-0.016 738 891 601 562 496 628 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 628(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
-0.016 738 891 601 562 496 628(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:

|-0.016 738 891 601 562 496 628| = 0.016 738 891 601 562 496 628


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: 0.016 738 891 601 562 496 628.

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) 0.016 738 891 601 562 496 628 × 2 = 0 + 0.033 477 783 203 124 993 256;
  • 2) 0.033 477 783 203 124 993 256 × 2 = 0 + 0.066 955 566 406 249 986 512;
  • 3) 0.066 955 566 406 249 986 512 × 2 = 0 + 0.133 911 132 812 499 973 024;
  • 4) 0.133 911 132 812 499 973 024 × 2 = 0 + 0.267 822 265 624 999 946 048;
  • 5) 0.267 822 265 624 999 946 048 × 2 = 0 + 0.535 644 531 249 999 892 096;
  • 6) 0.535 644 531 249 999 892 096 × 2 = 1 + 0.071 289 062 499 999 784 192;
  • 7) 0.071 289 062 499 999 784 192 × 2 = 0 + 0.142 578 124 999 999 568 384;
  • 8) 0.142 578 124 999 999 568 384 × 2 = 0 + 0.285 156 249 999 999 136 768;
  • 9) 0.285 156 249 999 999 136 768 × 2 = 0 + 0.570 312 499 999 998 273 536;
  • 10) 0.570 312 499 999 998 273 536 × 2 = 1 + 0.140 624 999 999 996 547 072;
  • 11) 0.140 624 999 999 996 547 072 × 2 = 0 + 0.281 249 999 999 993 094 144;
  • 12) 0.281 249 999 999 993 094 144 × 2 = 0 + 0.562 499 999 999 986 188 288;
  • 13) 0.562 499 999 999 986 188 288 × 2 = 1 + 0.124 999 999 999 972 376 576;
  • 14) 0.124 999 999 999 972 376 576 × 2 = 0 + 0.249 999 999 999 944 753 152;
  • 15) 0.249 999 999 999 944 753 152 × 2 = 0 + 0.499 999 999 999 889 506 304;
  • 16) 0.499 999 999 999 889 506 304 × 2 = 0 + 0.999 999 999 999 779 012 608;
  • 17) 0.999 999 999 999 779 012 608 × 2 = 1 + 0.999 999 999 999 558 025 216;
  • 18) 0.999 999 999 999 558 025 216 × 2 = 1 + 0.999 999 999 999 116 050 432;
  • 19) 0.999 999 999 999 116 050 432 × 2 = 1 + 0.999 999 999 998 232 100 864;
  • 20) 0.999 999 999 998 232 100 864 × 2 = 1 + 0.999 999 999 996 464 201 728;
  • 21) 0.999 999 999 996 464 201 728 × 2 = 1 + 0.999 999 999 992 928 403 456;
  • 22) 0.999 999 999 992 928 403 456 × 2 = 1 + 0.999 999 999 985 856 806 912;
  • 23) 0.999 999 999 985 856 806 912 × 2 = 1 + 0.999 999 999 971 713 613 824;
  • 24) 0.999 999 999 971 713 613 824 × 2 = 1 + 0.999 999 999 943 427 227 648;
  • 25) 0.999 999 999 943 427 227 648 × 2 = 1 + 0.999 999 999 886 854 455 296;
  • 26) 0.999 999 999 886 854 455 296 × 2 = 1 + 0.999 999 999 773 708 910 592;
  • 27) 0.999 999 999 773 708 910 592 × 2 = 1 + 0.999 999 999 547 417 821 184;
  • 28) 0.999 999 999 547 417 821 184 × 2 = 1 + 0.999 999 999 094 835 642 368;
  • 29) 0.999 999 999 094 835 642 368 × 2 = 1 + 0.999 999 998 189 671 284 736;
  • 30) 0.999 999 998 189 671 284 736 × 2 = 1 + 0.999 999 996 379 342 569 472;
  • 31) 0.999 999 996 379 342 569 472 × 2 = 1 + 0.999 999 992 758 685 138 944;
  • 32) 0.999 999 992 758 685 138 944 × 2 = 1 + 0.999 999 985 517 370 277 888;
  • 33) 0.999 999 985 517 370 277 888 × 2 = 1 + 0.999 999 971 034 740 555 776;
  • 34) 0.999 999 971 034 740 555 776 × 2 = 1 + 0.999 999 942 069 481 111 552;
  • 35) 0.999 999 942 069 481 111 552 × 2 = 1 + 0.999 999 884 138 962 223 104;
  • 36) 0.999 999 884 138 962 223 104 × 2 = 1 + 0.999 999 768 277 924 446 208;
  • 37) 0.999 999 768 277 924 446 208 × 2 = 1 + 0.999 999 536 555 848 892 416;
  • 38) 0.999 999 536 555 848 892 416 × 2 = 1 + 0.999 999 073 111 697 784 832;
  • 39) 0.999 999 073 111 697 784 832 × 2 = 1 + 0.999 998 146 223 395 569 664;
  • 40) 0.999 998 146 223 395 569 664 × 2 = 1 + 0.999 996 292 446 791 139 328;
  • 41) 0.999 996 292 446 791 139 328 × 2 = 1 + 0.999 992 584 893 582 278 656;
  • 42) 0.999 992 584 893 582 278 656 × 2 = 1 + 0.999 985 169 787 164 557 312;
  • 43) 0.999 985 169 787 164 557 312 × 2 = 1 + 0.999 970 339 574 329 114 624;
  • 44) 0.999 970 339 574 329 114 624 × 2 = 1 + 0.999 940 679 148 658 229 248;
  • 45) 0.999 940 679 148 658 229 248 × 2 = 1 + 0.999 881 358 297 316 458 496;
  • 46) 0.999 881 358 297 316 458 496 × 2 = 1 + 0.999 762 716 594 632 916 992;
  • 47) 0.999 762 716 594 632 916 992 × 2 = 1 + 0.999 525 433 189 265 833 984;
  • 48) 0.999 525 433 189 265 833 984 × 2 = 1 + 0.999 050 866 378 531 667 968;
  • 49) 0.999 050 866 378 531 667 968 × 2 = 1 + 0.998 101 732 757 063 335 936;
  • 50) 0.998 101 732 757 063 335 936 × 2 = 1 + 0.996 203 465 514 126 671 872;
  • 51) 0.996 203 465 514 126 671 872 × 2 = 1 + 0.992 406 931 028 253 343 744;
  • 52) 0.992 406 931 028 253 343 744 × 2 = 1 + 0.984 813 862 056 506 687 488;
  • 53) 0.984 813 862 056 506 687 488 × 2 = 1 + 0.969 627 724 113 013 374 976;
  • 54) 0.969 627 724 113 013 374 976 × 2 = 1 + 0.939 255 448 226 026 749 952;
  • 55) 0.939 255 448 226 026 749 952 × 2 = 1 + 0.878 510 896 452 053 499 904;
  • 56) 0.878 510 896 452 053 499 904 × 2 = 1 + 0.757 021 792 904 106 999 808;
  • 57) 0.757 021 792 904 106 999 808 × 2 = 1 + 0.514 043 585 808 213 999 616;
  • 58) 0.514 043 585 808 213 999 616 × 2 = 1 + 0.028 087 171 616 427 999 232;

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:


0.016 738 891 601 562 496 628(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 628(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:


0.016 738 891 601 562 496 628(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 -0.016 738 891 601 562 496 628 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