-0.016 738 891 601 562 496 557 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 557(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
-0.016 738 891 601 562 496 557(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:

|-0.016 738 891 601 562 496 557| = 0.016 738 891 601 562 496 557


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: 0.016 738 891 601 562 496 557.

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) 0.016 738 891 601 562 496 557 × 2 = 0 + 0.033 477 783 203 124 993 114;
  • 2) 0.033 477 783 203 124 993 114 × 2 = 0 + 0.066 955 566 406 249 986 228;
  • 3) 0.066 955 566 406 249 986 228 × 2 = 0 + 0.133 911 132 812 499 972 456;
  • 4) 0.133 911 132 812 499 972 456 × 2 = 0 + 0.267 822 265 624 999 944 912;
  • 5) 0.267 822 265 624 999 944 912 × 2 = 0 + 0.535 644 531 249 999 889 824;
  • 6) 0.535 644 531 249 999 889 824 × 2 = 1 + 0.071 289 062 499 999 779 648;
  • 7) 0.071 289 062 499 999 779 648 × 2 = 0 + 0.142 578 124 999 999 559 296;
  • 8) 0.142 578 124 999 999 559 296 × 2 = 0 + 0.285 156 249 999 999 118 592;
  • 9) 0.285 156 249 999 999 118 592 × 2 = 0 + 0.570 312 499 999 998 237 184;
  • 10) 0.570 312 499 999 998 237 184 × 2 = 1 + 0.140 624 999 999 996 474 368;
  • 11) 0.140 624 999 999 996 474 368 × 2 = 0 + 0.281 249 999 999 992 948 736;
  • 12) 0.281 249 999 999 992 948 736 × 2 = 0 + 0.562 499 999 999 985 897 472;
  • 13) 0.562 499 999 999 985 897 472 × 2 = 1 + 0.124 999 999 999 971 794 944;
  • 14) 0.124 999 999 999 971 794 944 × 2 = 0 + 0.249 999 999 999 943 589 888;
  • 15) 0.249 999 999 999 943 589 888 × 2 = 0 + 0.499 999 999 999 887 179 776;
  • 16) 0.499 999 999 999 887 179 776 × 2 = 0 + 0.999 999 999 999 774 359 552;
  • 17) 0.999 999 999 999 774 359 552 × 2 = 1 + 0.999 999 999 999 548 719 104;
  • 18) 0.999 999 999 999 548 719 104 × 2 = 1 + 0.999 999 999 999 097 438 208;
  • 19) 0.999 999 999 999 097 438 208 × 2 = 1 + 0.999 999 999 998 194 876 416;
  • 20) 0.999 999 999 998 194 876 416 × 2 = 1 + 0.999 999 999 996 389 752 832;
  • 21) 0.999 999 999 996 389 752 832 × 2 = 1 + 0.999 999 999 992 779 505 664;
  • 22) 0.999 999 999 992 779 505 664 × 2 = 1 + 0.999 999 999 985 559 011 328;
  • 23) 0.999 999 999 985 559 011 328 × 2 = 1 + 0.999 999 999 971 118 022 656;
  • 24) 0.999 999 999 971 118 022 656 × 2 = 1 + 0.999 999 999 942 236 045 312;
  • 25) 0.999 999 999 942 236 045 312 × 2 = 1 + 0.999 999 999 884 472 090 624;
  • 26) 0.999 999 999 884 472 090 624 × 2 = 1 + 0.999 999 999 768 944 181 248;
  • 27) 0.999 999 999 768 944 181 248 × 2 = 1 + 0.999 999 999 537 888 362 496;
  • 28) 0.999 999 999 537 888 362 496 × 2 = 1 + 0.999 999 999 075 776 724 992;
  • 29) 0.999 999 999 075 776 724 992 × 2 = 1 + 0.999 999 998 151 553 449 984;
  • 30) 0.999 999 998 151 553 449 984 × 2 = 1 + 0.999 999 996 303 106 899 968;
  • 31) 0.999 999 996 303 106 899 968 × 2 = 1 + 0.999 999 992 606 213 799 936;
  • 32) 0.999 999 992 606 213 799 936 × 2 = 1 + 0.999 999 985 212 427 599 872;
  • 33) 0.999 999 985 212 427 599 872 × 2 = 1 + 0.999 999 970 424 855 199 744;
  • 34) 0.999 999 970 424 855 199 744 × 2 = 1 + 0.999 999 940 849 710 399 488;
  • 35) 0.999 999 940 849 710 399 488 × 2 = 1 + 0.999 999 881 699 420 798 976;
  • 36) 0.999 999 881 699 420 798 976 × 2 = 1 + 0.999 999 763 398 841 597 952;
  • 37) 0.999 999 763 398 841 597 952 × 2 = 1 + 0.999 999 526 797 683 195 904;
  • 38) 0.999 999 526 797 683 195 904 × 2 = 1 + 0.999 999 053 595 366 391 808;
  • 39) 0.999 999 053 595 366 391 808 × 2 = 1 + 0.999 998 107 190 732 783 616;
  • 40) 0.999 998 107 190 732 783 616 × 2 = 1 + 0.999 996 214 381 465 567 232;
  • 41) 0.999 996 214 381 465 567 232 × 2 = 1 + 0.999 992 428 762 931 134 464;
  • 42) 0.999 992 428 762 931 134 464 × 2 = 1 + 0.999 984 857 525 862 268 928;
  • 43) 0.999 984 857 525 862 268 928 × 2 = 1 + 0.999 969 715 051 724 537 856;
  • 44) 0.999 969 715 051 724 537 856 × 2 = 1 + 0.999 939 430 103 449 075 712;
  • 45) 0.999 939 430 103 449 075 712 × 2 = 1 + 0.999 878 860 206 898 151 424;
  • 46) 0.999 878 860 206 898 151 424 × 2 = 1 + 0.999 757 720 413 796 302 848;
  • 47) 0.999 757 720 413 796 302 848 × 2 = 1 + 0.999 515 440 827 592 605 696;
  • 48) 0.999 515 440 827 592 605 696 × 2 = 1 + 0.999 030 881 655 185 211 392;
  • 49) 0.999 030 881 655 185 211 392 × 2 = 1 + 0.998 061 763 310 370 422 784;
  • 50) 0.998 061 763 310 370 422 784 × 2 = 1 + 0.996 123 526 620 740 845 568;
  • 51) 0.996 123 526 620 740 845 568 × 2 = 1 + 0.992 247 053 241 481 691 136;
  • 52) 0.992 247 053 241 481 691 136 × 2 = 1 + 0.984 494 106 482 963 382 272;
  • 53) 0.984 494 106 482 963 382 272 × 2 = 1 + 0.968 988 212 965 926 764 544;
  • 54) 0.968 988 212 965 926 764 544 × 2 = 1 + 0.937 976 425 931 853 529 088;
  • 55) 0.937 976 425 931 853 529 088 × 2 = 1 + 0.875 952 851 863 707 058 176;
  • 56) 0.875 952 851 863 707 058 176 × 2 = 1 + 0.751 905 703 727 414 116 352;
  • 57) 0.751 905 703 727 414 116 352 × 2 = 1 + 0.503 811 407 454 828 232 704;
  • 58) 0.503 811 407 454 828 232 704 × 2 = 1 + 0.007 622 814 909 656 465 408;

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:


0.016 738 891 601 562 496 557(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 557(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:


0.016 738 891 601 562 496 557(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 -0.016 738 891 601 562 496 557 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