-0.016 738 891 601 562 496 530 442 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 442(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
-0.016 738 891 601 562 496 530 442(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:

|-0.016 738 891 601 562 496 530 442| = 0.016 738 891 601 562 496 530 442


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: 0.016 738 891 601 562 496 530 442.

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) 0.016 738 891 601 562 496 530 442 × 2 = 0 + 0.033 477 783 203 124 993 060 884;
  • 2) 0.033 477 783 203 124 993 060 884 × 2 = 0 + 0.066 955 566 406 249 986 121 768;
  • 3) 0.066 955 566 406 249 986 121 768 × 2 = 0 + 0.133 911 132 812 499 972 243 536;
  • 4) 0.133 911 132 812 499 972 243 536 × 2 = 0 + 0.267 822 265 624 999 944 487 072;
  • 5) 0.267 822 265 624 999 944 487 072 × 2 = 0 + 0.535 644 531 249 999 888 974 144;
  • 6) 0.535 644 531 249 999 888 974 144 × 2 = 1 + 0.071 289 062 499 999 777 948 288;
  • 7) 0.071 289 062 499 999 777 948 288 × 2 = 0 + 0.142 578 124 999 999 555 896 576;
  • 8) 0.142 578 124 999 999 555 896 576 × 2 = 0 + 0.285 156 249 999 999 111 793 152;
  • 9) 0.285 156 249 999 999 111 793 152 × 2 = 0 + 0.570 312 499 999 998 223 586 304;
  • 10) 0.570 312 499 999 998 223 586 304 × 2 = 1 + 0.140 624 999 999 996 447 172 608;
  • 11) 0.140 624 999 999 996 447 172 608 × 2 = 0 + 0.281 249 999 999 992 894 345 216;
  • 12) 0.281 249 999 999 992 894 345 216 × 2 = 0 + 0.562 499 999 999 985 788 690 432;
  • 13) 0.562 499 999 999 985 788 690 432 × 2 = 1 + 0.124 999 999 999 971 577 380 864;
  • 14) 0.124 999 999 999 971 577 380 864 × 2 = 0 + 0.249 999 999 999 943 154 761 728;
  • 15) 0.249 999 999 999 943 154 761 728 × 2 = 0 + 0.499 999 999 999 886 309 523 456;
  • 16) 0.499 999 999 999 886 309 523 456 × 2 = 0 + 0.999 999 999 999 772 619 046 912;
  • 17) 0.999 999 999 999 772 619 046 912 × 2 = 1 + 0.999 999 999 999 545 238 093 824;
  • 18) 0.999 999 999 999 545 238 093 824 × 2 = 1 + 0.999 999 999 999 090 476 187 648;
  • 19) 0.999 999 999 999 090 476 187 648 × 2 = 1 + 0.999 999 999 998 180 952 375 296;
  • 20) 0.999 999 999 998 180 952 375 296 × 2 = 1 + 0.999 999 999 996 361 904 750 592;
  • 21) 0.999 999 999 996 361 904 750 592 × 2 = 1 + 0.999 999 999 992 723 809 501 184;
  • 22) 0.999 999 999 992 723 809 501 184 × 2 = 1 + 0.999 999 999 985 447 619 002 368;
  • 23) 0.999 999 999 985 447 619 002 368 × 2 = 1 + 0.999 999 999 970 895 238 004 736;
  • 24) 0.999 999 999 970 895 238 004 736 × 2 = 1 + 0.999 999 999 941 790 476 009 472;
  • 25) 0.999 999 999 941 790 476 009 472 × 2 = 1 + 0.999 999 999 883 580 952 018 944;
  • 26) 0.999 999 999 883 580 952 018 944 × 2 = 1 + 0.999 999 999 767 161 904 037 888;
  • 27) 0.999 999 999 767 161 904 037 888 × 2 = 1 + 0.999 999 999 534 323 808 075 776;
  • 28) 0.999 999 999 534 323 808 075 776 × 2 = 1 + 0.999 999 999 068 647 616 151 552;
  • 29) 0.999 999 999 068 647 616 151 552 × 2 = 1 + 0.999 999 998 137 295 232 303 104;
  • 30) 0.999 999 998 137 295 232 303 104 × 2 = 1 + 0.999 999 996 274 590 464 606 208;
  • 31) 0.999 999 996 274 590 464 606 208 × 2 = 1 + 0.999 999 992 549 180 929 212 416;
  • 32) 0.999 999 992 549 180 929 212 416 × 2 = 1 + 0.999 999 985 098 361 858 424 832;
  • 33) 0.999 999 985 098 361 858 424 832 × 2 = 1 + 0.999 999 970 196 723 716 849 664;
  • 34) 0.999 999 970 196 723 716 849 664 × 2 = 1 + 0.999 999 940 393 447 433 699 328;
  • 35) 0.999 999 940 393 447 433 699 328 × 2 = 1 + 0.999 999 880 786 894 867 398 656;
  • 36) 0.999 999 880 786 894 867 398 656 × 2 = 1 + 0.999 999 761 573 789 734 797 312;
  • 37) 0.999 999 761 573 789 734 797 312 × 2 = 1 + 0.999 999 523 147 579 469 594 624;
  • 38) 0.999 999 523 147 579 469 594 624 × 2 = 1 + 0.999 999 046 295 158 939 189 248;
  • 39) 0.999 999 046 295 158 939 189 248 × 2 = 1 + 0.999 998 092 590 317 878 378 496;
  • 40) 0.999 998 092 590 317 878 378 496 × 2 = 1 + 0.999 996 185 180 635 756 756 992;
  • 41) 0.999 996 185 180 635 756 756 992 × 2 = 1 + 0.999 992 370 361 271 513 513 984;
  • 42) 0.999 992 370 361 271 513 513 984 × 2 = 1 + 0.999 984 740 722 543 027 027 968;
  • 43) 0.999 984 740 722 543 027 027 968 × 2 = 1 + 0.999 969 481 445 086 054 055 936;
  • 44) 0.999 969 481 445 086 054 055 936 × 2 = 1 + 0.999 938 962 890 172 108 111 872;
  • 45) 0.999 938 962 890 172 108 111 872 × 2 = 1 + 0.999 877 925 780 344 216 223 744;
  • 46) 0.999 877 925 780 344 216 223 744 × 2 = 1 + 0.999 755 851 560 688 432 447 488;
  • 47) 0.999 755 851 560 688 432 447 488 × 2 = 1 + 0.999 511 703 121 376 864 894 976;
  • 48) 0.999 511 703 121 376 864 894 976 × 2 = 1 + 0.999 023 406 242 753 729 789 952;
  • 49) 0.999 023 406 242 753 729 789 952 × 2 = 1 + 0.998 046 812 485 507 459 579 904;
  • 50) 0.998 046 812 485 507 459 579 904 × 2 = 1 + 0.996 093 624 971 014 919 159 808;
  • 51) 0.996 093 624 971 014 919 159 808 × 2 = 1 + 0.992 187 249 942 029 838 319 616;
  • 52) 0.992 187 249 942 029 838 319 616 × 2 = 1 + 0.984 374 499 884 059 676 639 232;
  • 53) 0.984 374 499 884 059 676 639 232 × 2 = 1 + 0.968 748 999 768 119 353 278 464;
  • 54) 0.968 748 999 768 119 353 278 464 × 2 = 1 + 0.937 497 999 536 238 706 556 928;
  • 55) 0.937 497 999 536 238 706 556 928 × 2 = 1 + 0.874 995 999 072 477 413 113 856;
  • 56) 0.874 995 999 072 477 413 113 856 × 2 = 1 + 0.749 991 998 144 954 826 227 712;
  • 57) 0.749 991 998 144 954 826 227 712 × 2 = 1 + 0.499 983 996 289 909 652 455 424;
  • 58) 0.499 983 996 289 909 652 455 424 × 2 = 0 + 0.999 967 992 579 819 304 910 848;

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:


0.016 738 891 601 562 496 530 442(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 442(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:


0.016 738 891 601 562 496 530 442(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 -0.016 738 891 601 562 496 530 442 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