-0.016 738 891 601 562 496 530 893 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 893(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
-0.016 738 891 601 562 496 530 893(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:

|-0.016 738 891 601 562 496 530 893| = 0.016 738 891 601 562 496 530 893


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: 0.016 738 891 601 562 496 530 893.

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) 0.016 738 891 601 562 496 530 893 × 2 = 0 + 0.033 477 783 203 124 993 061 786;
  • 2) 0.033 477 783 203 124 993 061 786 × 2 = 0 + 0.066 955 566 406 249 986 123 572;
  • 3) 0.066 955 566 406 249 986 123 572 × 2 = 0 + 0.133 911 132 812 499 972 247 144;
  • 4) 0.133 911 132 812 499 972 247 144 × 2 = 0 + 0.267 822 265 624 999 944 494 288;
  • 5) 0.267 822 265 624 999 944 494 288 × 2 = 0 + 0.535 644 531 249 999 888 988 576;
  • 6) 0.535 644 531 249 999 888 988 576 × 2 = 1 + 0.071 289 062 499 999 777 977 152;
  • 7) 0.071 289 062 499 999 777 977 152 × 2 = 0 + 0.142 578 124 999 999 555 954 304;
  • 8) 0.142 578 124 999 999 555 954 304 × 2 = 0 + 0.285 156 249 999 999 111 908 608;
  • 9) 0.285 156 249 999 999 111 908 608 × 2 = 0 + 0.570 312 499 999 998 223 817 216;
  • 10) 0.570 312 499 999 998 223 817 216 × 2 = 1 + 0.140 624 999 999 996 447 634 432;
  • 11) 0.140 624 999 999 996 447 634 432 × 2 = 0 + 0.281 249 999 999 992 895 268 864;
  • 12) 0.281 249 999 999 992 895 268 864 × 2 = 0 + 0.562 499 999 999 985 790 537 728;
  • 13) 0.562 499 999 999 985 790 537 728 × 2 = 1 + 0.124 999 999 999 971 581 075 456;
  • 14) 0.124 999 999 999 971 581 075 456 × 2 = 0 + 0.249 999 999 999 943 162 150 912;
  • 15) 0.249 999 999 999 943 162 150 912 × 2 = 0 + 0.499 999 999 999 886 324 301 824;
  • 16) 0.499 999 999 999 886 324 301 824 × 2 = 0 + 0.999 999 999 999 772 648 603 648;
  • 17) 0.999 999 999 999 772 648 603 648 × 2 = 1 + 0.999 999 999 999 545 297 207 296;
  • 18) 0.999 999 999 999 545 297 207 296 × 2 = 1 + 0.999 999 999 999 090 594 414 592;
  • 19) 0.999 999 999 999 090 594 414 592 × 2 = 1 + 0.999 999 999 998 181 188 829 184;
  • 20) 0.999 999 999 998 181 188 829 184 × 2 = 1 + 0.999 999 999 996 362 377 658 368;
  • 21) 0.999 999 999 996 362 377 658 368 × 2 = 1 + 0.999 999 999 992 724 755 316 736;
  • 22) 0.999 999 999 992 724 755 316 736 × 2 = 1 + 0.999 999 999 985 449 510 633 472;
  • 23) 0.999 999 999 985 449 510 633 472 × 2 = 1 + 0.999 999 999 970 899 021 266 944;
  • 24) 0.999 999 999 970 899 021 266 944 × 2 = 1 + 0.999 999 999 941 798 042 533 888;
  • 25) 0.999 999 999 941 798 042 533 888 × 2 = 1 + 0.999 999 999 883 596 085 067 776;
  • 26) 0.999 999 999 883 596 085 067 776 × 2 = 1 + 0.999 999 999 767 192 170 135 552;
  • 27) 0.999 999 999 767 192 170 135 552 × 2 = 1 + 0.999 999 999 534 384 340 271 104;
  • 28) 0.999 999 999 534 384 340 271 104 × 2 = 1 + 0.999 999 999 068 768 680 542 208;
  • 29) 0.999 999 999 068 768 680 542 208 × 2 = 1 + 0.999 999 998 137 537 361 084 416;
  • 30) 0.999 999 998 137 537 361 084 416 × 2 = 1 + 0.999 999 996 275 074 722 168 832;
  • 31) 0.999 999 996 275 074 722 168 832 × 2 = 1 + 0.999 999 992 550 149 444 337 664;
  • 32) 0.999 999 992 550 149 444 337 664 × 2 = 1 + 0.999 999 985 100 298 888 675 328;
  • 33) 0.999 999 985 100 298 888 675 328 × 2 = 1 + 0.999 999 970 200 597 777 350 656;
  • 34) 0.999 999 970 200 597 777 350 656 × 2 = 1 + 0.999 999 940 401 195 554 701 312;
  • 35) 0.999 999 940 401 195 554 701 312 × 2 = 1 + 0.999 999 880 802 391 109 402 624;
  • 36) 0.999 999 880 802 391 109 402 624 × 2 = 1 + 0.999 999 761 604 782 218 805 248;
  • 37) 0.999 999 761 604 782 218 805 248 × 2 = 1 + 0.999 999 523 209 564 437 610 496;
  • 38) 0.999 999 523 209 564 437 610 496 × 2 = 1 + 0.999 999 046 419 128 875 220 992;
  • 39) 0.999 999 046 419 128 875 220 992 × 2 = 1 + 0.999 998 092 838 257 750 441 984;
  • 40) 0.999 998 092 838 257 750 441 984 × 2 = 1 + 0.999 996 185 676 515 500 883 968;
  • 41) 0.999 996 185 676 515 500 883 968 × 2 = 1 + 0.999 992 371 353 031 001 767 936;
  • 42) 0.999 992 371 353 031 001 767 936 × 2 = 1 + 0.999 984 742 706 062 003 535 872;
  • 43) 0.999 984 742 706 062 003 535 872 × 2 = 1 + 0.999 969 485 412 124 007 071 744;
  • 44) 0.999 969 485 412 124 007 071 744 × 2 = 1 + 0.999 938 970 824 248 014 143 488;
  • 45) 0.999 938 970 824 248 014 143 488 × 2 = 1 + 0.999 877 941 648 496 028 286 976;
  • 46) 0.999 877 941 648 496 028 286 976 × 2 = 1 + 0.999 755 883 296 992 056 573 952;
  • 47) 0.999 755 883 296 992 056 573 952 × 2 = 1 + 0.999 511 766 593 984 113 147 904;
  • 48) 0.999 511 766 593 984 113 147 904 × 2 = 1 + 0.999 023 533 187 968 226 295 808;
  • 49) 0.999 023 533 187 968 226 295 808 × 2 = 1 + 0.998 047 066 375 936 452 591 616;
  • 50) 0.998 047 066 375 936 452 591 616 × 2 = 1 + 0.996 094 132 751 872 905 183 232;
  • 51) 0.996 094 132 751 872 905 183 232 × 2 = 1 + 0.992 188 265 503 745 810 366 464;
  • 52) 0.992 188 265 503 745 810 366 464 × 2 = 1 + 0.984 376 531 007 491 620 732 928;
  • 53) 0.984 376 531 007 491 620 732 928 × 2 = 1 + 0.968 753 062 014 983 241 465 856;
  • 54) 0.968 753 062 014 983 241 465 856 × 2 = 1 + 0.937 506 124 029 966 482 931 712;
  • 55) 0.937 506 124 029 966 482 931 712 × 2 = 1 + 0.875 012 248 059 932 965 863 424;
  • 56) 0.875 012 248 059 932 965 863 424 × 2 = 1 + 0.750 024 496 119 865 931 726 848;
  • 57) 0.750 024 496 119 865 931 726 848 × 2 = 1 + 0.500 048 992 239 731 863 453 696;
  • 58) 0.500 048 992 239 731 863 453 696 × 2 = 1 + 0.000 097 984 479 463 726 907 392;

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:


0.016 738 891 601 562 496 530 893(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 893(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:


0.016 738 891 601 562 496 530 893(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 -0.016 738 891 601 562 496 530 893 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