-0.016 738 891 601 562 496 530 552 832 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 832(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
-0.016 738 891 601 562 496 530 552 832(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:

|-0.016 738 891 601 562 496 530 552 832| = 0.016 738 891 601 562 496 530 552 832


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

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 496 530 552 832.

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) 0.016 738 891 601 562 496 530 552 832 × 2 = 0 + 0.033 477 783 203 124 993 061 105 664;
  • 2) 0.033 477 783 203 124 993 061 105 664 × 2 = 0 + 0.066 955 566 406 249 986 122 211 328;
  • 3) 0.066 955 566 406 249 986 122 211 328 × 2 = 0 + 0.133 911 132 812 499 972 244 422 656;
  • 4) 0.133 911 132 812 499 972 244 422 656 × 2 = 0 + 0.267 822 265 624 999 944 488 845 312;
  • 5) 0.267 822 265 624 999 944 488 845 312 × 2 = 0 + 0.535 644 531 249 999 888 977 690 624;
  • 6) 0.535 644 531 249 999 888 977 690 624 × 2 = 1 + 0.071 289 062 499 999 777 955 381 248;
  • 7) 0.071 289 062 499 999 777 955 381 248 × 2 = 0 + 0.142 578 124 999 999 555 910 762 496;
  • 8) 0.142 578 124 999 999 555 910 762 496 × 2 = 0 + 0.285 156 249 999 999 111 821 524 992;
  • 9) 0.285 156 249 999 999 111 821 524 992 × 2 = 0 + 0.570 312 499 999 998 223 643 049 984;
  • 10) 0.570 312 499 999 998 223 643 049 984 × 2 = 1 + 0.140 624 999 999 996 447 286 099 968;
  • 11) 0.140 624 999 999 996 447 286 099 968 × 2 = 0 + 0.281 249 999 999 992 894 572 199 936;
  • 12) 0.281 249 999 999 992 894 572 199 936 × 2 = 0 + 0.562 499 999 999 985 789 144 399 872;
  • 13) 0.562 499 999 999 985 789 144 399 872 × 2 = 1 + 0.124 999 999 999 971 578 288 799 744;
  • 14) 0.124 999 999 999 971 578 288 799 744 × 2 = 0 + 0.249 999 999 999 943 156 577 599 488;
  • 15) 0.249 999 999 999 943 156 577 599 488 × 2 = 0 + 0.499 999 999 999 886 313 155 198 976;
  • 16) 0.499 999 999 999 886 313 155 198 976 × 2 = 0 + 0.999 999 999 999 772 626 310 397 952;
  • 17) 0.999 999 999 999 772 626 310 397 952 × 2 = 1 + 0.999 999 999 999 545 252 620 795 904;
  • 18) 0.999 999 999 999 545 252 620 795 904 × 2 = 1 + 0.999 999 999 999 090 505 241 591 808;
  • 19) 0.999 999 999 999 090 505 241 591 808 × 2 = 1 + 0.999 999 999 998 181 010 483 183 616;
  • 20) 0.999 999 999 998 181 010 483 183 616 × 2 = 1 + 0.999 999 999 996 362 020 966 367 232;
  • 21) 0.999 999 999 996 362 020 966 367 232 × 2 = 1 + 0.999 999 999 992 724 041 932 734 464;
  • 22) 0.999 999 999 992 724 041 932 734 464 × 2 = 1 + 0.999 999 999 985 448 083 865 468 928;
  • 23) 0.999 999 999 985 448 083 865 468 928 × 2 = 1 + 0.999 999 999 970 896 167 730 937 856;
  • 24) 0.999 999 999 970 896 167 730 937 856 × 2 = 1 + 0.999 999 999 941 792 335 461 875 712;
  • 25) 0.999 999 999 941 792 335 461 875 712 × 2 = 1 + 0.999 999 999 883 584 670 923 751 424;
  • 26) 0.999 999 999 883 584 670 923 751 424 × 2 = 1 + 0.999 999 999 767 169 341 847 502 848;
  • 27) 0.999 999 999 767 169 341 847 502 848 × 2 = 1 + 0.999 999 999 534 338 683 695 005 696;
  • 28) 0.999 999 999 534 338 683 695 005 696 × 2 = 1 + 0.999 999 999 068 677 367 390 011 392;
  • 29) 0.999 999 999 068 677 367 390 011 392 × 2 = 1 + 0.999 999 998 137 354 734 780 022 784;
  • 30) 0.999 999 998 137 354 734 780 022 784 × 2 = 1 + 0.999 999 996 274 709 469 560 045 568;
  • 31) 0.999 999 996 274 709 469 560 045 568 × 2 = 1 + 0.999 999 992 549 418 939 120 091 136;
  • 32) 0.999 999 992 549 418 939 120 091 136 × 2 = 1 + 0.999 999 985 098 837 878 240 182 272;
  • 33) 0.999 999 985 098 837 878 240 182 272 × 2 = 1 + 0.999 999 970 197 675 756 480 364 544;
  • 34) 0.999 999 970 197 675 756 480 364 544 × 2 = 1 + 0.999 999 940 395 351 512 960 729 088;
  • 35) 0.999 999 940 395 351 512 960 729 088 × 2 = 1 + 0.999 999 880 790 703 025 921 458 176;
  • 36) 0.999 999 880 790 703 025 921 458 176 × 2 = 1 + 0.999 999 761 581 406 051 842 916 352;
  • 37) 0.999 999 761 581 406 051 842 916 352 × 2 = 1 + 0.999 999 523 162 812 103 685 832 704;
  • 38) 0.999 999 523 162 812 103 685 832 704 × 2 = 1 + 0.999 999 046 325 624 207 371 665 408;
  • 39) 0.999 999 046 325 624 207 371 665 408 × 2 = 1 + 0.999 998 092 651 248 414 743 330 816;
  • 40) 0.999 998 092 651 248 414 743 330 816 × 2 = 1 + 0.999 996 185 302 496 829 486 661 632;
  • 41) 0.999 996 185 302 496 829 486 661 632 × 2 = 1 + 0.999 992 370 604 993 658 973 323 264;
  • 42) 0.999 992 370 604 993 658 973 323 264 × 2 = 1 + 0.999 984 741 209 987 317 946 646 528;
  • 43) 0.999 984 741 209 987 317 946 646 528 × 2 = 1 + 0.999 969 482 419 974 635 893 293 056;
  • 44) 0.999 969 482 419 974 635 893 293 056 × 2 = 1 + 0.999 938 964 839 949 271 786 586 112;
  • 45) 0.999 938 964 839 949 271 786 586 112 × 2 = 1 + 0.999 877 929 679 898 543 573 172 224;
  • 46) 0.999 877 929 679 898 543 573 172 224 × 2 = 1 + 0.999 755 859 359 797 087 146 344 448;
  • 47) 0.999 755 859 359 797 087 146 344 448 × 2 = 1 + 0.999 511 718 719 594 174 292 688 896;
  • 48) 0.999 511 718 719 594 174 292 688 896 × 2 = 1 + 0.999 023 437 439 188 348 585 377 792;
  • 49) 0.999 023 437 439 188 348 585 377 792 × 2 = 1 + 0.998 046 874 878 376 697 170 755 584;
  • 50) 0.998 046 874 878 376 697 170 755 584 × 2 = 1 + 0.996 093 749 756 753 394 341 511 168;
  • 51) 0.996 093 749 756 753 394 341 511 168 × 2 = 1 + 0.992 187 499 513 506 788 683 022 336;
  • 52) 0.992 187 499 513 506 788 683 022 336 × 2 = 1 + 0.984 374 999 027 013 577 366 044 672;
  • 53) 0.984 374 999 027 013 577 366 044 672 × 2 = 1 + 0.968 749 998 054 027 154 732 089 344;
  • 54) 0.968 749 998 054 027 154 732 089 344 × 2 = 1 + 0.937 499 996 108 054 309 464 178 688;
  • 55) 0.937 499 996 108 054 309 464 178 688 × 2 = 1 + 0.874 999 992 216 108 618 928 357 376;
  • 56) 0.874 999 992 216 108 618 928 357 376 × 2 = 1 + 0.749 999 984 432 217 237 856 714 752;
  • 57) 0.749 999 984 432 217 237 856 714 752 × 2 = 1 + 0.499 999 968 864 434 475 713 429 504;
  • 58) 0.499 999 968 864 434 475 713 429 504 × 2 = 0 + 0.999 999 937 728 868 951 426 859 008;

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:


0.016 738 891 601 562 496 530 552 832(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 832(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 552 832(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1017(10) =

011 1111 1001(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 552 832 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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