-0.016 738 891 601 562 496 525 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 525 9(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
-0.016 738 891 601 562 496 525 9(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:

|-0.016 738 891 601 562 496 525 9| = 0.016 738 891 601 562 496 525 9


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: 0.016 738 891 601 562 496 525 9.

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) 0.016 738 891 601 562 496 525 9 × 2 = 0 + 0.033 477 783 203 124 993 051 8;
  • 2) 0.033 477 783 203 124 993 051 8 × 2 = 0 + 0.066 955 566 406 249 986 103 6;
  • 3) 0.066 955 566 406 249 986 103 6 × 2 = 0 + 0.133 911 132 812 499 972 207 2;
  • 4) 0.133 911 132 812 499 972 207 2 × 2 = 0 + 0.267 822 265 624 999 944 414 4;
  • 5) 0.267 822 265 624 999 944 414 4 × 2 = 0 + 0.535 644 531 249 999 888 828 8;
  • 6) 0.535 644 531 249 999 888 828 8 × 2 = 1 + 0.071 289 062 499 999 777 657 6;
  • 7) 0.071 289 062 499 999 777 657 6 × 2 = 0 + 0.142 578 124 999 999 555 315 2;
  • 8) 0.142 578 124 999 999 555 315 2 × 2 = 0 + 0.285 156 249 999 999 110 630 4;
  • 9) 0.285 156 249 999 999 110 630 4 × 2 = 0 + 0.570 312 499 999 998 221 260 8;
  • 10) 0.570 312 499 999 998 221 260 8 × 2 = 1 + 0.140 624 999 999 996 442 521 6;
  • 11) 0.140 624 999 999 996 442 521 6 × 2 = 0 + 0.281 249 999 999 992 885 043 2;
  • 12) 0.281 249 999 999 992 885 043 2 × 2 = 0 + 0.562 499 999 999 985 770 086 4;
  • 13) 0.562 499 999 999 985 770 086 4 × 2 = 1 + 0.124 999 999 999 971 540 172 8;
  • 14) 0.124 999 999 999 971 540 172 8 × 2 = 0 + 0.249 999 999 999 943 080 345 6;
  • 15) 0.249 999 999 999 943 080 345 6 × 2 = 0 + 0.499 999 999 999 886 160 691 2;
  • 16) 0.499 999 999 999 886 160 691 2 × 2 = 0 + 0.999 999 999 999 772 321 382 4;
  • 17) 0.999 999 999 999 772 321 382 4 × 2 = 1 + 0.999 999 999 999 544 642 764 8;
  • 18) 0.999 999 999 999 544 642 764 8 × 2 = 1 + 0.999 999 999 999 089 285 529 6;
  • 19) 0.999 999 999 999 089 285 529 6 × 2 = 1 + 0.999 999 999 998 178 571 059 2;
  • 20) 0.999 999 999 998 178 571 059 2 × 2 = 1 + 0.999 999 999 996 357 142 118 4;
  • 21) 0.999 999 999 996 357 142 118 4 × 2 = 1 + 0.999 999 999 992 714 284 236 8;
  • 22) 0.999 999 999 992 714 284 236 8 × 2 = 1 + 0.999 999 999 985 428 568 473 6;
  • 23) 0.999 999 999 985 428 568 473 6 × 2 = 1 + 0.999 999 999 970 857 136 947 2;
  • 24) 0.999 999 999 970 857 136 947 2 × 2 = 1 + 0.999 999 999 941 714 273 894 4;
  • 25) 0.999 999 999 941 714 273 894 4 × 2 = 1 + 0.999 999 999 883 428 547 788 8;
  • 26) 0.999 999 999 883 428 547 788 8 × 2 = 1 + 0.999 999 999 766 857 095 577 6;
  • 27) 0.999 999 999 766 857 095 577 6 × 2 = 1 + 0.999 999 999 533 714 191 155 2;
  • 28) 0.999 999 999 533 714 191 155 2 × 2 = 1 + 0.999 999 999 067 428 382 310 4;
  • 29) 0.999 999 999 067 428 382 310 4 × 2 = 1 + 0.999 999 998 134 856 764 620 8;
  • 30) 0.999 999 998 134 856 764 620 8 × 2 = 1 + 0.999 999 996 269 713 529 241 6;
  • 31) 0.999 999 996 269 713 529 241 6 × 2 = 1 + 0.999 999 992 539 427 058 483 2;
  • 32) 0.999 999 992 539 427 058 483 2 × 2 = 1 + 0.999 999 985 078 854 116 966 4;
  • 33) 0.999 999 985 078 854 116 966 4 × 2 = 1 + 0.999 999 970 157 708 233 932 8;
  • 34) 0.999 999 970 157 708 233 932 8 × 2 = 1 + 0.999 999 940 315 416 467 865 6;
  • 35) 0.999 999 940 315 416 467 865 6 × 2 = 1 + 0.999 999 880 630 832 935 731 2;
  • 36) 0.999 999 880 630 832 935 731 2 × 2 = 1 + 0.999 999 761 261 665 871 462 4;
  • 37) 0.999 999 761 261 665 871 462 4 × 2 = 1 + 0.999 999 522 523 331 742 924 8;
  • 38) 0.999 999 522 523 331 742 924 8 × 2 = 1 + 0.999 999 045 046 663 485 849 6;
  • 39) 0.999 999 045 046 663 485 849 6 × 2 = 1 + 0.999 998 090 093 326 971 699 2;
  • 40) 0.999 998 090 093 326 971 699 2 × 2 = 1 + 0.999 996 180 186 653 943 398 4;
  • 41) 0.999 996 180 186 653 943 398 4 × 2 = 1 + 0.999 992 360 373 307 886 796 8;
  • 42) 0.999 992 360 373 307 886 796 8 × 2 = 1 + 0.999 984 720 746 615 773 593 6;
  • 43) 0.999 984 720 746 615 773 593 6 × 2 = 1 + 0.999 969 441 493 231 547 187 2;
  • 44) 0.999 969 441 493 231 547 187 2 × 2 = 1 + 0.999 938 882 986 463 094 374 4;
  • 45) 0.999 938 882 986 463 094 374 4 × 2 = 1 + 0.999 877 765 972 926 188 748 8;
  • 46) 0.999 877 765 972 926 188 748 8 × 2 = 1 + 0.999 755 531 945 852 377 497 6;
  • 47) 0.999 755 531 945 852 377 497 6 × 2 = 1 + 0.999 511 063 891 704 754 995 2;
  • 48) 0.999 511 063 891 704 754 995 2 × 2 = 1 + 0.999 022 127 783 409 509 990 4;
  • 49) 0.999 022 127 783 409 509 990 4 × 2 = 1 + 0.998 044 255 566 819 019 980 8;
  • 50) 0.998 044 255 566 819 019 980 8 × 2 = 1 + 0.996 088 511 133 638 039 961 6;
  • 51) 0.996 088 511 133 638 039 961 6 × 2 = 1 + 0.992 177 022 267 276 079 923 2;
  • 52) 0.992 177 022 267 276 079 923 2 × 2 = 1 + 0.984 354 044 534 552 159 846 4;
  • 53) 0.984 354 044 534 552 159 846 4 × 2 = 1 + 0.968 708 089 069 104 319 692 8;
  • 54) 0.968 708 089 069 104 319 692 8 × 2 = 1 + 0.937 416 178 138 208 639 385 6;
  • 55) 0.937 416 178 138 208 639 385 6 × 2 = 1 + 0.874 832 356 276 417 278 771 2;
  • 56) 0.874 832 356 276 417 278 771 2 × 2 = 1 + 0.749 664 712 552 834 557 542 4;
  • 57) 0.749 664 712 552 834 557 542 4 × 2 = 1 + 0.499 329 425 105 669 115 084 8;
  • 58) 0.499 329 425 105 669 115 084 8 × 2 = 0 + 0.998 658 850 211 338 230 169 6;

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:


0.016 738 891 601 562 496 525 9(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 525 9(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:


0.016 738 891 601 562 496 525 9(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 -0.016 738 891 601 562 496 525 9 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