-0.016 738 891 601 562 496 368 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 368(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
-0.016 738 891 601 562 496 368(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:

|-0.016 738 891 601 562 496 368| = 0.016 738 891 601 562 496 368


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: 0.016 738 891 601 562 496 368.

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) 0.016 738 891 601 562 496 368 × 2 = 0 + 0.033 477 783 203 124 992 736;
  • 2) 0.033 477 783 203 124 992 736 × 2 = 0 + 0.066 955 566 406 249 985 472;
  • 3) 0.066 955 566 406 249 985 472 × 2 = 0 + 0.133 911 132 812 499 970 944;
  • 4) 0.133 911 132 812 499 970 944 × 2 = 0 + 0.267 822 265 624 999 941 888;
  • 5) 0.267 822 265 624 999 941 888 × 2 = 0 + 0.535 644 531 249 999 883 776;
  • 6) 0.535 644 531 249 999 883 776 × 2 = 1 + 0.071 289 062 499 999 767 552;
  • 7) 0.071 289 062 499 999 767 552 × 2 = 0 + 0.142 578 124 999 999 535 104;
  • 8) 0.142 578 124 999 999 535 104 × 2 = 0 + 0.285 156 249 999 999 070 208;
  • 9) 0.285 156 249 999 999 070 208 × 2 = 0 + 0.570 312 499 999 998 140 416;
  • 10) 0.570 312 499 999 998 140 416 × 2 = 1 + 0.140 624 999 999 996 280 832;
  • 11) 0.140 624 999 999 996 280 832 × 2 = 0 + 0.281 249 999 999 992 561 664;
  • 12) 0.281 249 999 999 992 561 664 × 2 = 0 + 0.562 499 999 999 985 123 328;
  • 13) 0.562 499 999 999 985 123 328 × 2 = 1 + 0.124 999 999 999 970 246 656;
  • 14) 0.124 999 999 999 970 246 656 × 2 = 0 + 0.249 999 999 999 940 493 312;
  • 15) 0.249 999 999 999 940 493 312 × 2 = 0 + 0.499 999 999 999 880 986 624;
  • 16) 0.499 999 999 999 880 986 624 × 2 = 0 + 0.999 999 999 999 761 973 248;
  • 17) 0.999 999 999 999 761 973 248 × 2 = 1 + 0.999 999 999 999 523 946 496;
  • 18) 0.999 999 999 999 523 946 496 × 2 = 1 + 0.999 999 999 999 047 892 992;
  • 19) 0.999 999 999 999 047 892 992 × 2 = 1 + 0.999 999 999 998 095 785 984;
  • 20) 0.999 999 999 998 095 785 984 × 2 = 1 + 0.999 999 999 996 191 571 968;
  • 21) 0.999 999 999 996 191 571 968 × 2 = 1 + 0.999 999 999 992 383 143 936;
  • 22) 0.999 999 999 992 383 143 936 × 2 = 1 + 0.999 999 999 984 766 287 872;
  • 23) 0.999 999 999 984 766 287 872 × 2 = 1 + 0.999 999 999 969 532 575 744;
  • 24) 0.999 999 999 969 532 575 744 × 2 = 1 + 0.999 999 999 939 065 151 488;
  • 25) 0.999 999 999 939 065 151 488 × 2 = 1 + 0.999 999 999 878 130 302 976;
  • 26) 0.999 999 999 878 130 302 976 × 2 = 1 + 0.999 999 999 756 260 605 952;
  • 27) 0.999 999 999 756 260 605 952 × 2 = 1 + 0.999 999 999 512 521 211 904;
  • 28) 0.999 999 999 512 521 211 904 × 2 = 1 + 0.999 999 999 025 042 423 808;
  • 29) 0.999 999 999 025 042 423 808 × 2 = 1 + 0.999 999 998 050 084 847 616;
  • 30) 0.999 999 998 050 084 847 616 × 2 = 1 + 0.999 999 996 100 169 695 232;
  • 31) 0.999 999 996 100 169 695 232 × 2 = 1 + 0.999 999 992 200 339 390 464;
  • 32) 0.999 999 992 200 339 390 464 × 2 = 1 + 0.999 999 984 400 678 780 928;
  • 33) 0.999 999 984 400 678 780 928 × 2 = 1 + 0.999 999 968 801 357 561 856;
  • 34) 0.999 999 968 801 357 561 856 × 2 = 1 + 0.999 999 937 602 715 123 712;
  • 35) 0.999 999 937 602 715 123 712 × 2 = 1 + 0.999 999 875 205 430 247 424;
  • 36) 0.999 999 875 205 430 247 424 × 2 = 1 + 0.999 999 750 410 860 494 848;
  • 37) 0.999 999 750 410 860 494 848 × 2 = 1 + 0.999 999 500 821 720 989 696;
  • 38) 0.999 999 500 821 720 989 696 × 2 = 1 + 0.999 999 001 643 441 979 392;
  • 39) 0.999 999 001 643 441 979 392 × 2 = 1 + 0.999 998 003 286 883 958 784;
  • 40) 0.999 998 003 286 883 958 784 × 2 = 1 + 0.999 996 006 573 767 917 568;
  • 41) 0.999 996 006 573 767 917 568 × 2 = 1 + 0.999 992 013 147 535 835 136;
  • 42) 0.999 992 013 147 535 835 136 × 2 = 1 + 0.999 984 026 295 071 670 272;
  • 43) 0.999 984 026 295 071 670 272 × 2 = 1 + 0.999 968 052 590 143 340 544;
  • 44) 0.999 968 052 590 143 340 544 × 2 = 1 + 0.999 936 105 180 286 681 088;
  • 45) 0.999 936 105 180 286 681 088 × 2 = 1 + 0.999 872 210 360 573 362 176;
  • 46) 0.999 872 210 360 573 362 176 × 2 = 1 + 0.999 744 420 721 146 724 352;
  • 47) 0.999 744 420 721 146 724 352 × 2 = 1 + 0.999 488 841 442 293 448 704;
  • 48) 0.999 488 841 442 293 448 704 × 2 = 1 + 0.998 977 682 884 586 897 408;
  • 49) 0.998 977 682 884 586 897 408 × 2 = 1 + 0.997 955 365 769 173 794 816;
  • 50) 0.997 955 365 769 173 794 816 × 2 = 1 + 0.995 910 731 538 347 589 632;
  • 51) 0.995 910 731 538 347 589 632 × 2 = 1 + 0.991 821 463 076 695 179 264;
  • 52) 0.991 821 463 076 695 179 264 × 2 = 1 + 0.983 642 926 153 390 358 528;
  • 53) 0.983 642 926 153 390 358 528 × 2 = 1 + 0.967 285 852 306 780 717 056;
  • 54) 0.967 285 852 306 780 717 056 × 2 = 1 + 0.934 571 704 613 561 434 112;
  • 55) 0.934 571 704 613 561 434 112 × 2 = 1 + 0.869 143 409 227 122 868 224;
  • 56) 0.869 143 409 227 122 868 224 × 2 = 1 + 0.738 286 818 454 245 736 448;
  • 57) 0.738 286 818 454 245 736 448 × 2 = 1 + 0.476 573 636 908 491 472 896;
  • 58) 0.476 573 636 908 491 472 896 × 2 = 0 + 0.953 147 273 816 982 945 792;

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:


0.016 738 891 601 562 496 368(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 368(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:


0.016 738 891 601 562 496 368(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 -0.016 738 891 601 562 496 368 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