-0.016 738 891 601 562 496 530 614 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 614(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
-0.016 738 891 601 562 496 530 614(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:

|-0.016 738 891 601 562 496 530 614| = 0.016 738 891 601 562 496 530 614


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: 0.016 738 891 601 562 496 530 614.

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) 0.016 738 891 601 562 496 530 614 × 2 = 0 + 0.033 477 783 203 124 993 061 228;
  • 2) 0.033 477 783 203 124 993 061 228 × 2 = 0 + 0.066 955 566 406 249 986 122 456;
  • 3) 0.066 955 566 406 249 986 122 456 × 2 = 0 + 0.133 911 132 812 499 972 244 912;
  • 4) 0.133 911 132 812 499 972 244 912 × 2 = 0 + 0.267 822 265 624 999 944 489 824;
  • 5) 0.267 822 265 624 999 944 489 824 × 2 = 0 + 0.535 644 531 249 999 888 979 648;
  • 6) 0.535 644 531 249 999 888 979 648 × 2 = 1 + 0.071 289 062 499 999 777 959 296;
  • 7) 0.071 289 062 499 999 777 959 296 × 2 = 0 + 0.142 578 124 999 999 555 918 592;
  • 8) 0.142 578 124 999 999 555 918 592 × 2 = 0 + 0.285 156 249 999 999 111 837 184;
  • 9) 0.285 156 249 999 999 111 837 184 × 2 = 0 + 0.570 312 499 999 998 223 674 368;
  • 10) 0.570 312 499 999 998 223 674 368 × 2 = 1 + 0.140 624 999 999 996 447 348 736;
  • 11) 0.140 624 999 999 996 447 348 736 × 2 = 0 + 0.281 249 999 999 992 894 697 472;
  • 12) 0.281 249 999 999 992 894 697 472 × 2 = 0 + 0.562 499 999 999 985 789 394 944;
  • 13) 0.562 499 999 999 985 789 394 944 × 2 = 1 + 0.124 999 999 999 971 578 789 888;
  • 14) 0.124 999 999 999 971 578 789 888 × 2 = 0 + 0.249 999 999 999 943 157 579 776;
  • 15) 0.249 999 999 999 943 157 579 776 × 2 = 0 + 0.499 999 999 999 886 315 159 552;
  • 16) 0.499 999 999 999 886 315 159 552 × 2 = 0 + 0.999 999 999 999 772 630 319 104;
  • 17) 0.999 999 999 999 772 630 319 104 × 2 = 1 + 0.999 999 999 999 545 260 638 208;
  • 18) 0.999 999 999 999 545 260 638 208 × 2 = 1 + 0.999 999 999 999 090 521 276 416;
  • 19) 0.999 999 999 999 090 521 276 416 × 2 = 1 + 0.999 999 999 998 181 042 552 832;
  • 20) 0.999 999 999 998 181 042 552 832 × 2 = 1 + 0.999 999 999 996 362 085 105 664;
  • 21) 0.999 999 999 996 362 085 105 664 × 2 = 1 + 0.999 999 999 992 724 170 211 328;
  • 22) 0.999 999 999 992 724 170 211 328 × 2 = 1 + 0.999 999 999 985 448 340 422 656;
  • 23) 0.999 999 999 985 448 340 422 656 × 2 = 1 + 0.999 999 999 970 896 680 845 312;
  • 24) 0.999 999 999 970 896 680 845 312 × 2 = 1 + 0.999 999 999 941 793 361 690 624;
  • 25) 0.999 999 999 941 793 361 690 624 × 2 = 1 + 0.999 999 999 883 586 723 381 248;
  • 26) 0.999 999 999 883 586 723 381 248 × 2 = 1 + 0.999 999 999 767 173 446 762 496;
  • 27) 0.999 999 999 767 173 446 762 496 × 2 = 1 + 0.999 999 999 534 346 893 524 992;
  • 28) 0.999 999 999 534 346 893 524 992 × 2 = 1 + 0.999 999 999 068 693 787 049 984;
  • 29) 0.999 999 999 068 693 787 049 984 × 2 = 1 + 0.999 999 998 137 387 574 099 968;
  • 30) 0.999 999 998 137 387 574 099 968 × 2 = 1 + 0.999 999 996 274 775 148 199 936;
  • 31) 0.999 999 996 274 775 148 199 936 × 2 = 1 + 0.999 999 992 549 550 296 399 872;
  • 32) 0.999 999 992 549 550 296 399 872 × 2 = 1 + 0.999 999 985 099 100 592 799 744;
  • 33) 0.999 999 985 099 100 592 799 744 × 2 = 1 + 0.999 999 970 198 201 185 599 488;
  • 34) 0.999 999 970 198 201 185 599 488 × 2 = 1 + 0.999 999 940 396 402 371 198 976;
  • 35) 0.999 999 940 396 402 371 198 976 × 2 = 1 + 0.999 999 880 792 804 742 397 952;
  • 36) 0.999 999 880 792 804 742 397 952 × 2 = 1 + 0.999 999 761 585 609 484 795 904;
  • 37) 0.999 999 761 585 609 484 795 904 × 2 = 1 + 0.999 999 523 171 218 969 591 808;
  • 38) 0.999 999 523 171 218 969 591 808 × 2 = 1 + 0.999 999 046 342 437 939 183 616;
  • 39) 0.999 999 046 342 437 939 183 616 × 2 = 1 + 0.999 998 092 684 875 878 367 232;
  • 40) 0.999 998 092 684 875 878 367 232 × 2 = 1 + 0.999 996 185 369 751 756 734 464;
  • 41) 0.999 996 185 369 751 756 734 464 × 2 = 1 + 0.999 992 370 739 503 513 468 928;
  • 42) 0.999 992 370 739 503 513 468 928 × 2 = 1 + 0.999 984 741 479 007 026 937 856;
  • 43) 0.999 984 741 479 007 026 937 856 × 2 = 1 + 0.999 969 482 958 014 053 875 712;
  • 44) 0.999 969 482 958 014 053 875 712 × 2 = 1 + 0.999 938 965 916 028 107 751 424;
  • 45) 0.999 938 965 916 028 107 751 424 × 2 = 1 + 0.999 877 931 832 056 215 502 848;
  • 46) 0.999 877 931 832 056 215 502 848 × 2 = 1 + 0.999 755 863 664 112 431 005 696;
  • 47) 0.999 755 863 664 112 431 005 696 × 2 = 1 + 0.999 511 727 328 224 862 011 392;
  • 48) 0.999 511 727 328 224 862 011 392 × 2 = 1 + 0.999 023 454 656 449 724 022 784;
  • 49) 0.999 023 454 656 449 724 022 784 × 2 = 1 + 0.998 046 909 312 899 448 045 568;
  • 50) 0.998 046 909 312 899 448 045 568 × 2 = 1 + 0.996 093 818 625 798 896 091 136;
  • 51) 0.996 093 818 625 798 896 091 136 × 2 = 1 + 0.992 187 637 251 597 792 182 272;
  • 52) 0.992 187 637 251 597 792 182 272 × 2 = 1 + 0.984 375 274 503 195 584 364 544;
  • 53) 0.984 375 274 503 195 584 364 544 × 2 = 1 + 0.968 750 549 006 391 168 729 088;
  • 54) 0.968 750 549 006 391 168 729 088 × 2 = 1 + 0.937 501 098 012 782 337 458 176;
  • 55) 0.937 501 098 012 782 337 458 176 × 2 = 1 + 0.875 002 196 025 564 674 916 352;
  • 56) 0.875 002 196 025 564 674 916 352 × 2 = 1 + 0.750 004 392 051 129 349 832 704;
  • 57) 0.750 004 392 051 129 349 832 704 × 2 = 1 + 0.500 008 784 102 258 699 665 408;
  • 58) 0.500 008 784 102 258 699 665 408 × 2 = 1 + 0.000 017 568 204 517 399 330 816;

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:


0.016 738 891 601 562 496 530 614(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 614(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 530 614(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 496 530 614 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 1111

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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