-0.016 738 891 601 562 496 861 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 861(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
-0.016 738 891 601 562 496 861(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:

|-0.016 738 891 601 562 496 861| = 0.016 738 891 601 562 496 861


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: 0.016 738 891 601 562 496 861.

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) 0.016 738 891 601 562 496 861 × 2 = 0 + 0.033 477 783 203 124 993 722;
  • 2) 0.033 477 783 203 124 993 722 × 2 = 0 + 0.066 955 566 406 249 987 444;
  • 3) 0.066 955 566 406 249 987 444 × 2 = 0 + 0.133 911 132 812 499 974 888;
  • 4) 0.133 911 132 812 499 974 888 × 2 = 0 + 0.267 822 265 624 999 949 776;
  • 5) 0.267 822 265 624 999 949 776 × 2 = 0 + 0.535 644 531 249 999 899 552;
  • 6) 0.535 644 531 249 999 899 552 × 2 = 1 + 0.071 289 062 499 999 799 104;
  • 7) 0.071 289 062 499 999 799 104 × 2 = 0 + 0.142 578 124 999 999 598 208;
  • 8) 0.142 578 124 999 999 598 208 × 2 = 0 + 0.285 156 249 999 999 196 416;
  • 9) 0.285 156 249 999 999 196 416 × 2 = 0 + 0.570 312 499 999 998 392 832;
  • 10) 0.570 312 499 999 998 392 832 × 2 = 1 + 0.140 624 999 999 996 785 664;
  • 11) 0.140 624 999 999 996 785 664 × 2 = 0 + 0.281 249 999 999 993 571 328;
  • 12) 0.281 249 999 999 993 571 328 × 2 = 0 + 0.562 499 999 999 987 142 656;
  • 13) 0.562 499 999 999 987 142 656 × 2 = 1 + 0.124 999 999 999 974 285 312;
  • 14) 0.124 999 999 999 974 285 312 × 2 = 0 + 0.249 999 999 999 948 570 624;
  • 15) 0.249 999 999 999 948 570 624 × 2 = 0 + 0.499 999 999 999 897 141 248;
  • 16) 0.499 999 999 999 897 141 248 × 2 = 0 + 0.999 999 999 999 794 282 496;
  • 17) 0.999 999 999 999 794 282 496 × 2 = 1 + 0.999 999 999 999 588 564 992;
  • 18) 0.999 999 999 999 588 564 992 × 2 = 1 + 0.999 999 999 999 177 129 984;
  • 19) 0.999 999 999 999 177 129 984 × 2 = 1 + 0.999 999 999 998 354 259 968;
  • 20) 0.999 999 999 998 354 259 968 × 2 = 1 + 0.999 999 999 996 708 519 936;
  • 21) 0.999 999 999 996 708 519 936 × 2 = 1 + 0.999 999 999 993 417 039 872;
  • 22) 0.999 999 999 993 417 039 872 × 2 = 1 + 0.999 999 999 986 834 079 744;
  • 23) 0.999 999 999 986 834 079 744 × 2 = 1 + 0.999 999 999 973 668 159 488;
  • 24) 0.999 999 999 973 668 159 488 × 2 = 1 + 0.999 999 999 947 336 318 976;
  • 25) 0.999 999 999 947 336 318 976 × 2 = 1 + 0.999 999 999 894 672 637 952;
  • 26) 0.999 999 999 894 672 637 952 × 2 = 1 + 0.999 999 999 789 345 275 904;
  • 27) 0.999 999 999 789 345 275 904 × 2 = 1 + 0.999 999 999 578 690 551 808;
  • 28) 0.999 999 999 578 690 551 808 × 2 = 1 + 0.999 999 999 157 381 103 616;
  • 29) 0.999 999 999 157 381 103 616 × 2 = 1 + 0.999 999 998 314 762 207 232;
  • 30) 0.999 999 998 314 762 207 232 × 2 = 1 + 0.999 999 996 629 524 414 464;
  • 31) 0.999 999 996 629 524 414 464 × 2 = 1 + 0.999 999 993 259 048 828 928;
  • 32) 0.999 999 993 259 048 828 928 × 2 = 1 + 0.999 999 986 518 097 657 856;
  • 33) 0.999 999 986 518 097 657 856 × 2 = 1 + 0.999 999 973 036 195 315 712;
  • 34) 0.999 999 973 036 195 315 712 × 2 = 1 + 0.999 999 946 072 390 631 424;
  • 35) 0.999 999 946 072 390 631 424 × 2 = 1 + 0.999 999 892 144 781 262 848;
  • 36) 0.999 999 892 144 781 262 848 × 2 = 1 + 0.999 999 784 289 562 525 696;
  • 37) 0.999 999 784 289 562 525 696 × 2 = 1 + 0.999 999 568 579 125 051 392;
  • 38) 0.999 999 568 579 125 051 392 × 2 = 1 + 0.999 999 137 158 250 102 784;
  • 39) 0.999 999 137 158 250 102 784 × 2 = 1 + 0.999 998 274 316 500 205 568;
  • 40) 0.999 998 274 316 500 205 568 × 2 = 1 + 0.999 996 548 633 000 411 136;
  • 41) 0.999 996 548 633 000 411 136 × 2 = 1 + 0.999 993 097 266 000 822 272;
  • 42) 0.999 993 097 266 000 822 272 × 2 = 1 + 0.999 986 194 532 001 644 544;
  • 43) 0.999 986 194 532 001 644 544 × 2 = 1 + 0.999 972 389 064 003 289 088;
  • 44) 0.999 972 389 064 003 289 088 × 2 = 1 + 0.999 944 778 128 006 578 176;
  • 45) 0.999 944 778 128 006 578 176 × 2 = 1 + 0.999 889 556 256 013 156 352;
  • 46) 0.999 889 556 256 013 156 352 × 2 = 1 + 0.999 779 112 512 026 312 704;
  • 47) 0.999 779 112 512 026 312 704 × 2 = 1 + 0.999 558 225 024 052 625 408;
  • 48) 0.999 558 225 024 052 625 408 × 2 = 1 + 0.999 116 450 048 105 250 816;
  • 49) 0.999 116 450 048 105 250 816 × 2 = 1 + 0.998 232 900 096 210 501 632;
  • 50) 0.998 232 900 096 210 501 632 × 2 = 1 + 0.996 465 800 192 421 003 264;
  • 51) 0.996 465 800 192 421 003 264 × 2 = 1 + 0.992 931 600 384 842 006 528;
  • 52) 0.992 931 600 384 842 006 528 × 2 = 1 + 0.985 863 200 769 684 013 056;
  • 53) 0.985 863 200 769 684 013 056 × 2 = 1 + 0.971 726 401 539 368 026 112;
  • 54) 0.971 726 401 539 368 026 112 × 2 = 1 + 0.943 452 803 078 736 052 224;
  • 55) 0.943 452 803 078 736 052 224 × 2 = 1 + 0.886 905 606 157 472 104 448;
  • 56) 0.886 905 606 157 472 104 448 × 2 = 1 + 0.773 811 212 314 944 208 896;
  • 57) 0.773 811 212 314 944 208 896 × 2 = 1 + 0.547 622 424 629 888 417 792;
  • 58) 0.547 622 424 629 888 417 792 × 2 = 1 + 0.095 244 849 259 776 835 584;

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:


0.016 738 891 601 562 496 861(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 861(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:


0.016 738 891 601 562 496 861(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 -0.016 738 891 601 562 496 861 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