-0.016 738 891 601 562 496 842 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 842(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
-0.016 738 891 601 562 496 842(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:

|-0.016 738 891 601 562 496 842| = 0.016 738 891 601 562 496 842


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: 0.016 738 891 601 562 496 842.

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) 0.016 738 891 601 562 496 842 × 2 = 0 + 0.033 477 783 203 124 993 684;
  • 2) 0.033 477 783 203 124 993 684 × 2 = 0 + 0.066 955 566 406 249 987 368;
  • 3) 0.066 955 566 406 249 987 368 × 2 = 0 + 0.133 911 132 812 499 974 736;
  • 4) 0.133 911 132 812 499 974 736 × 2 = 0 + 0.267 822 265 624 999 949 472;
  • 5) 0.267 822 265 624 999 949 472 × 2 = 0 + 0.535 644 531 249 999 898 944;
  • 6) 0.535 644 531 249 999 898 944 × 2 = 1 + 0.071 289 062 499 999 797 888;
  • 7) 0.071 289 062 499 999 797 888 × 2 = 0 + 0.142 578 124 999 999 595 776;
  • 8) 0.142 578 124 999 999 595 776 × 2 = 0 + 0.285 156 249 999 999 191 552;
  • 9) 0.285 156 249 999 999 191 552 × 2 = 0 + 0.570 312 499 999 998 383 104;
  • 10) 0.570 312 499 999 998 383 104 × 2 = 1 + 0.140 624 999 999 996 766 208;
  • 11) 0.140 624 999 999 996 766 208 × 2 = 0 + 0.281 249 999 999 993 532 416;
  • 12) 0.281 249 999 999 993 532 416 × 2 = 0 + 0.562 499 999 999 987 064 832;
  • 13) 0.562 499 999 999 987 064 832 × 2 = 1 + 0.124 999 999 999 974 129 664;
  • 14) 0.124 999 999 999 974 129 664 × 2 = 0 + 0.249 999 999 999 948 259 328;
  • 15) 0.249 999 999 999 948 259 328 × 2 = 0 + 0.499 999 999 999 896 518 656;
  • 16) 0.499 999 999 999 896 518 656 × 2 = 0 + 0.999 999 999 999 793 037 312;
  • 17) 0.999 999 999 999 793 037 312 × 2 = 1 + 0.999 999 999 999 586 074 624;
  • 18) 0.999 999 999 999 586 074 624 × 2 = 1 + 0.999 999 999 999 172 149 248;
  • 19) 0.999 999 999 999 172 149 248 × 2 = 1 + 0.999 999 999 998 344 298 496;
  • 20) 0.999 999 999 998 344 298 496 × 2 = 1 + 0.999 999 999 996 688 596 992;
  • 21) 0.999 999 999 996 688 596 992 × 2 = 1 + 0.999 999 999 993 377 193 984;
  • 22) 0.999 999 999 993 377 193 984 × 2 = 1 + 0.999 999 999 986 754 387 968;
  • 23) 0.999 999 999 986 754 387 968 × 2 = 1 + 0.999 999 999 973 508 775 936;
  • 24) 0.999 999 999 973 508 775 936 × 2 = 1 + 0.999 999 999 947 017 551 872;
  • 25) 0.999 999 999 947 017 551 872 × 2 = 1 + 0.999 999 999 894 035 103 744;
  • 26) 0.999 999 999 894 035 103 744 × 2 = 1 + 0.999 999 999 788 070 207 488;
  • 27) 0.999 999 999 788 070 207 488 × 2 = 1 + 0.999 999 999 576 140 414 976;
  • 28) 0.999 999 999 576 140 414 976 × 2 = 1 + 0.999 999 999 152 280 829 952;
  • 29) 0.999 999 999 152 280 829 952 × 2 = 1 + 0.999 999 998 304 561 659 904;
  • 30) 0.999 999 998 304 561 659 904 × 2 = 1 + 0.999 999 996 609 123 319 808;
  • 31) 0.999 999 996 609 123 319 808 × 2 = 1 + 0.999 999 993 218 246 639 616;
  • 32) 0.999 999 993 218 246 639 616 × 2 = 1 + 0.999 999 986 436 493 279 232;
  • 33) 0.999 999 986 436 493 279 232 × 2 = 1 + 0.999 999 972 872 986 558 464;
  • 34) 0.999 999 972 872 986 558 464 × 2 = 1 + 0.999 999 945 745 973 116 928;
  • 35) 0.999 999 945 745 973 116 928 × 2 = 1 + 0.999 999 891 491 946 233 856;
  • 36) 0.999 999 891 491 946 233 856 × 2 = 1 + 0.999 999 782 983 892 467 712;
  • 37) 0.999 999 782 983 892 467 712 × 2 = 1 + 0.999 999 565 967 784 935 424;
  • 38) 0.999 999 565 967 784 935 424 × 2 = 1 + 0.999 999 131 935 569 870 848;
  • 39) 0.999 999 131 935 569 870 848 × 2 = 1 + 0.999 998 263 871 139 741 696;
  • 40) 0.999 998 263 871 139 741 696 × 2 = 1 + 0.999 996 527 742 279 483 392;
  • 41) 0.999 996 527 742 279 483 392 × 2 = 1 + 0.999 993 055 484 558 966 784;
  • 42) 0.999 993 055 484 558 966 784 × 2 = 1 + 0.999 986 110 969 117 933 568;
  • 43) 0.999 986 110 969 117 933 568 × 2 = 1 + 0.999 972 221 938 235 867 136;
  • 44) 0.999 972 221 938 235 867 136 × 2 = 1 + 0.999 944 443 876 471 734 272;
  • 45) 0.999 944 443 876 471 734 272 × 2 = 1 + 0.999 888 887 752 943 468 544;
  • 46) 0.999 888 887 752 943 468 544 × 2 = 1 + 0.999 777 775 505 886 937 088;
  • 47) 0.999 777 775 505 886 937 088 × 2 = 1 + 0.999 555 551 011 773 874 176;
  • 48) 0.999 555 551 011 773 874 176 × 2 = 1 + 0.999 111 102 023 547 748 352;
  • 49) 0.999 111 102 023 547 748 352 × 2 = 1 + 0.998 222 204 047 095 496 704;
  • 50) 0.998 222 204 047 095 496 704 × 2 = 1 + 0.996 444 408 094 190 993 408;
  • 51) 0.996 444 408 094 190 993 408 × 2 = 1 + 0.992 888 816 188 381 986 816;
  • 52) 0.992 888 816 188 381 986 816 × 2 = 1 + 0.985 777 632 376 763 973 632;
  • 53) 0.985 777 632 376 763 973 632 × 2 = 1 + 0.971 555 264 753 527 947 264;
  • 54) 0.971 555 264 753 527 947 264 × 2 = 1 + 0.943 110 529 507 055 894 528;
  • 55) 0.943 110 529 507 055 894 528 × 2 = 1 + 0.886 221 059 014 111 789 056;
  • 56) 0.886 221 059 014 111 789 056 × 2 = 1 + 0.772 442 118 028 223 578 112;
  • 57) 0.772 442 118 028 223 578 112 × 2 = 1 + 0.544 884 236 056 447 156 224;
  • 58) 0.544 884 236 056 447 156 224 × 2 = 1 + 0.089 768 472 112 894 312 448;

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:


0.016 738 891 601 562 496 842(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 842(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:


0.016 738 891 601 562 496 842(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 -0.016 738 891 601 562 496 842 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