-0.016 738 891 601 562 496 531 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 531 2(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
-0.016 738 891 601 562 496 531 2(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:

|-0.016 738 891 601 562 496 531 2| = 0.016 738 891 601 562 496 531 2


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: 0.016 738 891 601 562 496 531 2.

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) 0.016 738 891 601 562 496 531 2 × 2 = 0 + 0.033 477 783 203 124 993 062 4;
  • 2) 0.033 477 783 203 124 993 062 4 × 2 = 0 + 0.066 955 566 406 249 986 124 8;
  • 3) 0.066 955 566 406 249 986 124 8 × 2 = 0 + 0.133 911 132 812 499 972 249 6;
  • 4) 0.133 911 132 812 499 972 249 6 × 2 = 0 + 0.267 822 265 624 999 944 499 2;
  • 5) 0.267 822 265 624 999 944 499 2 × 2 = 0 + 0.535 644 531 249 999 888 998 4;
  • 6) 0.535 644 531 249 999 888 998 4 × 2 = 1 + 0.071 289 062 499 999 777 996 8;
  • 7) 0.071 289 062 499 999 777 996 8 × 2 = 0 + 0.142 578 124 999 999 555 993 6;
  • 8) 0.142 578 124 999 999 555 993 6 × 2 = 0 + 0.285 156 249 999 999 111 987 2;
  • 9) 0.285 156 249 999 999 111 987 2 × 2 = 0 + 0.570 312 499 999 998 223 974 4;
  • 10) 0.570 312 499 999 998 223 974 4 × 2 = 1 + 0.140 624 999 999 996 447 948 8;
  • 11) 0.140 624 999 999 996 447 948 8 × 2 = 0 + 0.281 249 999 999 992 895 897 6;
  • 12) 0.281 249 999 999 992 895 897 6 × 2 = 0 + 0.562 499 999 999 985 791 795 2;
  • 13) 0.562 499 999 999 985 791 795 2 × 2 = 1 + 0.124 999 999 999 971 583 590 4;
  • 14) 0.124 999 999 999 971 583 590 4 × 2 = 0 + 0.249 999 999 999 943 167 180 8;
  • 15) 0.249 999 999 999 943 167 180 8 × 2 = 0 + 0.499 999 999 999 886 334 361 6;
  • 16) 0.499 999 999 999 886 334 361 6 × 2 = 0 + 0.999 999 999 999 772 668 723 2;
  • 17) 0.999 999 999 999 772 668 723 2 × 2 = 1 + 0.999 999 999 999 545 337 446 4;
  • 18) 0.999 999 999 999 545 337 446 4 × 2 = 1 + 0.999 999 999 999 090 674 892 8;
  • 19) 0.999 999 999 999 090 674 892 8 × 2 = 1 + 0.999 999 999 998 181 349 785 6;
  • 20) 0.999 999 999 998 181 349 785 6 × 2 = 1 + 0.999 999 999 996 362 699 571 2;
  • 21) 0.999 999 999 996 362 699 571 2 × 2 = 1 + 0.999 999 999 992 725 399 142 4;
  • 22) 0.999 999 999 992 725 399 142 4 × 2 = 1 + 0.999 999 999 985 450 798 284 8;
  • 23) 0.999 999 999 985 450 798 284 8 × 2 = 1 + 0.999 999 999 970 901 596 569 6;
  • 24) 0.999 999 999 970 901 596 569 6 × 2 = 1 + 0.999 999 999 941 803 193 139 2;
  • 25) 0.999 999 999 941 803 193 139 2 × 2 = 1 + 0.999 999 999 883 606 386 278 4;
  • 26) 0.999 999 999 883 606 386 278 4 × 2 = 1 + 0.999 999 999 767 212 772 556 8;
  • 27) 0.999 999 999 767 212 772 556 8 × 2 = 1 + 0.999 999 999 534 425 545 113 6;
  • 28) 0.999 999 999 534 425 545 113 6 × 2 = 1 + 0.999 999 999 068 851 090 227 2;
  • 29) 0.999 999 999 068 851 090 227 2 × 2 = 1 + 0.999 999 998 137 702 180 454 4;
  • 30) 0.999 999 998 137 702 180 454 4 × 2 = 1 + 0.999 999 996 275 404 360 908 8;
  • 31) 0.999 999 996 275 404 360 908 8 × 2 = 1 + 0.999 999 992 550 808 721 817 6;
  • 32) 0.999 999 992 550 808 721 817 6 × 2 = 1 + 0.999 999 985 101 617 443 635 2;
  • 33) 0.999 999 985 101 617 443 635 2 × 2 = 1 + 0.999 999 970 203 234 887 270 4;
  • 34) 0.999 999 970 203 234 887 270 4 × 2 = 1 + 0.999 999 940 406 469 774 540 8;
  • 35) 0.999 999 940 406 469 774 540 8 × 2 = 1 + 0.999 999 880 812 939 549 081 6;
  • 36) 0.999 999 880 812 939 549 081 6 × 2 = 1 + 0.999 999 761 625 879 098 163 2;
  • 37) 0.999 999 761 625 879 098 163 2 × 2 = 1 + 0.999 999 523 251 758 196 326 4;
  • 38) 0.999 999 523 251 758 196 326 4 × 2 = 1 + 0.999 999 046 503 516 392 652 8;
  • 39) 0.999 999 046 503 516 392 652 8 × 2 = 1 + 0.999 998 093 007 032 785 305 6;
  • 40) 0.999 998 093 007 032 785 305 6 × 2 = 1 + 0.999 996 186 014 065 570 611 2;
  • 41) 0.999 996 186 014 065 570 611 2 × 2 = 1 + 0.999 992 372 028 131 141 222 4;
  • 42) 0.999 992 372 028 131 141 222 4 × 2 = 1 + 0.999 984 744 056 262 282 444 8;
  • 43) 0.999 984 744 056 262 282 444 8 × 2 = 1 + 0.999 969 488 112 524 564 889 6;
  • 44) 0.999 969 488 112 524 564 889 6 × 2 = 1 + 0.999 938 976 225 049 129 779 2;
  • 45) 0.999 938 976 225 049 129 779 2 × 2 = 1 + 0.999 877 952 450 098 259 558 4;
  • 46) 0.999 877 952 450 098 259 558 4 × 2 = 1 + 0.999 755 904 900 196 519 116 8;
  • 47) 0.999 755 904 900 196 519 116 8 × 2 = 1 + 0.999 511 809 800 393 038 233 6;
  • 48) 0.999 511 809 800 393 038 233 6 × 2 = 1 + 0.999 023 619 600 786 076 467 2;
  • 49) 0.999 023 619 600 786 076 467 2 × 2 = 1 + 0.998 047 239 201 572 152 934 4;
  • 50) 0.998 047 239 201 572 152 934 4 × 2 = 1 + 0.996 094 478 403 144 305 868 8;
  • 51) 0.996 094 478 403 144 305 868 8 × 2 = 1 + 0.992 188 956 806 288 611 737 6;
  • 52) 0.992 188 956 806 288 611 737 6 × 2 = 1 + 0.984 377 913 612 577 223 475 2;
  • 53) 0.984 377 913 612 577 223 475 2 × 2 = 1 + 0.968 755 827 225 154 446 950 4;
  • 54) 0.968 755 827 225 154 446 950 4 × 2 = 1 + 0.937 511 654 450 308 893 900 8;
  • 55) 0.937 511 654 450 308 893 900 8 × 2 = 1 + 0.875 023 308 900 617 787 801 6;
  • 56) 0.875 023 308 900 617 787 801 6 × 2 = 1 + 0.750 046 617 801 235 575 603 2;
  • 57) 0.750 046 617 801 235 575 603 2 × 2 = 1 + 0.500 093 235 602 471 151 206 4;
  • 58) 0.500 093 235 602 471 151 206 4 × 2 = 1 + 0.000 186 471 204 942 302 412 8;

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:


0.016 738 891 601 562 496 531 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 531 2(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:


0.016 738 891 601 562 496 531 2(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 -0.016 738 891 601 562 496 531 2 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