-0.016 738 891 601 562 496 529 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 91(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
-0.016 738 891 601 562 496 529 91(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:

|-0.016 738 891 601 562 496 529 91| = 0.016 738 891 601 562 496 529 91


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: 0.016 738 891 601 562 496 529 91.

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) 0.016 738 891 601 562 496 529 91 × 2 = 0 + 0.033 477 783 203 124 993 059 82;
  • 2) 0.033 477 783 203 124 993 059 82 × 2 = 0 + 0.066 955 566 406 249 986 119 64;
  • 3) 0.066 955 566 406 249 986 119 64 × 2 = 0 + 0.133 911 132 812 499 972 239 28;
  • 4) 0.133 911 132 812 499 972 239 28 × 2 = 0 + 0.267 822 265 624 999 944 478 56;
  • 5) 0.267 822 265 624 999 944 478 56 × 2 = 0 + 0.535 644 531 249 999 888 957 12;
  • 6) 0.535 644 531 249 999 888 957 12 × 2 = 1 + 0.071 289 062 499 999 777 914 24;
  • 7) 0.071 289 062 499 999 777 914 24 × 2 = 0 + 0.142 578 124 999 999 555 828 48;
  • 8) 0.142 578 124 999 999 555 828 48 × 2 = 0 + 0.285 156 249 999 999 111 656 96;
  • 9) 0.285 156 249 999 999 111 656 96 × 2 = 0 + 0.570 312 499 999 998 223 313 92;
  • 10) 0.570 312 499 999 998 223 313 92 × 2 = 1 + 0.140 624 999 999 996 446 627 84;
  • 11) 0.140 624 999 999 996 446 627 84 × 2 = 0 + 0.281 249 999 999 992 893 255 68;
  • 12) 0.281 249 999 999 992 893 255 68 × 2 = 0 + 0.562 499 999 999 985 786 511 36;
  • 13) 0.562 499 999 999 985 786 511 36 × 2 = 1 + 0.124 999 999 999 971 573 022 72;
  • 14) 0.124 999 999 999 971 573 022 72 × 2 = 0 + 0.249 999 999 999 943 146 045 44;
  • 15) 0.249 999 999 999 943 146 045 44 × 2 = 0 + 0.499 999 999 999 886 292 090 88;
  • 16) 0.499 999 999 999 886 292 090 88 × 2 = 0 + 0.999 999 999 999 772 584 181 76;
  • 17) 0.999 999 999 999 772 584 181 76 × 2 = 1 + 0.999 999 999 999 545 168 363 52;
  • 18) 0.999 999 999 999 545 168 363 52 × 2 = 1 + 0.999 999 999 999 090 336 727 04;
  • 19) 0.999 999 999 999 090 336 727 04 × 2 = 1 + 0.999 999 999 998 180 673 454 08;
  • 20) 0.999 999 999 998 180 673 454 08 × 2 = 1 + 0.999 999 999 996 361 346 908 16;
  • 21) 0.999 999 999 996 361 346 908 16 × 2 = 1 + 0.999 999 999 992 722 693 816 32;
  • 22) 0.999 999 999 992 722 693 816 32 × 2 = 1 + 0.999 999 999 985 445 387 632 64;
  • 23) 0.999 999 999 985 445 387 632 64 × 2 = 1 + 0.999 999 999 970 890 775 265 28;
  • 24) 0.999 999 999 970 890 775 265 28 × 2 = 1 + 0.999 999 999 941 781 550 530 56;
  • 25) 0.999 999 999 941 781 550 530 56 × 2 = 1 + 0.999 999 999 883 563 101 061 12;
  • 26) 0.999 999 999 883 563 101 061 12 × 2 = 1 + 0.999 999 999 767 126 202 122 24;
  • 27) 0.999 999 999 767 126 202 122 24 × 2 = 1 + 0.999 999 999 534 252 404 244 48;
  • 28) 0.999 999 999 534 252 404 244 48 × 2 = 1 + 0.999 999 999 068 504 808 488 96;
  • 29) 0.999 999 999 068 504 808 488 96 × 2 = 1 + 0.999 999 998 137 009 616 977 92;
  • 30) 0.999 999 998 137 009 616 977 92 × 2 = 1 + 0.999 999 996 274 019 233 955 84;
  • 31) 0.999 999 996 274 019 233 955 84 × 2 = 1 + 0.999 999 992 548 038 467 911 68;
  • 32) 0.999 999 992 548 038 467 911 68 × 2 = 1 + 0.999 999 985 096 076 935 823 36;
  • 33) 0.999 999 985 096 076 935 823 36 × 2 = 1 + 0.999 999 970 192 153 871 646 72;
  • 34) 0.999 999 970 192 153 871 646 72 × 2 = 1 + 0.999 999 940 384 307 743 293 44;
  • 35) 0.999 999 940 384 307 743 293 44 × 2 = 1 + 0.999 999 880 768 615 486 586 88;
  • 36) 0.999 999 880 768 615 486 586 88 × 2 = 1 + 0.999 999 761 537 230 973 173 76;
  • 37) 0.999 999 761 537 230 973 173 76 × 2 = 1 + 0.999 999 523 074 461 946 347 52;
  • 38) 0.999 999 523 074 461 946 347 52 × 2 = 1 + 0.999 999 046 148 923 892 695 04;
  • 39) 0.999 999 046 148 923 892 695 04 × 2 = 1 + 0.999 998 092 297 847 785 390 08;
  • 40) 0.999 998 092 297 847 785 390 08 × 2 = 1 + 0.999 996 184 595 695 570 780 16;
  • 41) 0.999 996 184 595 695 570 780 16 × 2 = 1 + 0.999 992 369 191 391 141 560 32;
  • 42) 0.999 992 369 191 391 141 560 32 × 2 = 1 + 0.999 984 738 382 782 283 120 64;
  • 43) 0.999 984 738 382 782 283 120 64 × 2 = 1 + 0.999 969 476 765 564 566 241 28;
  • 44) 0.999 969 476 765 564 566 241 28 × 2 = 1 + 0.999 938 953 531 129 132 482 56;
  • 45) 0.999 938 953 531 129 132 482 56 × 2 = 1 + 0.999 877 907 062 258 264 965 12;
  • 46) 0.999 877 907 062 258 264 965 12 × 2 = 1 + 0.999 755 814 124 516 529 930 24;
  • 47) 0.999 755 814 124 516 529 930 24 × 2 = 1 + 0.999 511 628 249 033 059 860 48;
  • 48) 0.999 511 628 249 033 059 860 48 × 2 = 1 + 0.999 023 256 498 066 119 720 96;
  • 49) 0.999 023 256 498 066 119 720 96 × 2 = 1 + 0.998 046 512 996 132 239 441 92;
  • 50) 0.998 046 512 996 132 239 441 92 × 2 = 1 + 0.996 093 025 992 264 478 883 84;
  • 51) 0.996 093 025 992 264 478 883 84 × 2 = 1 + 0.992 186 051 984 528 957 767 68;
  • 52) 0.992 186 051 984 528 957 767 68 × 2 = 1 + 0.984 372 103 969 057 915 535 36;
  • 53) 0.984 372 103 969 057 915 535 36 × 2 = 1 + 0.968 744 207 938 115 831 070 72;
  • 54) 0.968 744 207 938 115 831 070 72 × 2 = 1 + 0.937 488 415 876 231 662 141 44;
  • 55) 0.937 488 415 876 231 662 141 44 × 2 = 1 + 0.874 976 831 752 463 324 282 88;
  • 56) 0.874 976 831 752 463 324 282 88 × 2 = 1 + 0.749 953 663 504 926 648 565 76;
  • 57) 0.749 953 663 504 926 648 565 76 × 2 = 1 + 0.499 907 327 009 853 297 131 52;
  • 58) 0.499 907 327 009 853 297 131 52 × 2 = 0 + 0.999 814 654 019 706 594 263 04;

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:


0.016 738 891 601 562 496 529 91(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 529 91(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:


0.016 738 891 601 562 496 529 91(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 -0.016 738 891 601 562 496 529 91 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