-2.447 581 244 357 921 521 399 255 989 065 644 470 729 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.447 581 244 357 921 521 399 255 989 065 644 470 729(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-2.447 581 244 357 921 521 399 255 989 065 644 470 729(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-2.447 581 244 357 921 521 399 255 989 065 644 470 729| = 2.447 581 244 357 921 521 399 255 989 065 644 470 729


2. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

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

2(10) =

10(2)


4. Convert to binary (base 2) the fractional part: 0.447 581 244 357 921 521 399 255 989 065 644 470 729.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.447 581 244 357 921 521 399 255 989 065 644 470 729 × 2 = 0 + 0.895 162 488 715 843 042 798 511 978 131 288 941 458;
  • 2) 0.895 162 488 715 843 042 798 511 978 131 288 941 458 × 2 = 1 + 0.790 324 977 431 686 085 597 023 956 262 577 882 916;
  • 3) 0.790 324 977 431 686 085 597 023 956 262 577 882 916 × 2 = 1 + 0.580 649 954 863 372 171 194 047 912 525 155 765 832;
  • 4) 0.580 649 954 863 372 171 194 047 912 525 155 765 832 × 2 = 1 + 0.161 299 909 726 744 342 388 095 825 050 311 531 664;
  • 5) 0.161 299 909 726 744 342 388 095 825 050 311 531 664 × 2 = 0 + 0.322 599 819 453 488 684 776 191 650 100 623 063 328;
  • 6) 0.322 599 819 453 488 684 776 191 650 100 623 063 328 × 2 = 0 + 0.645 199 638 906 977 369 552 383 300 201 246 126 656;
  • 7) 0.645 199 638 906 977 369 552 383 300 201 246 126 656 × 2 = 1 + 0.290 399 277 813 954 739 104 766 600 402 492 253 312;
  • 8) 0.290 399 277 813 954 739 104 766 600 402 492 253 312 × 2 = 0 + 0.580 798 555 627 909 478 209 533 200 804 984 506 624;
  • 9) 0.580 798 555 627 909 478 209 533 200 804 984 506 624 × 2 = 1 + 0.161 597 111 255 818 956 419 066 401 609 969 013 248;
  • 10) 0.161 597 111 255 818 956 419 066 401 609 969 013 248 × 2 = 0 + 0.323 194 222 511 637 912 838 132 803 219 938 026 496;
  • 11) 0.323 194 222 511 637 912 838 132 803 219 938 026 496 × 2 = 0 + 0.646 388 445 023 275 825 676 265 606 439 876 052 992;
  • 12) 0.646 388 445 023 275 825 676 265 606 439 876 052 992 × 2 = 1 + 0.292 776 890 046 551 651 352 531 212 879 752 105 984;
  • 13) 0.292 776 890 046 551 651 352 531 212 879 752 105 984 × 2 = 0 + 0.585 553 780 093 103 302 705 062 425 759 504 211 968;
  • 14) 0.585 553 780 093 103 302 705 062 425 759 504 211 968 × 2 = 1 + 0.171 107 560 186 206 605 410 124 851 519 008 423 936;
  • 15) 0.171 107 560 186 206 605 410 124 851 519 008 423 936 × 2 = 0 + 0.342 215 120 372 413 210 820 249 703 038 016 847 872;
  • 16) 0.342 215 120 372 413 210 820 249 703 038 016 847 872 × 2 = 0 + 0.684 430 240 744 826 421 640 499 406 076 033 695 744;
  • 17) 0.684 430 240 744 826 421 640 499 406 076 033 695 744 × 2 = 1 + 0.368 860 481 489 652 843 280 998 812 152 067 391 488;
  • 18) 0.368 860 481 489 652 843 280 998 812 152 067 391 488 × 2 = 0 + 0.737 720 962 979 305 686 561 997 624 304 134 782 976;
  • 19) 0.737 720 962 979 305 686 561 997 624 304 134 782 976 × 2 = 1 + 0.475 441 925 958 611 373 123 995 248 608 269 565 952;
  • 20) 0.475 441 925 958 611 373 123 995 248 608 269 565 952 × 2 = 0 + 0.950 883 851 917 222 746 247 990 497 216 539 131 904;
  • 21) 0.950 883 851 917 222 746 247 990 497 216 539 131 904 × 2 = 1 + 0.901 767 703 834 445 492 495 980 994 433 078 263 808;
  • 22) 0.901 767 703 834 445 492 495 980 994 433 078 263 808 × 2 = 1 + 0.803 535 407 668 890 984 991 961 988 866 156 527 616;
  • 23) 0.803 535 407 668 890 984 991 961 988 866 156 527 616 × 2 = 1 + 0.607 070 815 337 781 969 983 923 977 732 313 055 232;
  • 24) 0.607 070 815 337 781 969 983 923 977 732 313 055 232 × 2 = 1 + 0.214 141 630 675 563 939 967 847 955 464 626 110 464;
  • 25) 0.214 141 630 675 563 939 967 847 955 464 626 110 464 × 2 = 0 + 0.428 283 261 351 127 879 935 695 910 929 252 220 928;
  • 26) 0.428 283 261 351 127 879 935 695 910 929 252 220 928 × 2 = 0 + 0.856 566 522 702 255 759 871 391 821 858 504 441 856;
  • 27) 0.856 566 522 702 255 759 871 391 821 858 504 441 856 × 2 = 1 + 0.713 133 045 404 511 519 742 783 643 717 008 883 712;
  • 28) 0.713 133 045 404 511 519 742 783 643 717 008 883 712 × 2 = 1 + 0.426 266 090 809 023 039 485 567 287 434 017 767 424;
  • 29) 0.426 266 090 809 023 039 485 567 287 434 017 767 424 × 2 = 0 + 0.852 532 181 618 046 078 971 134 574 868 035 534 848;
  • 30) 0.852 532 181 618 046 078 971 134 574 868 035 534 848 × 2 = 1 + 0.705 064 363 236 092 157 942 269 149 736 071 069 696;
  • 31) 0.705 064 363 236 092 157 942 269 149 736 071 069 696 × 2 = 1 + 0.410 128 726 472 184 315 884 538 299 472 142 139 392;
  • 32) 0.410 128 726 472 184 315 884 538 299 472 142 139 392 × 2 = 0 + 0.820 257 452 944 368 631 769 076 598 944 284 278 784;
  • 33) 0.820 257 452 944 368 631 769 076 598 944 284 278 784 × 2 = 1 + 0.640 514 905 888 737 263 538 153 197 888 568 557 568;
  • 34) 0.640 514 905 888 737 263 538 153 197 888 568 557 568 × 2 = 1 + 0.281 029 811 777 474 527 076 306 395 777 137 115 136;
  • 35) 0.281 029 811 777 474 527 076 306 395 777 137 115 136 × 2 = 0 + 0.562 059 623 554 949 054 152 612 791 554 274 230 272;
  • 36) 0.562 059 623 554 949 054 152 612 791 554 274 230 272 × 2 = 1 + 0.124 119 247 109 898 108 305 225 583 108 548 460 544;
  • 37) 0.124 119 247 109 898 108 305 225 583 108 548 460 544 × 2 = 0 + 0.248 238 494 219 796 216 610 451 166 217 096 921 088;
  • 38) 0.248 238 494 219 796 216 610 451 166 217 096 921 088 × 2 = 0 + 0.496 476 988 439 592 433 220 902 332 434 193 842 176;
  • 39) 0.496 476 988 439 592 433 220 902 332 434 193 842 176 × 2 = 0 + 0.992 953 976 879 184 866 441 804 664 868 387 684 352;
  • 40) 0.992 953 976 879 184 866 441 804 664 868 387 684 352 × 2 = 1 + 0.985 907 953 758 369 732 883 609 329 736 775 368 704;
  • 41) 0.985 907 953 758 369 732 883 609 329 736 775 368 704 × 2 = 1 + 0.971 815 907 516 739 465 767 218 659 473 550 737 408;
  • 42) 0.971 815 907 516 739 465 767 218 659 473 550 737 408 × 2 = 1 + 0.943 631 815 033 478 931 534 437 318 947 101 474 816;
  • 43) 0.943 631 815 033 478 931 534 437 318 947 101 474 816 × 2 = 1 + 0.887 263 630 066 957 863 068 874 637 894 202 949 632;
  • 44) 0.887 263 630 066 957 863 068 874 637 894 202 949 632 × 2 = 1 + 0.774 527 260 133 915 726 137 749 275 788 405 899 264;
  • 45) 0.774 527 260 133 915 726 137 749 275 788 405 899 264 × 2 = 1 + 0.549 054 520 267 831 452 275 498 551 576 811 798 528;
  • 46) 0.549 054 520 267 831 452 275 498 551 576 811 798 528 × 2 = 1 + 0.098 109 040 535 662 904 550 997 103 153 623 597 056;
  • 47) 0.098 109 040 535 662 904 550 997 103 153 623 597 056 × 2 = 0 + 0.196 218 081 071 325 809 101 994 206 307 247 194 112;
  • 48) 0.196 218 081 071 325 809 101 994 206 307 247 194 112 × 2 = 0 + 0.392 436 162 142 651 618 203 988 412 614 494 388 224;
  • 49) 0.392 436 162 142 651 618 203 988 412 614 494 388 224 × 2 = 0 + 0.784 872 324 285 303 236 407 976 825 228 988 776 448;
  • 50) 0.784 872 324 285 303 236 407 976 825 228 988 776 448 × 2 = 1 + 0.569 744 648 570 606 472 815 953 650 457 977 552 896;
  • 51) 0.569 744 648 570 606 472 815 953 650 457 977 552 896 × 2 = 1 + 0.139 489 297 141 212 945 631 907 300 915 955 105 792;
  • 52) 0.139 489 297 141 212 945 631 907 300 915 955 105 792 × 2 = 0 + 0.278 978 594 282 425 891 263 814 601 831 910 211 584;
  • 53) 0.278 978 594 282 425 891 263 814 601 831 910 211 584 × 2 = 0 + 0.557 957 188 564 851 782 527 629 203 663 820 423 168;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.447 581 244 357 921 521 399 255 989 065 644 470 729(10) =

0.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2)

6. Positive number before normalization:

2.447 581 244 357 921 521 399 255 989 065 644 470 729(10) =

10.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.447 581 244 357 921 521 399 255 989 065 644 470 729(10) =

10.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2) =

10.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2) × 20 =

1.0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011 00(2) × 21


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011 00


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


12. 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 52 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. 0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011 00 =

0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011


Decimal number -2.447 581 244 357 921 521 399 255 989 065 644 470 729 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100