-0.016 738 891 601 562 496 530 466 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 466(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
-0.016 738 891 601 562 496 530 466(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:

|-0.016 738 891 601 562 496 530 466| = 0.016 738 891 601 562 496 530 466


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: 0.016 738 891 601 562 496 530 466.

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) 0.016 738 891 601 562 496 530 466 × 2 = 0 + 0.033 477 783 203 124 993 060 932;
  • 2) 0.033 477 783 203 124 993 060 932 × 2 = 0 + 0.066 955 566 406 249 986 121 864;
  • 3) 0.066 955 566 406 249 986 121 864 × 2 = 0 + 0.133 911 132 812 499 972 243 728;
  • 4) 0.133 911 132 812 499 972 243 728 × 2 = 0 + 0.267 822 265 624 999 944 487 456;
  • 5) 0.267 822 265 624 999 944 487 456 × 2 = 0 + 0.535 644 531 249 999 888 974 912;
  • 6) 0.535 644 531 249 999 888 974 912 × 2 = 1 + 0.071 289 062 499 999 777 949 824;
  • 7) 0.071 289 062 499 999 777 949 824 × 2 = 0 + 0.142 578 124 999 999 555 899 648;
  • 8) 0.142 578 124 999 999 555 899 648 × 2 = 0 + 0.285 156 249 999 999 111 799 296;
  • 9) 0.285 156 249 999 999 111 799 296 × 2 = 0 + 0.570 312 499 999 998 223 598 592;
  • 10) 0.570 312 499 999 998 223 598 592 × 2 = 1 + 0.140 624 999 999 996 447 197 184;
  • 11) 0.140 624 999 999 996 447 197 184 × 2 = 0 + 0.281 249 999 999 992 894 394 368;
  • 12) 0.281 249 999 999 992 894 394 368 × 2 = 0 + 0.562 499 999 999 985 788 788 736;
  • 13) 0.562 499 999 999 985 788 788 736 × 2 = 1 + 0.124 999 999 999 971 577 577 472;
  • 14) 0.124 999 999 999 971 577 577 472 × 2 = 0 + 0.249 999 999 999 943 155 154 944;
  • 15) 0.249 999 999 999 943 155 154 944 × 2 = 0 + 0.499 999 999 999 886 310 309 888;
  • 16) 0.499 999 999 999 886 310 309 888 × 2 = 0 + 0.999 999 999 999 772 620 619 776;
  • 17) 0.999 999 999 999 772 620 619 776 × 2 = 1 + 0.999 999 999 999 545 241 239 552;
  • 18) 0.999 999 999 999 545 241 239 552 × 2 = 1 + 0.999 999 999 999 090 482 479 104;
  • 19) 0.999 999 999 999 090 482 479 104 × 2 = 1 + 0.999 999 999 998 180 964 958 208;
  • 20) 0.999 999 999 998 180 964 958 208 × 2 = 1 + 0.999 999 999 996 361 929 916 416;
  • 21) 0.999 999 999 996 361 929 916 416 × 2 = 1 + 0.999 999 999 992 723 859 832 832;
  • 22) 0.999 999 999 992 723 859 832 832 × 2 = 1 + 0.999 999 999 985 447 719 665 664;
  • 23) 0.999 999 999 985 447 719 665 664 × 2 = 1 + 0.999 999 999 970 895 439 331 328;
  • 24) 0.999 999 999 970 895 439 331 328 × 2 = 1 + 0.999 999 999 941 790 878 662 656;
  • 25) 0.999 999 999 941 790 878 662 656 × 2 = 1 + 0.999 999 999 883 581 757 325 312;
  • 26) 0.999 999 999 883 581 757 325 312 × 2 = 1 + 0.999 999 999 767 163 514 650 624;
  • 27) 0.999 999 999 767 163 514 650 624 × 2 = 1 + 0.999 999 999 534 327 029 301 248;
  • 28) 0.999 999 999 534 327 029 301 248 × 2 = 1 + 0.999 999 999 068 654 058 602 496;
  • 29) 0.999 999 999 068 654 058 602 496 × 2 = 1 + 0.999 999 998 137 308 117 204 992;
  • 30) 0.999 999 998 137 308 117 204 992 × 2 = 1 + 0.999 999 996 274 616 234 409 984;
  • 31) 0.999 999 996 274 616 234 409 984 × 2 = 1 + 0.999 999 992 549 232 468 819 968;
  • 32) 0.999 999 992 549 232 468 819 968 × 2 = 1 + 0.999 999 985 098 464 937 639 936;
  • 33) 0.999 999 985 098 464 937 639 936 × 2 = 1 + 0.999 999 970 196 929 875 279 872;
  • 34) 0.999 999 970 196 929 875 279 872 × 2 = 1 + 0.999 999 940 393 859 750 559 744;
  • 35) 0.999 999 940 393 859 750 559 744 × 2 = 1 + 0.999 999 880 787 719 501 119 488;
  • 36) 0.999 999 880 787 719 501 119 488 × 2 = 1 + 0.999 999 761 575 439 002 238 976;
  • 37) 0.999 999 761 575 439 002 238 976 × 2 = 1 + 0.999 999 523 150 878 004 477 952;
  • 38) 0.999 999 523 150 878 004 477 952 × 2 = 1 + 0.999 999 046 301 756 008 955 904;
  • 39) 0.999 999 046 301 756 008 955 904 × 2 = 1 + 0.999 998 092 603 512 017 911 808;
  • 40) 0.999 998 092 603 512 017 911 808 × 2 = 1 + 0.999 996 185 207 024 035 823 616;
  • 41) 0.999 996 185 207 024 035 823 616 × 2 = 1 + 0.999 992 370 414 048 071 647 232;
  • 42) 0.999 992 370 414 048 071 647 232 × 2 = 1 + 0.999 984 740 828 096 143 294 464;
  • 43) 0.999 984 740 828 096 143 294 464 × 2 = 1 + 0.999 969 481 656 192 286 588 928;
  • 44) 0.999 969 481 656 192 286 588 928 × 2 = 1 + 0.999 938 963 312 384 573 177 856;
  • 45) 0.999 938 963 312 384 573 177 856 × 2 = 1 + 0.999 877 926 624 769 146 355 712;
  • 46) 0.999 877 926 624 769 146 355 712 × 2 = 1 + 0.999 755 853 249 538 292 711 424;
  • 47) 0.999 755 853 249 538 292 711 424 × 2 = 1 + 0.999 511 706 499 076 585 422 848;
  • 48) 0.999 511 706 499 076 585 422 848 × 2 = 1 + 0.999 023 412 998 153 170 845 696;
  • 49) 0.999 023 412 998 153 170 845 696 × 2 = 1 + 0.998 046 825 996 306 341 691 392;
  • 50) 0.998 046 825 996 306 341 691 392 × 2 = 1 + 0.996 093 651 992 612 683 382 784;
  • 51) 0.996 093 651 992 612 683 382 784 × 2 = 1 + 0.992 187 303 985 225 366 765 568;
  • 52) 0.992 187 303 985 225 366 765 568 × 2 = 1 + 0.984 374 607 970 450 733 531 136;
  • 53) 0.984 374 607 970 450 733 531 136 × 2 = 1 + 0.968 749 215 940 901 467 062 272;
  • 54) 0.968 749 215 940 901 467 062 272 × 2 = 1 + 0.937 498 431 881 802 934 124 544;
  • 55) 0.937 498 431 881 802 934 124 544 × 2 = 1 + 0.874 996 863 763 605 868 249 088;
  • 56) 0.874 996 863 763 605 868 249 088 × 2 = 1 + 0.749 993 727 527 211 736 498 176;
  • 57) 0.749 993 727 527 211 736 498 176 × 2 = 1 + 0.499 987 455 054 423 472 996 352;
  • 58) 0.499 987 455 054 423 472 996 352 × 2 = 0 + 0.999 974 910 108 846 945 992 704;

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:


0.016 738 891 601 562 496 530 466(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 466(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:


0.016 738 891 601 562 496 530 466(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 -0.016 738 891 601 562 496 530 466 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