-0.016 738 891 601 562 496 472 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 472(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
-0.016 738 891 601 562 496 472(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:

|-0.016 738 891 601 562 496 472| = 0.016 738 891 601 562 496 472


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: 0.016 738 891 601 562 496 472.

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) 0.016 738 891 601 562 496 472 × 2 = 0 + 0.033 477 783 203 124 992 944;
  • 2) 0.033 477 783 203 124 992 944 × 2 = 0 + 0.066 955 566 406 249 985 888;
  • 3) 0.066 955 566 406 249 985 888 × 2 = 0 + 0.133 911 132 812 499 971 776;
  • 4) 0.133 911 132 812 499 971 776 × 2 = 0 + 0.267 822 265 624 999 943 552;
  • 5) 0.267 822 265 624 999 943 552 × 2 = 0 + 0.535 644 531 249 999 887 104;
  • 6) 0.535 644 531 249 999 887 104 × 2 = 1 + 0.071 289 062 499 999 774 208;
  • 7) 0.071 289 062 499 999 774 208 × 2 = 0 + 0.142 578 124 999 999 548 416;
  • 8) 0.142 578 124 999 999 548 416 × 2 = 0 + 0.285 156 249 999 999 096 832;
  • 9) 0.285 156 249 999 999 096 832 × 2 = 0 + 0.570 312 499 999 998 193 664;
  • 10) 0.570 312 499 999 998 193 664 × 2 = 1 + 0.140 624 999 999 996 387 328;
  • 11) 0.140 624 999 999 996 387 328 × 2 = 0 + 0.281 249 999 999 992 774 656;
  • 12) 0.281 249 999 999 992 774 656 × 2 = 0 + 0.562 499 999 999 985 549 312;
  • 13) 0.562 499 999 999 985 549 312 × 2 = 1 + 0.124 999 999 999 971 098 624;
  • 14) 0.124 999 999 999 971 098 624 × 2 = 0 + 0.249 999 999 999 942 197 248;
  • 15) 0.249 999 999 999 942 197 248 × 2 = 0 + 0.499 999 999 999 884 394 496;
  • 16) 0.499 999 999 999 884 394 496 × 2 = 0 + 0.999 999 999 999 768 788 992;
  • 17) 0.999 999 999 999 768 788 992 × 2 = 1 + 0.999 999 999 999 537 577 984;
  • 18) 0.999 999 999 999 537 577 984 × 2 = 1 + 0.999 999 999 999 075 155 968;
  • 19) 0.999 999 999 999 075 155 968 × 2 = 1 + 0.999 999 999 998 150 311 936;
  • 20) 0.999 999 999 998 150 311 936 × 2 = 1 + 0.999 999 999 996 300 623 872;
  • 21) 0.999 999 999 996 300 623 872 × 2 = 1 + 0.999 999 999 992 601 247 744;
  • 22) 0.999 999 999 992 601 247 744 × 2 = 1 + 0.999 999 999 985 202 495 488;
  • 23) 0.999 999 999 985 202 495 488 × 2 = 1 + 0.999 999 999 970 404 990 976;
  • 24) 0.999 999 999 970 404 990 976 × 2 = 1 + 0.999 999 999 940 809 981 952;
  • 25) 0.999 999 999 940 809 981 952 × 2 = 1 + 0.999 999 999 881 619 963 904;
  • 26) 0.999 999 999 881 619 963 904 × 2 = 1 + 0.999 999 999 763 239 927 808;
  • 27) 0.999 999 999 763 239 927 808 × 2 = 1 + 0.999 999 999 526 479 855 616;
  • 28) 0.999 999 999 526 479 855 616 × 2 = 1 + 0.999 999 999 052 959 711 232;
  • 29) 0.999 999 999 052 959 711 232 × 2 = 1 + 0.999 999 998 105 919 422 464;
  • 30) 0.999 999 998 105 919 422 464 × 2 = 1 + 0.999 999 996 211 838 844 928;
  • 31) 0.999 999 996 211 838 844 928 × 2 = 1 + 0.999 999 992 423 677 689 856;
  • 32) 0.999 999 992 423 677 689 856 × 2 = 1 + 0.999 999 984 847 355 379 712;
  • 33) 0.999 999 984 847 355 379 712 × 2 = 1 + 0.999 999 969 694 710 759 424;
  • 34) 0.999 999 969 694 710 759 424 × 2 = 1 + 0.999 999 939 389 421 518 848;
  • 35) 0.999 999 939 389 421 518 848 × 2 = 1 + 0.999 999 878 778 843 037 696;
  • 36) 0.999 999 878 778 843 037 696 × 2 = 1 + 0.999 999 757 557 686 075 392;
  • 37) 0.999 999 757 557 686 075 392 × 2 = 1 + 0.999 999 515 115 372 150 784;
  • 38) 0.999 999 515 115 372 150 784 × 2 = 1 + 0.999 999 030 230 744 301 568;
  • 39) 0.999 999 030 230 744 301 568 × 2 = 1 + 0.999 998 060 461 488 603 136;
  • 40) 0.999 998 060 461 488 603 136 × 2 = 1 + 0.999 996 120 922 977 206 272;
  • 41) 0.999 996 120 922 977 206 272 × 2 = 1 + 0.999 992 241 845 954 412 544;
  • 42) 0.999 992 241 845 954 412 544 × 2 = 1 + 0.999 984 483 691 908 825 088;
  • 43) 0.999 984 483 691 908 825 088 × 2 = 1 + 0.999 968 967 383 817 650 176;
  • 44) 0.999 968 967 383 817 650 176 × 2 = 1 + 0.999 937 934 767 635 300 352;
  • 45) 0.999 937 934 767 635 300 352 × 2 = 1 + 0.999 875 869 535 270 600 704;
  • 46) 0.999 875 869 535 270 600 704 × 2 = 1 + 0.999 751 739 070 541 201 408;
  • 47) 0.999 751 739 070 541 201 408 × 2 = 1 + 0.999 503 478 141 082 402 816;
  • 48) 0.999 503 478 141 082 402 816 × 2 = 1 + 0.999 006 956 282 164 805 632;
  • 49) 0.999 006 956 282 164 805 632 × 2 = 1 + 0.998 013 912 564 329 611 264;
  • 50) 0.998 013 912 564 329 611 264 × 2 = 1 + 0.996 027 825 128 659 222 528;
  • 51) 0.996 027 825 128 659 222 528 × 2 = 1 + 0.992 055 650 257 318 445 056;
  • 52) 0.992 055 650 257 318 445 056 × 2 = 1 + 0.984 111 300 514 636 890 112;
  • 53) 0.984 111 300 514 636 890 112 × 2 = 1 + 0.968 222 601 029 273 780 224;
  • 54) 0.968 222 601 029 273 780 224 × 2 = 1 + 0.936 445 202 058 547 560 448;
  • 55) 0.936 445 202 058 547 560 448 × 2 = 1 + 0.872 890 404 117 095 120 896;
  • 56) 0.872 890 404 117 095 120 896 × 2 = 1 + 0.745 780 808 234 190 241 792;
  • 57) 0.745 780 808 234 190 241 792 × 2 = 1 + 0.491 561 616 468 380 483 584;
  • 58) 0.491 561 616 468 380 483 584 × 2 = 0 + 0.983 123 232 936 760 967 168;

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:


0.016 738 891 601 562 496 472(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 472(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:


0.016 738 891 601 562 496 472(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 -0.016 738 891 601 562 496 472 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