-0.016 738 891 601 562 496 514 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 514 8(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
-0.016 738 891 601 562 496 514 8(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:

|-0.016 738 891 601 562 496 514 8| = 0.016 738 891 601 562 496 514 8


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: 0.016 738 891 601 562 496 514 8.

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) 0.016 738 891 601 562 496 514 8 × 2 = 0 + 0.033 477 783 203 124 993 029 6;
  • 2) 0.033 477 783 203 124 993 029 6 × 2 = 0 + 0.066 955 566 406 249 986 059 2;
  • 3) 0.066 955 566 406 249 986 059 2 × 2 = 0 + 0.133 911 132 812 499 972 118 4;
  • 4) 0.133 911 132 812 499 972 118 4 × 2 = 0 + 0.267 822 265 624 999 944 236 8;
  • 5) 0.267 822 265 624 999 944 236 8 × 2 = 0 + 0.535 644 531 249 999 888 473 6;
  • 6) 0.535 644 531 249 999 888 473 6 × 2 = 1 + 0.071 289 062 499 999 776 947 2;
  • 7) 0.071 289 062 499 999 776 947 2 × 2 = 0 + 0.142 578 124 999 999 553 894 4;
  • 8) 0.142 578 124 999 999 553 894 4 × 2 = 0 + 0.285 156 249 999 999 107 788 8;
  • 9) 0.285 156 249 999 999 107 788 8 × 2 = 0 + 0.570 312 499 999 998 215 577 6;
  • 10) 0.570 312 499 999 998 215 577 6 × 2 = 1 + 0.140 624 999 999 996 431 155 2;
  • 11) 0.140 624 999 999 996 431 155 2 × 2 = 0 + 0.281 249 999 999 992 862 310 4;
  • 12) 0.281 249 999 999 992 862 310 4 × 2 = 0 + 0.562 499 999 999 985 724 620 8;
  • 13) 0.562 499 999 999 985 724 620 8 × 2 = 1 + 0.124 999 999 999 971 449 241 6;
  • 14) 0.124 999 999 999 971 449 241 6 × 2 = 0 + 0.249 999 999 999 942 898 483 2;
  • 15) 0.249 999 999 999 942 898 483 2 × 2 = 0 + 0.499 999 999 999 885 796 966 4;
  • 16) 0.499 999 999 999 885 796 966 4 × 2 = 0 + 0.999 999 999 999 771 593 932 8;
  • 17) 0.999 999 999 999 771 593 932 8 × 2 = 1 + 0.999 999 999 999 543 187 865 6;
  • 18) 0.999 999 999 999 543 187 865 6 × 2 = 1 + 0.999 999 999 999 086 375 731 2;
  • 19) 0.999 999 999 999 086 375 731 2 × 2 = 1 + 0.999 999 999 998 172 751 462 4;
  • 20) 0.999 999 999 998 172 751 462 4 × 2 = 1 + 0.999 999 999 996 345 502 924 8;
  • 21) 0.999 999 999 996 345 502 924 8 × 2 = 1 + 0.999 999 999 992 691 005 849 6;
  • 22) 0.999 999 999 992 691 005 849 6 × 2 = 1 + 0.999 999 999 985 382 011 699 2;
  • 23) 0.999 999 999 985 382 011 699 2 × 2 = 1 + 0.999 999 999 970 764 023 398 4;
  • 24) 0.999 999 999 970 764 023 398 4 × 2 = 1 + 0.999 999 999 941 528 046 796 8;
  • 25) 0.999 999 999 941 528 046 796 8 × 2 = 1 + 0.999 999 999 883 056 093 593 6;
  • 26) 0.999 999 999 883 056 093 593 6 × 2 = 1 + 0.999 999 999 766 112 187 187 2;
  • 27) 0.999 999 999 766 112 187 187 2 × 2 = 1 + 0.999 999 999 532 224 374 374 4;
  • 28) 0.999 999 999 532 224 374 374 4 × 2 = 1 + 0.999 999 999 064 448 748 748 8;
  • 29) 0.999 999 999 064 448 748 748 8 × 2 = 1 + 0.999 999 998 128 897 497 497 6;
  • 30) 0.999 999 998 128 897 497 497 6 × 2 = 1 + 0.999 999 996 257 794 994 995 2;
  • 31) 0.999 999 996 257 794 994 995 2 × 2 = 1 + 0.999 999 992 515 589 989 990 4;
  • 32) 0.999 999 992 515 589 989 990 4 × 2 = 1 + 0.999 999 985 031 179 979 980 8;
  • 33) 0.999 999 985 031 179 979 980 8 × 2 = 1 + 0.999 999 970 062 359 959 961 6;
  • 34) 0.999 999 970 062 359 959 961 6 × 2 = 1 + 0.999 999 940 124 719 919 923 2;
  • 35) 0.999 999 940 124 719 919 923 2 × 2 = 1 + 0.999 999 880 249 439 839 846 4;
  • 36) 0.999 999 880 249 439 839 846 4 × 2 = 1 + 0.999 999 760 498 879 679 692 8;
  • 37) 0.999 999 760 498 879 679 692 8 × 2 = 1 + 0.999 999 520 997 759 359 385 6;
  • 38) 0.999 999 520 997 759 359 385 6 × 2 = 1 + 0.999 999 041 995 518 718 771 2;
  • 39) 0.999 999 041 995 518 718 771 2 × 2 = 1 + 0.999 998 083 991 037 437 542 4;
  • 40) 0.999 998 083 991 037 437 542 4 × 2 = 1 + 0.999 996 167 982 074 875 084 8;
  • 41) 0.999 996 167 982 074 875 084 8 × 2 = 1 + 0.999 992 335 964 149 750 169 6;
  • 42) 0.999 992 335 964 149 750 169 6 × 2 = 1 + 0.999 984 671 928 299 500 339 2;
  • 43) 0.999 984 671 928 299 500 339 2 × 2 = 1 + 0.999 969 343 856 599 000 678 4;
  • 44) 0.999 969 343 856 599 000 678 4 × 2 = 1 + 0.999 938 687 713 198 001 356 8;
  • 45) 0.999 938 687 713 198 001 356 8 × 2 = 1 + 0.999 877 375 426 396 002 713 6;
  • 46) 0.999 877 375 426 396 002 713 6 × 2 = 1 + 0.999 754 750 852 792 005 427 2;
  • 47) 0.999 754 750 852 792 005 427 2 × 2 = 1 + 0.999 509 501 705 584 010 854 4;
  • 48) 0.999 509 501 705 584 010 854 4 × 2 = 1 + 0.999 019 003 411 168 021 708 8;
  • 49) 0.999 019 003 411 168 021 708 8 × 2 = 1 + 0.998 038 006 822 336 043 417 6;
  • 50) 0.998 038 006 822 336 043 417 6 × 2 = 1 + 0.996 076 013 644 672 086 835 2;
  • 51) 0.996 076 013 644 672 086 835 2 × 2 = 1 + 0.992 152 027 289 344 173 670 4;
  • 52) 0.992 152 027 289 344 173 670 4 × 2 = 1 + 0.984 304 054 578 688 347 340 8;
  • 53) 0.984 304 054 578 688 347 340 8 × 2 = 1 + 0.968 608 109 157 376 694 681 6;
  • 54) 0.968 608 109 157 376 694 681 6 × 2 = 1 + 0.937 216 218 314 753 389 363 2;
  • 55) 0.937 216 218 314 753 389 363 2 × 2 = 1 + 0.874 432 436 629 506 778 726 4;
  • 56) 0.874 432 436 629 506 778 726 4 × 2 = 1 + 0.748 864 873 259 013 557 452 8;
  • 57) 0.748 864 873 259 013 557 452 8 × 2 = 1 + 0.497 729 746 518 027 114 905 6;
  • 58) 0.497 729 746 518 027 114 905 6 × 2 = 0 + 0.995 459 493 036 054 229 811 2;

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:


0.016 738 891 601 562 496 514 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 514 8(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:


0.016 738 891 601 562 496 514 8(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 -0.016 738 891 601 562 496 514 8 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