-2.211 829 052 383 358 300 119 548 661 699 653 268 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 653 268 8(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
-2.211 829 052 383 358 300 119 548 661 699 653 268 8(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:

|-2.211 829 052 383 358 300 119 548 661 699 653 268 8| = 2.211 829 052 383 358 300 119 548 661 699 653 268 8


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: 0.211 829 052 383 358 300 119 548 661 699 653 268 8.

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) 0.211 829 052 383 358 300 119 548 661 699 653 268 8 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 306 537 6;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 306 537 6 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 798 613 075 2;
  • 3) 0.847 316 209 533 433 200 478 194 646 798 613 075 2 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 597 226 150 4;
  • 4) 0.694 632 419 066 866 400 956 389 293 597 226 150 4 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 194 452 300 8;
  • 5) 0.389 264 838 133 732 801 912 778 587 194 452 300 8 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 388 904 601 6;
  • 6) 0.778 529 676 267 465 603 825 557 174 388 904 601 6 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 777 809 203 2;
  • 7) 0.557 059 352 534 931 207 651 114 348 777 809 203 2 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 555 618 406 4;
  • 8) 0.114 118 705 069 862 415 302 228 697 555 618 406 4 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 111 236 812 8;
  • 9) 0.228 237 410 139 724 830 604 457 395 111 236 812 8 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 222 473 625 6;
  • 10) 0.456 474 820 279 449 661 208 914 790 222 473 625 6 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 444 947 251 2;
  • 11) 0.912 949 640 558 899 322 417 829 580 444 947 251 2 × 2 = 1 + 0.825 899 281 117 798 644 835 659 160 889 894 502 4;
  • 12) 0.825 899 281 117 798 644 835 659 160 889 894 502 4 × 2 = 1 + 0.651 798 562 235 597 289 671 318 321 779 789 004 8;
  • 13) 0.651 798 562 235 597 289 671 318 321 779 789 004 8 × 2 = 1 + 0.303 597 124 471 194 579 342 636 643 559 578 009 6;
  • 14) 0.303 597 124 471 194 579 342 636 643 559 578 009 6 × 2 = 0 + 0.607 194 248 942 389 158 685 273 287 119 156 019 2;
  • 15) 0.607 194 248 942 389 158 685 273 287 119 156 019 2 × 2 = 1 + 0.214 388 497 884 778 317 370 546 574 238 312 038 4;
  • 16) 0.214 388 497 884 778 317 370 546 574 238 312 038 4 × 2 = 0 + 0.428 776 995 769 556 634 741 093 148 476 624 076 8;
  • 17) 0.428 776 995 769 556 634 741 093 148 476 624 076 8 × 2 = 0 + 0.857 553 991 539 113 269 482 186 296 953 248 153 6;
  • 18) 0.857 553 991 539 113 269 482 186 296 953 248 153 6 × 2 = 1 + 0.715 107 983 078 226 538 964 372 593 906 496 307 2;
  • 19) 0.715 107 983 078 226 538 964 372 593 906 496 307 2 × 2 = 1 + 0.430 215 966 156 453 077 928 745 187 812 992 614 4;
  • 20) 0.430 215 966 156 453 077 928 745 187 812 992 614 4 × 2 = 0 + 0.860 431 932 312 906 155 857 490 375 625 985 228 8;
  • 21) 0.860 431 932 312 906 155 857 490 375 625 985 228 8 × 2 = 1 + 0.720 863 864 625 812 311 714 980 751 251 970 457 6;
  • 22) 0.720 863 864 625 812 311 714 980 751 251 970 457 6 × 2 = 1 + 0.441 727 729 251 624 623 429 961 502 503 940 915 2;
  • 23) 0.441 727 729 251 624 623 429 961 502 503 940 915 2 × 2 = 0 + 0.883 455 458 503 249 246 859 923 005 007 881 830 4;
  • 24) 0.883 455 458 503 249 246 859 923 005 007 881 830 4 × 2 = 1 + 0.766 910 917 006 498 493 719 846 010 015 763 660 8;
  • 25) 0.766 910 917 006 498 493 719 846 010 015 763 660 8 × 2 = 1 + 0.533 821 834 012 996 987 439 692 020 031 527 321 6;
  • 26) 0.533 821 834 012 996 987 439 692 020 031 527 321 6 × 2 = 1 + 0.067 643 668 025 993 974 879 384 040 063 054 643 2;
  • 27) 0.067 643 668 025 993 974 879 384 040 063 054 643 2 × 2 = 0 + 0.135 287 336 051 987 949 758 768 080 126 109 286 4;
  • 28) 0.135 287 336 051 987 949 758 768 080 126 109 286 4 × 2 = 0 + 0.270 574 672 103 975 899 517 536 160 252 218 572 8;
  • 29) 0.270 574 672 103 975 899 517 536 160 252 218 572 8 × 2 = 0 + 0.541 149 344 207 951 799 035 072 320 504 437 145 6;
  • 30) 0.541 149 344 207 951 799 035 072 320 504 437 145 6 × 2 = 1 + 0.082 298 688 415 903 598 070 144 641 008 874 291 2;
  • 31) 0.082 298 688 415 903 598 070 144 641 008 874 291 2 × 2 = 0 + 0.164 597 376 831 807 196 140 289 282 017 748 582 4;
  • 32) 0.164 597 376 831 807 196 140 289 282 017 748 582 4 × 2 = 0 + 0.329 194 753 663 614 392 280 578 564 035 497 164 8;
  • 33) 0.329 194 753 663 614 392 280 578 564 035 497 164 8 × 2 = 0 + 0.658 389 507 327 228 784 561 157 128 070 994 329 6;
  • 34) 0.658 389 507 327 228 784 561 157 128 070 994 329 6 × 2 = 1 + 0.316 779 014 654 457 569 122 314 256 141 988 659 2;
  • 35) 0.316 779 014 654 457 569 122 314 256 141 988 659 2 × 2 = 0 + 0.633 558 029 308 915 138 244 628 512 283 977 318 4;
  • 36) 0.633 558 029 308 915 138 244 628 512 283 977 318 4 × 2 = 1 + 0.267 116 058 617 830 276 489 257 024 567 954 636 8;
  • 37) 0.267 116 058 617 830 276 489 257 024 567 954 636 8 × 2 = 0 + 0.534 232 117 235 660 552 978 514 049 135 909 273 6;
  • 38) 0.534 232 117 235 660 552 978 514 049 135 909 273 6 × 2 = 1 + 0.068 464 234 471 321 105 957 028 098 271 818 547 2;
  • 39) 0.068 464 234 471 321 105 957 028 098 271 818 547 2 × 2 = 0 + 0.136 928 468 942 642 211 914 056 196 543 637 094 4;
  • 40) 0.136 928 468 942 642 211 914 056 196 543 637 094 4 × 2 = 0 + 0.273 856 937 885 284 423 828 112 393 087 274 188 8;
  • 41) 0.273 856 937 885 284 423 828 112 393 087 274 188 8 × 2 = 0 + 0.547 713 875 770 568 847 656 224 786 174 548 377 6;
  • 42) 0.547 713 875 770 568 847 656 224 786 174 548 377 6 × 2 = 1 + 0.095 427 751 541 137 695 312 449 572 349 096 755 2;
  • 43) 0.095 427 751 541 137 695 312 449 572 349 096 755 2 × 2 = 0 + 0.190 855 503 082 275 390 624 899 144 698 193 510 4;
  • 44) 0.190 855 503 082 275 390 624 899 144 698 193 510 4 × 2 = 0 + 0.381 711 006 164 550 781 249 798 289 396 387 020 8;
  • 45) 0.381 711 006 164 550 781 249 798 289 396 387 020 8 × 2 = 0 + 0.763 422 012 329 101 562 499 596 578 792 774 041 6;
  • 46) 0.763 422 012 329 101 562 499 596 578 792 774 041 6 × 2 = 1 + 0.526 844 024 658 203 124 999 193 157 585 548 083 2;
  • 47) 0.526 844 024 658 203 124 999 193 157 585 548 083 2 × 2 = 1 + 0.053 688 049 316 406 249 998 386 315 171 096 166 4;
  • 48) 0.053 688 049 316 406 249 998 386 315 171 096 166 4 × 2 = 0 + 0.107 376 098 632 812 499 996 772 630 342 192 332 8;
  • 49) 0.107 376 098 632 812 499 996 772 630 342 192 332 8 × 2 = 0 + 0.214 752 197 265 624 999 993 545 260 684 384 665 6;
  • 50) 0.214 752 197 265 624 999 993 545 260 684 384 665 6 × 2 = 0 + 0.429 504 394 531 249 999 987 090 521 368 769 331 2;
  • 51) 0.429 504 394 531 249 999 987 090 521 368 769 331 2 × 2 = 0 + 0.859 008 789 062 499 999 974 181 042 737 538 662 4;
  • 52) 0.859 008 789 062 499 999 974 181 042 737 538 662 4 × 2 = 1 + 0.718 017 578 124 999 999 948 362 085 475 077 324 8;
  • 53) 0.718 017 578 124 999 999 948 362 085 475 077 324 8 × 2 = 1 + 0.436 035 156 249 999 999 896 724 170 950 154 649 6;

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:


0.211 829 052 383 358 300 119 548 661 699 653 268 8(10) =

0.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 653 268 8(10) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(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:


2.211 829 052 383 358 300 119 548 661 699 653 268 8(10) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) × 20 =

1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11(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):
1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11


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. 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11 =

0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


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) =
0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


Decimal number -2.211 829 052 383 358 300 119 548 661 699 653 268 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000

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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