ÷ 2 = 29 107 061 302 376 969 251 + 1;
  • 29 107 061 302 376 969 251 ÷ 2 = 14 553 530 651 188 484 625 + 1;
  • 14 553 530 651 188 484 625 ÷ 2 = 7 276 765 325 594 242 312 + 1;
  • 7 276 765 325 594 242 312 ÷ 2 = 3 638 382 662 797 121 156 + 0;
  • 3 638 382 662 797 121 156 ÷ 2 = 1 819 191 331 398 560 578 + 0;
  • 1 819 191 331 398 560 578 ÷ 2 = 909 595 665 699 280 289 + 0;
  • 909 595 665 699 280 289 ÷ 2 = 454 797 832 849 640 144 + 1;
  • 454 797 832 849 640 144 ÷ 2 = 227 398 916 424 820 072 + 0;
  • 227 398 916 424 820 072 ÷ 2 = 113 699 458 212 410 036 + 0;
  • 113 699 458 212 410 036 ÷ 2 = 56 849 729 106 205 018 + 0;
  • 56 849 729 106 205 018 ÷ 2 = 28 424 864 553 102 509 + 0;
  • 28 424 864 553 102 509 ÷ 2 = 14 212 432 276 551 254 + 1;
  • 14 212 432 276 551 254 ÷ 2 = 7 106 216 138 275 627 + 0;
  • 7 106 216 138 275 627 ÷ 2 = 3 553 108 069 137 813 + 1;
  • 3 553 108 069 137 813 ÷ 2 = 1 776 554 034 568 906 + 1;
  • 1 776 554 034 568 906 ÷ 2 = 888 277 017 284 453 + 0;
  • 888 277 017 284 453 ÷ 2 = 444 138 508 642 226 + 1;
  • 444 138 508 642 226 ÷ 2 = 222 069 254 321 113 + 0;
  • 222 069 254 321 113 ÷ 2 = 111 034 627 160 556 + 1;
  • 111 034 627 160 556 ÷ 2 = 55 517 313 580 278 + 0;
  • 55 517 313 580 278 ÷ 2 = 27 758 656 790 139 + 0;
  • 27 758 656 790 139 ÷ 2 = 13 879 328 395 069 + 1;
  • 13 879 328 395 069 ÷ 2 = 6 939 664 197 534 + 1;
  • 6 939 664 197 534 ÷ 2 = 3 469 832 098 767 + 0;
  • 3 469 832 098 767 ÷ 2 = 1 734 916 049 383 + 1;
  • 1 734 916 049 383 ÷ 2 = 867 458 024 691 + 1;
  • 867 458 024 691 ÷ 2 = 433 729 012 345 + 1;
  • 433 729 012 345 ÷ 2 = 216 864 506 172 + 1;
  • 216 864 506 172 ÷ 2 = 108 432 253 086 + 0;
  • 108 432 253 086 ÷ 2 = 54 216 126 543 + 0;
  • 54 216 126 543 ÷ 2 = 27 108 063 271 + 1;
  • 27 108 063 271 ÷ 2 = 13 554 031 635 + 1;
  • 13 554 031 635 ÷ 2 = 6 777 015 817 + 1;
  • 6 777 015 817 ÷ 2 = 3 388 507 908 + 1;
  • 3 388 507 908 ÷ 2 = 1 694 253 954 + 0;
  • 1 694 253 954 ÷ 2 = 847 126 977 + 0;
  • 847 126 977 ÷ 2 = 423 563 488 + 1;
  • 423 563 488 ÷ 2 = 211 781 744 + 0;
  • 211 781 744 ÷ 2 = 105 890 872 + 0;
  • 105 890 872 ÷ 2 = 52 945 436 + 0;
  • 52 945 436 ÷ 2 = 26 472 718 + 0;
  • 26 472 718 ÷ 2 = 13 236 359 + 0;
  • 13 236 359 ÷ 2 = 6 618 179 + 1;
  • 6 618 179 ÷ 2 = 3 309 089 + 1;
  • 3 309 089 ÷ 2 = 1 654 544 + 1;
  • 1 654 544 ÷ 2 = 827 272 + 0;
  • 827 272 ÷ 2 = 413 636 + 0;
  • 413 636 ÷ 2 = 206 818 + 0;
  • 206 818 ÷ 2 = 103 409 + 0;
  • 103 409 ÷ 2 = 51 704 + 1;
  • 51 704 ÷ 2 = 25 852 + 0;
  • 25 852 ÷ 2 = 12 926 + 0;
  • 12 926 ÷ 2 = 6 463 + 0;
  • 6 463 ÷ 2 = 3 231 + 1;
  • 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.

1 000 111 011 011 009 999 999 999 999 994(10) =


1100 1001 1111 1000 1000 0111 0000 0100 1111 0011 1101 1001 0101 1010 0001 0001 1110 0100 0000 1110 1100 1111 1111 1111 1010(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:

1 000 111 011 011 009 999 999 999 999 994(10) =


1100 1001 1111 1000 1000 0111 0000 0100 1111 0011 1101 1001 0101 1010 0001 0001 1110 0100 0000 1110 1100 1111 1111 1111 1010(2) =


1100 1001 1111 1000 1000 0111 0000 0100 1111 0011 1101 1001 0101 1010 0001 0001 1110 0100 0000 1110 1100 1111 1111 1111 1010(2) × 20 =


1.1001 0011 1111 0001 0000 1110 0000 1001 1110 0111 1011 0010 1011 0100 0010 0011 1100 1000 0001 1101 1001 1111 1111 1111 010(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 1111 0001 0000 1110 0000 1001 1110 0111 1011 0010 1011 0100 0010 0011 1100 1000 0001 1101 1001 1111 1111 1111 010


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 1000 1000 0111 0000 0100 1111 0011 1101 1001 0101 1010 0001 0001 1110 0100 0000 1110 1100 1111 1111 1111 1010 =


100 1001 1111 1000 1000 0111


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 1000 1000 0111


Number 1 000 111 011 011 009 999 999 999 999 994 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1110 0010 - 100 1001 1111 1000 1000 0111

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 1

      1
    • 1

      0

More operations of this kind:

1 000 111 011 011 009 999 999 999 999 993 = ? ... 1 000 111 011 011 009 999 999 999 999 995 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

1 000 111 011 011 009 999 999 999 999 994 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
216.05 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
111 010 110 092 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
-18.562 6 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
105.2 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
0.550 382 121 2 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
-65 542 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
21.906 28 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
-106.812 7 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
-2 842 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
-77.37 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
23.079 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
-1.875 28 to 32 bit single precision IEEE 754 binary floating point = ? Mar 09 10:53 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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