-0.016 738 891 601 562 496 736 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 736(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
-0.016 738 891 601 562 496 736(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:

|-0.016 738 891 601 562 496 736| = 0.016 738 891 601 562 496 736


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: 0.016 738 891 601 562 496 736.

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) 0.016 738 891 601 562 496 736 × 2 = 0 + 0.033 477 783 203 124 993 472;
  • 2) 0.033 477 783 203 124 993 472 × 2 = 0 + 0.066 955 566 406 249 986 944;
  • 3) 0.066 955 566 406 249 986 944 × 2 = 0 + 0.133 911 132 812 499 973 888;
  • 4) 0.133 911 132 812 499 973 888 × 2 = 0 + 0.267 822 265 624 999 947 776;
  • 5) 0.267 822 265 624 999 947 776 × 2 = 0 + 0.535 644 531 249 999 895 552;
  • 6) 0.535 644 531 249 999 895 552 × 2 = 1 + 0.071 289 062 499 999 791 104;
  • 7) 0.071 289 062 499 999 791 104 × 2 = 0 + 0.142 578 124 999 999 582 208;
  • 8) 0.142 578 124 999 999 582 208 × 2 = 0 + 0.285 156 249 999 999 164 416;
  • 9) 0.285 156 249 999 999 164 416 × 2 = 0 + 0.570 312 499 999 998 328 832;
  • 10) 0.570 312 499 999 998 328 832 × 2 = 1 + 0.140 624 999 999 996 657 664;
  • 11) 0.140 624 999 999 996 657 664 × 2 = 0 + 0.281 249 999 999 993 315 328;
  • 12) 0.281 249 999 999 993 315 328 × 2 = 0 + 0.562 499 999 999 986 630 656;
  • 13) 0.562 499 999 999 986 630 656 × 2 = 1 + 0.124 999 999 999 973 261 312;
  • 14) 0.124 999 999 999 973 261 312 × 2 = 0 + 0.249 999 999 999 946 522 624;
  • 15) 0.249 999 999 999 946 522 624 × 2 = 0 + 0.499 999 999 999 893 045 248;
  • 16) 0.499 999 999 999 893 045 248 × 2 = 0 + 0.999 999 999 999 786 090 496;
  • 17) 0.999 999 999 999 786 090 496 × 2 = 1 + 0.999 999 999 999 572 180 992;
  • 18) 0.999 999 999 999 572 180 992 × 2 = 1 + 0.999 999 999 999 144 361 984;
  • 19) 0.999 999 999 999 144 361 984 × 2 = 1 + 0.999 999 999 998 288 723 968;
  • 20) 0.999 999 999 998 288 723 968 × 2 = 1 + 0.999 999 999 996 577 447 936;
  • 21) 0.999 999 999 996 577 447 936 × 2 = 1 + 0.999 999 999 993 154 895 872;
  • 22) 0.999 999 999 993 154 895 872 × 2 = 1 + 0.999 999 999 986 309 791 744;
  • 23) 0.999 999 999 986 309 791 744 × 2 = 1 + 0.999 999 999 972 619 583 488;
  • 24) 0.999 999 999 972 619 583 488 × 2 = 1 + 0.999 999 999 945 239 166 976;
  • 25) 0.999 999 999 945 239 166 976 × 2 = 1 + 0.999 999 999 890 478 333 952;
  • 26) 0.999 999 999 890 478 333 952 × 2 = 1 + 0.999 999 999 780 956 667 904;
  • 27) 0.999 999 999 780 956 667 904 × 2 = 1 + 0.999 999 999 561 913 335 808;
  • 28) 0.999 999 999 561 913 335 808 × 2 = 1 + 0.999 999 999 123 826 671 616;
  • 29) 0.999 999 999 123 826 671 616 × 2 = 1 + 0.999 999 998 247 653 343 232;
  • 30) 0.999 999 998 247 653 343 232 × 2 = 1 + 0.999 999 996 495 306 686 464;
  • 31) 0.999 999 996 495 306 686 464 × 2 = 1 + 0.999 999 992 990 613 372 928;
  • 32) 0.999 999 992 990 613 372 928 × 2 = 1 + 0.999 999 985 981 226 745 856;
  • 33) 0.999 999 985 981 226 745 856 × 2 = 1 + 0.999 999 971 962 453 491 712;
  • 34) 0.999 999 971 962 453 491 712 × 2 = 1 + 0.999 999 943 924 906 983 424;
  • 35) 0.999 999 943 924 906 983 424 × 2 = 1 + 0.999 999 887 849 813 966 848;
  • 36) 0.999 999 887 849 813 966 848 × 2 = 1 + 0.999 999 775 699 627 933 696;
  • 37) 0.999 999 775 699 627 933 696 × 2 = 1 + 0.999 999 551 399 255 867 392;
  • 38) 0.999 999 551 399 255 867 392 × 2 = 1 + 0.999 999 102 798 511 734 784;
  • 39) 0.999 999 102 798 511 734 784 × 2 = 1 + 0.999 998 205 597 023 469 568;
  • 40) 0.999 998 205 597 023 469 568 × 2 = 1 + 0.999 996 411 194 046 939 136;
  • 41) 0.999 996 411 194 046 939 136 × 2 = 1 + 0.999 992 822 388 093 878 272;
  • 42) 0.999 992 822 388 093 878 272 × 2 = 1 + 0.999 985 644 776 187 756 544;
  • 43) 0.999 985 644 776 187 756 544 × 2 = 1 + 0.999 971 289 552 375 513 088;
  • 44) 0.999 971 289 552 375 513 088 × 2 = 1 + 0.999 942 579 104 751 026 176;
  • 45) 0.999 942 579 104 751 026 176 × 2 = 1 + 0.999 885 158 209 502 052 352;
  • 46) 0.999 885 158 209 502 052 352 × 2 = 1 + 0.999 770 316 419 004 104 704;
  • 47) 0.999 770 316 419 004 104 704 × 2 = 1 + 0.999 540 632 838 008 209 408;
  • 48) 0.999 540 632 838 008 209 408 × 2 = 1 + 0.999 081 265 676 016 418 816;
  • 49) 0.999 081 265 676 016 418 816 × 2 = 1 + 0.998 162 531 352 032 837 632;
  • 50) 0.998 162 531 352 032 837 632 × 2 = 1 + 0.996 325 062 704 065 675 264;
  • 51) 0.996 325 062 704 065 675 264 × 2 = 1 + 0.992 650 125 408 131 350 528;
  • 52) 0.992 650 125 408 131 350 528 × 2 = 1 + 0.985 300 250 816 262 701 056;
  • 53) 0.985 300 250 816 262 701 056 × 2 = 1 + 0.970 600 501 632 525 402 112;
  • 54) 0.970 600 501 632 525 402 112 × 2 = 1 + 0.941 201 003 265 050 804 224;
  • 55) 0.941 201 003 265 050 804 224 × 2 = 1 + 0.882 402 006 530 101 608 448;
  • 56) 0.882 402 006 530 101 608 448 × 2 = 1 + 0.764 804 013 060 203 216 896;
  • 57) 0.764 804 013 060 203 216 896 × 2 = 1 + 0.529 608 026 120 406 433 792;
  • 58) 0.529 608 026 120 406 433 792 × 2 = 1 + 0.059 216 052 240 812 867 584;

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:


0.016 738 891 601 562 496 736(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 736(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:


0.016 738 891 601 562 496 736(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 -0.016 738 891 601 562 496 736 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