-0.016 738 891 601 562 496 530 472 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 472(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
-0.016 738 891 601 562 496 530 472(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:

|-0.016 738 891 601 562 496 530 472| = 0.016 738 891 601 562 496 530 472


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: 0.016 738 891 601 562 496 530 472.

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) 0.016 738 891 601 562 496 530 472 × 2 = 0 + 0.033 477 783 203 124 993 060 944;
  • 2) 0.033 477 783 203 124 993 060 944 × 2 = 0 + 0.066 955 566 406 249 986 121 888;
  • 3) 0.066 955 566 406 249 986 121 888 × 2 = 0 + 0.133 911 132 812 499 972 243 776;
  • 4) 0.133 911 132 812 499 972 243 776 × 2 = 0 + 0.267 822 265 624 999 944 487 552;
  • 5) 0.267 822 265 624 999 944 487 552 × 2 = 0 + 0.535 644 531 249 999 888 975 104;
  • 6) 0.535 644 531 249 999 888 975 104 × 2 = 1 + 0.071 289 062 499 999 777 950 208;
  • 7) 0.071 289 062 499 999 777 950 208 × 2 = 0 + 0.142 578 124 999 999 555 900 416;
  • 8) 0.142 578 124 999 999 555 900 416 × 2 = 0 + 0.285 156 249 999 999 111 800 832;
  • 9) 0.285 156 249 999 999 111 800 832 × 2 = 0 + 0.570 312 499 999 998 223 601 664;
  • 10) 0.570 312 499 999 998 223 601 664 × 2 = 1 + 0.140 624 999 999 996 447 203 328;
  • 11) 0.140 624 999 999 996 447 203 328 × 2 = 0 + 0.281 249 999 999 992 894 406 656;
  • 12) 0.281 249 999 999 992 894 406 656 × 2 = 0 + 0.562 499 999 999 985 788 813 312;
  • 13) 0.562 499 999 999 985 788 813 312 × 2 = 1 + 0.124 999 999 999 971 577 626 624;
  • 14) 0.124 999 999 999 971 577 626 624 × 2 = 0 + 0.249 999 999 999 943 155 253 248;
  • 15) 0.249 999 999 999 943 155 253 248 × 2 = 0 + 0.499 999 999 999 886 310 506 496;
  • 16) 0.499 999 999 999 886 310 506 496 × 2 = 0 + 0.999 999 999 999 772 621 012 992;
  • 17) 0.999 999 999 999 772 621 012 992 × 2 = 1 + 0.999 999 999 999 545 242 025 984;
  • 18) 0.999 999 999 999 545 242 025 984 × 2 = 1 + 0.999 999 999 999 090 484 051 968;
  • 19) 0.999 999 999 999 090 484 051 968 × 2 = 1 + 0.999 999 999 998 180 968 103 936;
  • 20) 0.999 999 999 998 180 968 103 936 × 2 = 1 + 0.999 999 999 996 361 936 207 872;
  • 21) 0.999 999 999 996 361 936 207 872 × 2 = 1 + 0.999 999 999 992 723 872 415 744;
  • 22) 0.999 999 999 992 723 872 415 744 × 2 = 1 + 0.999 999 999 985 447 744 831 488;
  • 23) 0.999 999 999 985 447 744 831 488 × 2 = 1 + 0.999 999 999 970 895 489 662 976;
  • 24) 0.999 999 999 970 895 489 662 976 × 2 = 1 + 0.999 999 999 941 790 979 325 952;
  • 25) 0.999 999 999 941 790 979 325 952 × 2 = 1 + 0.999 999 999 883 581 958 651 904;
  • 26) 0.999 999 999 883 581 958 651 904 × 2 = 1 + 0.999 999 999 767 163 917 303 808;
  • 27) 0.999 999 999 767 163 917 303 808 × 2 = 1 + 0.999 999 999 534 327 834 607 616;
  • 28) 0.999 999 999 534 327 834 607 616 × 2 = 1 + 0.999 999 999 068 655 669 215 232;
  • 29) 0.999 999 999 068 655 669 215 232 × 2 = 1 + 0.999 999 998 137 311 338 430 464;
  • 30) 0.999 999 998 137 311 338 430 464 × 2 = 1 + 0.999 999 996 274 622 676 860 928;
  • 31) 0.999 999 996 274 622 676 860 928 × 2 = 1 + 0.999 999 992 549 245 353 721 856;
  • 32) 0.999 999 992 549 245 353 721 856 × 2 = 1 + 0.999 999 985 098 490 707 443 712;
  • 33) 0.999 999 985 098 490 707 443 712 × 2 = 1 + 0.999 999 970 196 981 414 887 424;
  • 34) 0.999 999 970 196 981 414 887 424 × 2 = 1 + 0.999 999 940 393 962 829 774 848;
  • 35) 0.999 999 940 393 962 829 774 848 × 2 = 1 + 0.999 999 880 787 925 659 549 696;
  • 36) 0.999 999 880 787 925 659 549 696 × 2 = 1 + 0.999 999 761 575 851 319 099 392;
  • 37) 0.999 999 761 575 851 319 099 392 × 2 = 1 + 0.999 999 523 151 702 638 198 784;
  • 38) 0.999 999 523 151 702 638 198 784 × 2 = 1 + 0.999 999 046 303 405 276 397 568;
  • 39) 0.999 999 046 303 405 276 397 568 × 2 = 1 + 0.999 998 092 606 810 552 795 136;
  • 40) 0.999 998 092 606 810 552 795 136 × 2 = 1 + 0.999 996 185 213 621 105 590 272;
  • 41) 0.999 996 185 213 621 105 590 272 × 2 = 1 + 0.999 992 370 427 242 211 180 544;
  • 42) 0.999 992 370 427 242 211 180 544 × 2 = 1 + 0.999 984 740 854 484 422 361 088;
  • 43) 0.999 984 740 854 484 422 361 088 × 2 = 1 + 0.999 969 481 708 968 844 722 176;
  • 44) 0.999 969 481 708 968 844 722 176 × 2 = 1 + 0.999 938 963 417 937 689 444 352;
  • 45) 0.999 938 963 417 937 689 444 352 × 2 = 1 + 0.999 877 926 835 875 378 888 704;
  • 46) 0.999 877 926 835 875 378 888 704 × 2 = 1 + 0.999 755 853 671 750 757 777 408;
  • 47) 0.999 755 853 671 750 757 777 408 × 2 = 1 + 0.999 511 707 343 501 515 554 816;
  • 48) 0.999 511 707 343 501 515 554 816 × 2 = 1 + 0.999 023 414 687 003 031 109 632;
  • 49) 0.999 023 414 687 003 031 109 632 × 2 = 1 + 0.998 046 829 374 006 062 219 264;
  • 50) 0.998 046 829 374 006 062 219 264 × 2 = 1 + 0.996 093 658 748 012 124 438 528;
  • 51) 0.996 093 658 748 012 124 438 528 × 2 = 1 + 0.992 187 317 496 024 248 877 056;
  • 52) 0.992 187 317 496 024 248 877 056 × 2 = 1 + 0.984 374 634 992 048 497 754 112;
  • 53) 0.984 374 634 992 048 497 754 112 × 2 = 1 + 0.968 749 269 984 096 995 508 224;
  • 54) 0.968 749 269 984 096 995 508 224 × 2 = 1 + 0.937 498 539 968 193 991 016 448;
  • 55) 0.937 498 539 968 193 991 016 448 × 2 = 1 + 0.874 997 079 936 387 982 032 896;
  • 56) 0.874 997 079 936 387 982 032 896 × 2 = 1 + 0.749 994 159 872 775 964 065 792;
  • 57) 0.749 994 159 872 775 964 065 792 × 2 = 1 + 0.499 988 319 745 551 928 131 584;
  • 58) 0.499 988 319 745 551 928 131 584 × 2 = 0 + 0.999 976 639 491 103 856 263 168;

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:


0.016 738 891 601 562 496 530 472(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 472(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:


0.016 738 891 601 562 496 530 472(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 -0.016 738 891 601 562 496 530 472 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