-0.016 738 891 601 562 496 591 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 591(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
-0.016 738 891 601 562 496 591(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:

|-0.016 738 891 601 562 496 591| = 0.016 738 891 601 562 496 591


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: 0.016 738 891 601 562 496 591.

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) 0.016 738 891 601 562 496 591 × 2 = 0 + 0.033 477 783 203 124 993 182;
  • 2) 0.033 477 783 203 124 993 182 × 2 = 0 + 0.066 955 566 406 249 986 364;
  • 3) 0.066 955 566 406 249 986 364 × 2 = 0 + 0.133 911 132 812 499 972 728;
  • 4) 0.133 911 132 812 499 972 728 × 2 = 0 + 0.267 822 265 624 999 945 456;
  • 5) 0.267 822 265 624 999 945 456 × 2 = 0 + 0.535 644 531 249 999 890 912;
  • 6) 0.535 644 531 249 999 890 912 × 2 = 1 + 0.071 289 062 499 999 781 824;
  • 7) 0.071 289 062 499 999 781 824 × 2 = 0 + 0.142 578 124 999 999 563 648;
  • 8) 0.142 578 124 999 999 563 648 × 2 = 0 + 0.285 156 249 999 999 127 296;
  • 9) 0.285 156 249 999 999 127 296 × 2 = 0 + 0.570 312 499 999 998 254 592;
  • 10) 0.570 312 499 999 998 254 592 × 2 = 1 + 0.140 624 999 999 996 509 184;
  • 11) 0.140 624 999 999 996 509 184 × 2 = 0 + 0.281 249 999 999 993 018 368;
  • 12) 0.281 249 999 999 993 018 368 × 2 = 0 + 0.562 499 999 999 986 036 736;
  • 13) 0.562 499 999 999 986 036 736 × 2 = 1 + 0.124 999 999 999 972 073 472;
  • 14) 0.124 999 999 999 972 073 472 × 2 = 0 + 0.249 999 999 999 944 146 944;
  • 15) 0.249 999 999 999 944 146 944 × 2 = 0 + 0.499 999 999 999 888 293 888;
  • 16) 0.499 999 999 999 888 293 888 × 2 = 0 + 0.999 999 999 999 776 587 776;
  • 17) 0.999 999 999 999 776 587 776 × 2 = 1 + 0.999 999 999 999 553 175 552;
  • 18) 0.999 999 999 999 553 175 552 × 2 = 1 + 0.999 999 999 999 106 351 104;
  • 19) 0.999 999 999 999 106 351 104 × 2 = 1 + 0.999 999 999 998 212 702 208;
  • 20) 0.999 999 999 998 212 702 208 × 2 = 1 + 0.999 999 999 996 425 404 416;
  • 21) 0.999 999 999 996 425 404 416 × 2 = 1 + 0.999 999 999 992 850 808 832;
  • 22) 0.999 999 999 992 850 808 832 × 2 = 1 + 0.999 999 999 985 701 617 664;
  • 23) 0.999 999 999 985 701 617 664 × 2 = 1 + 0.999 999 999 971 403 235 328;
  • 24) 0.999 999 999 971 403 235 328 × 2 = 1 + 0.999 999 999 942 806 470 656;
  • 25) 0.999 999 999 942 806 470 656 × 2 = 1 + 0.999 999 999 885 612 941 312;
  • 26) 0.999 999 999 885 612 941 312 × 2 = 1 + 0.999 999 999 771 225 882 624;
  • 27) 0.999 999 999 771 225 882 624 × 2 = 1 + 0.999 999 999 542 451 765 248;
  • 28) 0.999 999 999 542 451 765 248 × 2 = 1 + 0.999 999 999 084 903 530 496;
  • 29) 0.999 999 999 084 903 530 496 × 2 = 1 + 0.999 999 998 169 807 060 992;
  • 30) 0.999 999 998 169 807 060 992 × 2 = 1 + 0.999 999 996 339 614 121 984;
  • 31) 0.999 999 996 339 614 121 984 × 2 = 1 + 0.999 999 992 679 228 243 968;
  • 32) 0.999 999 992 679 228 243 968 × 2 = 1 + 0.999 999 985 358 456 487 936;
  • 33) 0.999 999 985 358 456 487 936 × 2 = 1 + 0.999 999 970 716 912 975 872;
  • 34) 0.999 999 970 716 912 975 872 × 2 = 1 + 0.999 999 941 433 825 951 744;
  • 35) 0.999 999 941 433 825 951 744 × 2 = 1 + 0.999 999 882 867 651 903 488;
  • 36) 0.999 999 882 867 651 903 488 × 2 = 1 + 0.999 999 765 735 303 806 976;
  • 37) 0.999 999 765 735 303 806 976 × 2 = 1 + 0.999 999 531 470 607 613 952;
  • 38) 0.999 999 531 470 607 613 952 × 2 = 1 + 0.999 999 062 941 215 227 904;
  • 39) 0.999 999 062 941 215 227 904 × 2 = 1 + 0.999 998 125 882 430 455 808;
  • 40) 0.999 998 125 882 430 455 808 × 2 = 1 + 0.999 996 251 764 860 911 616;
  • 41) 0.999 996 251 764 860 911 616 × 2 = 1 + 0.999 992 503 529 721 823 232;
  • 42) 0.999 992 503 529 721 823 232 × 2 = 1 + 0.999 985 007 059 443 646 464;
  • 43) 0.999 985 007 059 443 646 464 × 2 = 1 + 0.999 970 014 118 887 292 928;
  • 44) 0.999 970 014 118 887 292 928 × 2 = 1 + 0.999 940 028 237 774 585 856;
  • 45) 0.999 940 028 237 774 585 856 × 2 = 1 + 0.999 880 056 475 549 171 712;
  • 46) 0.999 880 056 475 549 171 712 × 2 = 1 + 0.999 760 112 951 098 343 424;
  • 47) 0.999 760 112 951 098 343 424 × 2 = 1 + 0.999 520 225 902 196 686 848;
  • 48) 0.999 520 225 902 196 686 848 × 2 = 1 + 0.999 040 451 804 393 373 696;
  • 49) 0.999 040 451 804 393 373 696 × 2 = 1 + 0.998 080 903 608 786 747 392;
  • 50) 0.998 080 903 608 786 747 392 × 2 = 1 + 0.996 161 807 217 573 494 784;
  • 51) 0.996 161 807 217 573 494 784 × 2 = 1 + 0.992 323 614 435 146 989 568;
  • 52) 0.992 323 614 435 146 989 568 × 2 = 1 + 0.984 647 228 870 293 979 136;
  • 53) 0.984 647 228 870 293 979 136 × 2 = 1 + 0.969 294 457 740 587 958 272;
  • 54) 0.969 294 457 740 587 958 272 × 2 = 1 + 0.938 588 915 481 175 916 544;
  • 55) 0.938 588 915 481 175 916 544 × 2 = 1 + 0.877 177 830 962 351 833 088;
  • 56) 0.877 177 830 962 351 833 088 × 2 = 1 + 0.754 355 661 924 703 666 176;
  • 57) 0.754 355 661 924 703 666 176 × 2 = 1 + 0.508 711 323 849 407 332 352;
  • 58) 0.508 711 323 849 407 332 352 × 2 = 1 + 0.017 422 647 698 814 664 704;

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:


0.016 738 891 601 562 496 591(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 591(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:


0.016 738 891 601 562 496 591(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 -0.016 738 891 601 562 496 591 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