-0.016 738 891 601 562 496 530 561 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 561 6(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
-0.016 738 891 601 562 496 530 561 6(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:

|-0.016 738 891 601 562 496 530 561 6| = 0.016 738 891 601 562 496 530 561 6


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: 0.016 738 891 601 562 496 530 561 6.

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) 0.016 738 891 601 562 496 530 561 6 × 2 = 0 + 0.033 477 783 203 124 993 061 123 2;
  • 2) 0.033 477 783 203 124 993 061 123 2 × 2 = 0 + 0.066 955 566 406 249 986 122 246 4;
  • 3) 0.066 955 566 406 249 986 122 246 4 × 2 = 0 + 0.133 911 132 812 499 972 244 492 8;
  • 4) 0.133 911 132 812 499 972 244 492 8 × 2 = 0 + 0.267 822 265 624 999 944 488 985 6;
  • 5) 0.267 822 265 624 999 944 488 985 6 × 2 = 0 + 0.535 644 531 249 999 888 977 971 2;
  • 6) 0.535 644 531 249 999 888 977 971 2 × 2 = 1 + 0.071 289 062 499 999 777 955 942 4;
  • 7) 0.071 289 062 499 999 777 955 942 4 × 2 = 0 + 0.142 578 124 999 999 555 911 884 8;
  • 8) 0.142 578 124 999 999 555 911 884 8 × 2 = 0 + 0.285 156 249 999 999 111 823 769 6;
  • 9) 0.285 156 249 999 999 111 823 769 6 × 2 = 0 + 0.570 312 499 999 998 223 647 539 2;
  • 10) 0.570 312 499 999 998 223 647 539 2 × 2 = 1 + 0.140 624 999 999 996 447 295 078 4;
  • 11) 0.140 624 999 999 996 447 295 078 4 × 2 = 0 + 0.281 249 999 999 992 894 590 156 8;
  • 12) 0.281 249 999 999 992 894 590 156 8 × 2 = 0 + 0.562 499 999 999 985 789 180 313 6;
  • 13) 0.562 499 999 999 985 789 180 313 6 × 2 = 1 + 0.124 999 999 999 971 578 360 627 2;
  • 14) 0.124 999 999 999 971 578 360 627 2 × 2 = 0 + 0.249 999 999 999 943 156 721 254 4;
  • 15) 0.249 999 999 999 943 156 721 254 4 × 2 = 0 + 0.499 999 999 999 886 313 442 508 8;
  • 16) 0.499 999 999 999 886 313 442 508 8 × 2 = 0 + 0.999 999 999 999 772 626 885 017 6;
  • 17) 0.999 999 999 999 772 626 885 017 6 × 2 = 1 + 0.999 999 999 999 545 253 770 035 2;
  • 18) 0.999 999 999 999 545 253 770 035 2 × 2 = 1 + 0.999 999 999 999 090 507 540 070 4;
  • 19) 0.999 999 999 999 090 507 540 070 4 × 2 = 1 + 0.999 999 999 998 181 015 080 140 8;
  • 20) 0.999 999 999 998 181 015 080 140 8 × 2 = 1 + 0.999 999 999 996 362 030 160 281 6;
  • 21) 0.999 999 999 996 362 030 160 281 6 × 2 = 1 + 0.999 999 999 992 724 060 320 563 2;
  • 22) 0.999 999 999 992 724 060 320 563 2 × 2 = 1 + 0.999 999 999 985 448 120 641 126 4;
  • 23) 0.999 999 999 985 448 120 641 126 4 × 2 = 1 + 0.999 999 999 970 896 241 282 252 8;
  • 24) 0.999 999 999 970 896 241 282 252 8 × 2 = 1 + 0.999 999 999 941 792 482 564 505 6;
  • 25) 0.999 999 999 941 792 482 564 505 6 × 2 = 1 + 0.999 999 999 883 584 965 129 011 2;
  • 26) 0.999 999 999 883 584 965 129 011 2 × 2 = 1 + 0.999 999 999 767 169 930 258 022 4;
  • 27) 0.999 999 999 767 169 930 258 022 4 × 2 = 1 + 0.999 999 999 534 339 860 516 044 8;
  • 28) 0.999 999 999 534 339 860 516 044 8 × 2 = 1 + 0.999 999 999 068 679 721 032 089 6;
  • 29) 0.999 999 999 068 679 721 032 089 6 × 2 = 1 + 0.999 999 998 137 359 442 064 179 2;
  • 30) 0.999 999 998 137 359 442 064 179 2 × 2 = 1 + 0.999 999 996 274 718 884 128 358 4;
  • 31) 0.999 999 996 274 718 884 128 358 4 × 2 = 1 + 0.999 999 992 549 437 768 256 716 8;
  • 32) 0.999 999 992 549 437 768 256 716 8 × 2 = 1 + 0.999 999 985 098 875 536 513 433 6;
  • 33) 0.999 999 985 098 875 536 513 433 6 × 2 = 1 + 0.999 999 970 197 751 073 026 867 2;
  • 34) 0.999 999 970 197 751 073 026 867 2 × 2 = 1 + 0.999 999 940 395 502 146 053 734 4;
  • 35) 0.999 999 940 395 502 146 053 734 4 × 2 = 1 + 0.999 999 880 791 004 292 107 468 8;
  • 36) 0.999 999 880 791 004 292 107 468 8 × 2 = 1 + 0.999 999 761 582 008 584 214 937 6;
  • 37) 0.999 999 761 582 008 584 214 937 6 × 2 = 1 + 0.999 999 523 164 017 168 429 875 2;
  • 38) 0.999 999 523 164 017 168 429 875 2 × 2 = 1 + 0.999 999 046 328 034 336 859 750 4;
  • 39) 0.999 999 046 328 034 336 859 750 4 × 2 = 1 + 0.999 998 092 656 068 673 719 500 8;
  • 40) 0.999 998 092 656 068 673 719 500 8 × 2 = 1 + 0.999 996 185 312 137 347 439 001 6;
  • 41) 0.999 996 185 312 137 347 439 001 6 × 2 = 1 + 0.999 992 370 624 274 694 878 003 2;
  • 42) 0.999 992 370 624 274 694 878 003 2 × 2 = 1 + 0.999 984 741 248 549 389 756 006 4;
  • 43) 0.999 984 741 248 549 389 756 006 4 × 2 = 1 + 0.999 969 482 497 098 779 512 012 8;
  • 44) 0.999 969 482 497 098 779 512 012 8 × 2 = 1 + 0.999 938 964 994 197 559 024 025 6;
  • 45) 0.999 938 964 994 197 559 024 025 6 × 2 = 1 + 0.999 877 929 988 395 118 048 051 2;
  • 46) 0.999 877 929 988 395 118 048 051 2 × 2 = 1 + 0.999 755 859 976 790 236 096 102 4;
  • 47) 0.999 755 859 976 790 236 096 102 4 × 2 = 1 + 0.999 511 719 953 580 472 192 204 8;
  • 48) 0.999 511 719 953 580 472 192 204 8 × 2 = 1 + 0.999 023 439 907 160 944 384 409 6;
  • 49) 0.999 023 439 907 160 944 384 409 6 × 2 = 1 + 0.998 046 879 814 321 888 768 819 2;
  • 50) 0.998 046 879 814 321 888 768 819 2 × 2 = 1 + 0.996 093 759 628 643 777 537 638 4;
  • 51) 0.996 093 759 628 643 777 537 638 4 × 2 = 1 + 0.992 187 519 257 287 555 075 276 8;
  • 52) 0.992 187 519 257 287 555 075 276 8 × 2 = 1 + 0.984 375 038 514 575 110 150 553 6;
  • 53) 0.984 375 038 514 575 110 150 553 6 × 2 = 1 + 0.968 750 077 029 150 220 301 107 2;
  • 54) 0.968 750 077 029 150 220 301 107 2 × 2 = 1 + 0.937 500 154 058 300 440 602 214 4;
  • 55) 0.937 500 154 058 300 440 602 214 4 × 2 = 1 + 0.875 000 308 116 600 881 204 428 8;
  • 56) 0.875 000 308 116 600 881 204 428 8 × 2 = 1 + 0.750 000 616 233 201 762 408 857 6;
  • 57) 0.750 000 616 233 201 762 408 857 6 × 2 = 1 + 0.500 001 232 466 403 524 817 715 2;
  • 58) 0.500 001 232 466 403 524 817 715 2 × 2 = 1 + 0.000 002 464 932 807 049 635 430 4;

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:


0.016 738 891 601 562 496 530 561 6(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 561 6(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:


0.016 738 891 601 562 496 530 561 6(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 -0.016 738 891 601 562 496 530 561 6 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