-0.016 738 891 601 562 496 530 597 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 597(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
-0.016 738 891 601 562 496 530 597(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:

|-0.016 738 891 601 562 496 530 597| = 0.016 738 891 601 562 496 530 597


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: 0.016 738 891 601 562 496 530 597.

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) 0.016 738 891 601 562 496 530 597 × 2 = 0 + 0.033 477 783 203 124 993 061 194;
  • 2) 0.033 477 783 203 124 993 061 194 × 2 = 0 + 0.066 955 566 406 249 986 122 388;
  • 3) 0.066 955 566 406 249 986 122 388 × 2 = 0 + 0.133 911 132 812 499 972 244 776;
  • 4) 0.133 911 132 812 499 972 244 776 × 2 = 0 + 0.267 822 265 624 999 944 489 552;
  • 5) 0.267 822 265 624 999 944 489 552 × 2 = 0 + 0.535 644 531 249 999 888 979 104;
  • 6) 0.535 644 531 249 999 888 979 104 × 2 = 1 + 0.071 289 062 499 999 777 958 208;
  • 7) 0.071 289 062 499 999 777 958 208 × 2 = 0 + 0.142 578 124 999 999 555 916 416;
  • 8) 0.142 578 124 999 999 555 916 416 × 2 = 0 + 0.285 156 249 999 999 111 832 832;
  • 9) 0.285 156 249 999 999 111 832 832 × 2 = 0 + 0.570 312 499 999 998 223 665 664;
  • 10) 0.570 312 499 999 998 223 665 664 × 2 = 1 + 0.140 624 999 999 996 447 331 328;
  • 11) 0.140 624 999 999 996 447 331 328 × 2 = 0 + 0.281 249 999 999 992 894 662 656;
  • 12) 0.281 249 999 999 992 894 662 656 × 2 = 0 + 0.562 499 999 999 985 789 325 312;
  • 13) 0.562 499 999 999 985 789 325 312 × 2 = 1 + 0.124 999 999 999 971 578 650 624;
  • 14) 0.124 999 999 999 971 578 650 624 × 2 = 0 + 0.249 999 999 999 943 157 301 248;
  • 15) 0.249 999 999 999 943 157 301 248 × 2 = 0 + 0.499 999 999 999 886 314 602 496;
  • 16) 0.499 999 999 999 886 314 602 496 × 2 = 0 + 0.999 999 999 999 772 629 204 992;
  • 17) 0.999 999 999 999 772 629 204 992 × 2 = 1 + 0.999 999 999 999 545 258 409 984;
  • 18) 0.999 999 999 999 545 258 409 984 × 2 = 1 + 0.999 999 999 999 090 516 819 968;
  • 19) 0.999 999 999 999 090 516 819 968 × 2 = 1 + 0.999 999 999 998 181 033 639 936;
  • 20) 0.999 999 999 998 181 033 639 936 × 2 = 1 + 0.999 999 999 996 362 067 279 872;
  • 21) 0.999 999 999 996 362 067 279 872 × 2 = 1 + 0.999 999 999 992 724 134 559 744;
  • 22) 0.999 999 999 992 724 134 559 744 × 2 = 1 + 0.999 999 999 985 448 269 119 488;
  • 23) 0.999 999 999 985 448 269 119 488 × 2 = 1 + 0.999 999 999 970 896 538 238 976;
  • 24) 0.999 999 999 970 896 538 238 976 × 2 = 1 + 0.999 999 999 941 793 076 477 952;
  • 25) 0.999 999 999 941 793 076 477 952 × 2 = 1 + 0.999 999 999 883 586 152 955 904;
  • 26) 0.999 999 999 883 586 152 955 904 × 2 = 1 + 0.999 999 999 767 172 305 911 808;
  • 27) 0.999 999 999 767 172 305 911 808 × 2 = 1 + 0.999 999 999 534 344 611 823 616;
  • 28) 0.999 999 999 534 344 611 823 616 × 2 = 1 + 0.999 999 999 068 689 223 647 232;
  • 29) 0.999 999 999 068 689 223 647 232 × 2 = 1 + 0.999 999 998 137 378 447 294 464;
  • 30) 0.999 999 998 137 378 447 294 464 × 2 = 1 + 0.999 999 996 274 756 894 588 928;
  • 31) 0.999 999 996 274 756 894 588 928 × 2 = 1 + 0.999 999 992 549 513 789 177 856;
  • 32) 0.999 999 992 549 513 789 177 856 × 2 = 1 + 0.999 999 985 099 027 578 355 712;
  • 33) 0.999 999 985 099 027 578 355 712 × 2 = 1 + 0.999 999 970 198 055 156 711 424;
  • 34) 0.999 999 970 198 055 156 711 424 × 2 = 1 + 0.999 999 940 396 110 313 422 848;
  • 35) 0.999 999 940 396 110 313 422 848 × 2 = 1 + 0.999 999 880 792 220 626 845 696;
  • 36) 0.999 999 880 792 220 626 845 696 × 2 = 1 + 0.999 999 761 584 441 253 691 392;
  • 37) 0.999 999 761 584 441 253 691 392 × 2 = 1 + 0.999 999 523 168 882 507 382 784;
  • 38) 0.999 999 523 168 882 507 382 784 × 2 = 1 + 0.999 999 046 337 765 014 765 568;
  • 39) 0.999 999 046 337 765 014 765 568 × 2 = 1 + 0.999 998 092 675 530 029 531 136;
  • 40) 0.999 998 092 675 530 029 531 136 × 2 = 1 + 0.999 996 185 351 060 059 062 272;
  • 41) 0.999 996 185 351 060 059 062 272 × 2 = 1 + 0.999 992 370 702 120 118 124 544;
  • 42) 0.999 992 370 702 120 118 124 544 × 2 = 1 + 0.999 984 741 404 240 236 249 088;
  • 43) 0.999 984 741 404 240 236 249 088 × 2 = 1 + 0.999 969 482 808 480 472 498 176;
  • 44) 0.999 969 482 808 480 472 498 176 × 2 = 1 + 0.999 938 965 616 960 944 996 352;
  • 45) 0.999 938 965 616 960 944 996 352 × 2 = 1 + 0.999 877 931 233 921 889 992 704;
  • 46) 0.999 877 931 233 921 889 992 704 × 2 = 1 + 0.999 755 862 467 843 779 985 408;
  • 47) 0.999 755 862 467 843 779 985 408 × 2 = 1 + 0.999 511 724 935 687 559 970 816;
  • 48) 0.999 511 724 935 687 559 970 816 × 2 = 1 + 0.999 023 449 871 375 119 941 632;
  • 49) 0.999 023 449 871 375 119 941 632 × 2 = 1 + 0.998 046 899 742 750 239 883 264;
  • 50) 0.998 046 899 742 750 239 883 264 × 2 = 1 + 0.996 093 799 485 500 479 766 528;
  • 51) 0.996 093 799 485 500 479 766 528 × 2 = 1 + 0.992 187 598 971 000 959 533 056;
  • 52) 0.992 187 598 971 000 959 533 056 × 2 = 1 + 0.984 375 197 942 001 919 066 112;
  • 53) 0.984 375 197 942 001 919 066 112 × 2 = 1 + 0.968 750 395 884 003 838 132 224;
  • 54) 0.968 750 395 884 003 838 132 224 × 2 = 1 + 0.937 500 791 768 007 676 264 448;
  • 55) 0.937 500 791 768 007 676 264 448 × 2 = 1 + 0.875 001 583 536 015 352 528 896;
  • 56) 0.875 001 583 536 015 352 528 896 × 2 = 1 + 0.750 003 167 072 030 705 057 792;
  • 57) 0.750 003 167 072 030 705 057 792 × 2 = 1 + 0.500 006 334 144 061 410 115 584;
  • 58) 0.500 006 334 144 061 410 115 584 × 2 = 1 + 0.000 012 668 288 122 820 231 168;

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:


0.016 738 891 601 562 496 530 597(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 597(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:


0.016 738 891 601 562 496 530 597(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 -0.016 738 891 601 562 496 530 597 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