-0.016 738 891 601 562 496 530 554 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 554 47(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
-0.016 738 891 601 562 496 530 554 47(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:

|-0.016 738 891 601 562 496 530 554 47| = 0.016 738 891 601 562 496 530 554 47


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: 0.016 738 891 601 562 496 530 554 47.

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) 0.016 738 891 601 562 496 530 554 47 × 2 = 0 + 0.033 477 783 203 124 993 061 108 94;
  • 2) 0.033 477 783 203 124 993 061 108 94 × 2 = 0 + 0.066 955 566 406 249 986 122 217 88;
  • 3) 0.066 955 566 406 249 986 122 217 88 × 2 = 0 + 0.133 911 132 812 499 972 244 435 76;
  • 4) 0.133 911 132 812 499 972 244 435 76 × 2 = 0 + 0.267 822 265 624 999 944 488 871 52;
  • 5) 0.267 822 265 624 999 944 488 871 52 × 2 = 0 + 0.535 644 531 249 999 888 977 743 04;
  • 6) 0.535 644 531 249 999 888 977 743 04 × 2 = 1 + 0.071 289 062 499 999 777 955 486 08;
  • 7) 0.071 289 062 499 999 777 955 486 08 × 2 = 0 + 0.142 578 124 999 999 555 910 972 16;
  • 8) 0.142 578 124 999 999 555 910 972 16 × 2 = 0 + 0.285 156 249 999 999 111 821 944 32;
  • 9) 0.285 156 249 999 999 111 821 944 32 × 2 = 0 + 0.570 312 499 999 998 223 643 888 64;
  • 10) 0.570 312 499 999 998 223 643 888 64 × 2 = 1 + 0.140 624 999 999 996 447 287 777 28;
  • 11) 0.140 624 999 999 996 447 287 777 28 × 2 = 0 + 0.281 249 999 999 992 894 575 554 56;
  • 12) 0.281 249 999 999 992 894 575 554 56 × 2 = 0 + 0.562 499 999 999 985 789 151 109 12;
  • 13) 0.562 499 999 999 985 789 151 109 12 × 2 = 1 + 0.124 999 999 999 971 578 302 218 24;
  • 14) 0.124 999 999 999 971 578 302 218 24 × 2 = 0 + 0.249 999 999 999 943 156 604 436 48;
  • 15) 0.249 999 999 999 943 156 604 436 48 × 2 = 0 + 0.499 999 999 999 886 313 208 872 96;
  • 16) 0.499 999 999 999 886 313 208 872 96 × 2 = 0 + 0.999 999 999 999 772 626 417 745 92;
  • 17) 0.999 999 999 999 772 626 417 745 92 × 2 = 1 + 0.999 999 999 999 545 252 835 491 84;
  • 18) 0.999 999 999 999 545 252 835 491 84 × 2 = 1 + 0.999 999 999 999 090 505 670 983 68;
  • 19) 0.999 999 999 999 090 505 670 983 68 × 2 = 1 + 0.999 999 999 998 181 011 341 967 36;
  • 20) 0.999 999 999 998 181 011 341 967 36 × 2 = 1 + 0.999 999 999 996 362 022 683 934 72;
  • 21) 0.999 999 999 996 362 022 683 934 72 × 2 = 1 + 0.999 999 999 992 724 045 367 869 44;
  • 22) 0.999 999 999 992 724 045 367 869 44 × 2 = 1 + 0.999 999 999 985 448 090 735 738 88;
  • 23) 0.999 999 999 985 448 090 735 738 88 × 2 = 1 + 0.999 999 999 970 896 181 471 477 76;
  • 24) 0.999 999 999 970 896 181 471 477 76 × 2 = 1 + 0.999 999 999 941 792 362 942 955 52;
  • 25) 0.999 999 999 941 792 362 942 955 52 × 2 = 1 + 0.999 999 999 883 584 725 885 911 04;
  • 26) 0.999 999 999 883 584 725 885 911 04 × 2 = 1 + 0.999 999 999 767 169 451 771 822 08;
  • 27) 0.999 999 999 767 169 451 771 822 08 × 2 = 1 + 0.999 999 999 534 338 903 543 644 16;
  • 28) 0.999 999 999 534 338 903 543 644 16 × 2 = 1 + 0.999 999 999 068 677 807 087 288 32;
  • 29) 0.999 999 999 068 677 807 087 288 32 × 2 = 1 + 0.999 999 998 137 355 614 174 576 64;
  • 30) 0.999 999 998 137 355 614 174 576 64 × 2 = 1 + 0.999 999 996 274 711 228 349 153 28;
  • 31) 0.999 999 996 274 711 228 349 153 28 × 2 = 1 + 0.999 999 992 549 422 456 698 306 56;
  • 32) 0.999 999 992 549 422 456 698 306 56 × 2 = 1 + 0.999 999 985 098 844 913 396 613 12;
  • 33) 0.999 999 985 098 844 913 396 613 12 × 2 = 1 + 0.999 999 970 197 689 826 793 226 24;
  • 34) 0.999 999 970 197 689 826 793 226 24 × 2 = 1 + 0.999 999 940 395 379 653 586 452 48;
  • 35) 0.999 999 940 395 379 653 586 452 48 × 2 = 1 + 0.999 999 880 790 759 307 172 904 96;
  • 36) 0.999 999 880 790 759 307 172 904 96 × 2 = 1 + 0.999 999 761 581 518 614 345 809 92;
  • 37) 0.999 999 761 581 518 614 345 809 92 × 2 = 1 + 0.999 999 523 163 037 228 691 619 84;
  • 38) 0.999 999 523 163 037 228 691 619 84 × 2 = 1 + 0.999 999 046 326 074 457 383 239 68;
  • 39) 0.999 999 046 326 074 457 383 239 68 × 2 = 1 + 0.999 998 092 652 148 914 766 479 36;
  • 40) 0.999 998 092 652 148 914 766 479 36 × 2 = 1 + 0.999 996 185 304 297 829 532 958 72;
  • 41) 0.999 996 185 304 297 829 532 958 72 × 2 = 1 + 0.999 992 370 608 595 659 065 917 44;
  • 42) 0.999 992 370 608 595 659 065 917 44 × 2 = 1 + 0.999 984 741 217 191 318 131 834 88;
  • 43) 0.999 984 741 217 191 318 131 834 88 × 2 = 1 + 0.999 969 482 434 382 636 263 669 76;
  • 44) 0.999 969 482 434 382 636 263 669 76 × 2 = 1 + 0.999 938 964 868 765 272 527 339 52;
  • 45) 0.999 938 964 868 765 272 527 339 52 × 2 = 1 + 0.999 877 929 737 530 545 054 679 04;
  • 46) 0.999 877 929 737 530 545 054 679 04 × 2 = 1 + 0.999 755 859 475 061 090 109 358 08;
  • 47) 0.999 755 859 475 061 090 109 358 08 × 2 = 1 + 0.999 511 718 950 122 180 218 716 16;
  • 48) 0.999 511 718 950 122 180 218 716 16 × 2 = 1 + 0.999 023 437 900 244 360 437 432 32;
  • 49) 0.999 023 437 900 244 360 437 432 32 × 2 = 1 + 0.998 046 875 800 488 720 874 864 64;
  • 50) 0.998 046 875 800 488 720 874 864 64 × 2 = 1 + 0.996 093 751 600 977 441 749 729 28;
  • 51) 0.996 093 751 600 977 441 749 729 28 × 2 = 1 + 0.992 187 503 201 954 883 499 458 56;
  • 52) 0.992 187 503 201 954 883 499 458 56 × 2 = 1 + 0.984 375 006 403 909 766 998 917 12;
  • 53) 0.984 375 006 403 909 766 998 917 12 × 2 = 1 + 0.968 750 012 807 819 533 997 834 24;
  • 54) 0.968 750 012 807 819 533 997 834 24 × 2 = 1 + 0.937 500 025 615 639 067 995 668 48;
  • 55) 0.937 500 025 615 639 067 995 668 48 × 2 = 1 + 0.875 000 051 231 278 135 991 336 96;
  • 56) 0.875 000 051 231 278 135 991 336 96 × 2 = 1 + 0.750 000 102 462 556 271 982 673 92;
  • 57) 0.750 000 102 462 556 271 982 673 92 × 2 = 1 + 0.500 000 204 925 112 543 965 347 84;
  • 58) 0.500 000 204 925 112 543 965 347 84 × 2 = 1 + 0.000 000 409 850 225 087 930 695 68;

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:


0.016 738 891 601 562 496 530 554 47(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 554 47(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:


0.016 738 891 601 562 496 530 554 47(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 -0.016 738 891 601 562 496 530 554 47 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