-0.016 738 891 601 562 496 530 548 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 548 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
-0.016 738 891 601 562 496 530 548 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:

|-0.016 738 891 601 562 496 530 548 6| = 0.016 738 891 601 562 496 530 548 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: 0.016 738 891 601 562 496 530 548 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) 0.016 738 891 601 562 496 530 548 6 × 2 = 0 + 0.033 477 783 203 124 993 061 097 2;
  • 2) 0.033 477 783 203 124 993 061 097 2 × 2 = 0 + 0.066 955 566 406 249 986 122 194 4;
  • 3) 0.066 955 566 406 249 986 122 194 4 × 2 = 0 + 0.133 911 132 812 499 972 244 388 8;
  • 4) 0.133 911 132 812 499 972 244 388 8 × 2 = 0 + 0.267 822 265 624 999 944 488 777 6;
  • 5) 0.267 822 265 624 999 944 488 777 6 × 2 = 0 + 0.535 644 531 249 999 888 977 555 2;
  • 6) 0.535 644 531 249 999 888 977 555 2 × 2 = 1 + 0.071 289 062 499 999 777 955 110 4;
  • 7) 0.071 289 062 499 999 777 955 110 4 × 2 = 0 + 0.142 578 124 999 999 555 910 220 8;
  • 8) 0.142 578 124 999 999 555 910 220 8 × 2 = 0 + 0.285 156 249 999 999 111 820 441 6;
  • 9) 0.285 156 249 999 999 111 820 441 6 × 2 = 0 + 0.570 312 499 999 998 223 640 883 2;
  • 10) 0.570 312 499 999 998 223 640 883 2 × 2 = 1 + 0.140 624 999 999 996 447 281 766 4;
  • 11) 0.140 624 999 999 996 447 281 766 4 × 2 = 0 + 0.281 249 999 999 992 894 563 532 8;
  • 12) 0.281 249 999 999 992 894 563 532 8 × 2 = 0 + 0.562 499 999 999 985 789 127 065 6;
  • 13) 0.562 499 999 999 985 789 127 065 6 × 2 = 1 + 0.124 999 999 999 971 578 254 131 2;
  • 14) 0.124 999 999 999 971 578 254 131 2 × 2 = 0 + 0.249 999 999 999 943 156 508 262 4;
  • 15) 0.249 999 999 999 943 156 508 262 4 × 2 = 0 + 0.499 999 999 999 886 313 016 524 8;
  • 16) 0.499 999 999 999 886 313 016 524 8 × 2 = 0 + 0.999 999 999 999 772 626 033 049 6;
  • 17) 0.999 999 999 999 772 626 033 049 6 × 2 = 1 + 0.999 999 999 999 545 252 066 099 2;
  • 18) 0.999 999 999 999 545 252 066 099 2 × 2 = 1 + 0.999 999 999 999 090 504 132 198 4;
  • 19) 0.999 999 999 999 090 504 132 198 4 × 2 = 1 + 0.999 999 999 998 181 008 264 396 8;
  • 20) 0.999 999 999 998 181 008 264 396 8 × 2 = 1 + 0.999 999 999 996 362 016 528 793 6;
  • 21) 0.999 999 999 996 362 016 528 793 6 × 2 = 1 + 0.999 999 999 992 724 033 057 587 2;
  • 22) 0.999 999 999 992 724 033 057 587 2 × 2 = 1 + 0.999 999 999 985 448 066 115 174 4;
  • 23) 0.999 999 999 985 448 066 115 174 4 × 2 = 1 + 0.999 999 999 970 896 132 230 348 8;
  • 24) 0.999 999 999 970 896 132 230 348 8 × 2 = 1 + 0.999 999 999 941 792 264 460 697 6;
  • 25) 0.999 999 999 941 792 264 460 697 6 × 2 = 1 + 0.999 999 999 883 584 528 921 395 2;
  • 26) 0.999 999 999 883 584 528 921 395 2 × 2 = 1 + 0.999 999 999 767 169 057 842 790 4;
  • 27) 0.999 999 999 767 169 057 842 790 4 × 2 = 1 + 0.999 999 999 534 338 115 685 580 8;
  • 28) 0.999 999 999 534 338 115 685 580 8 × 2 = 1 + 0.999 999 999 068 676 231 371 161 6;
  • 29) 0.999 999 999 068 676 231 371 161 6 × 2 = 1 + 0.999 999 998 137 352 462 742 323 2;
  • 30) 0.999 999 998 137 352 462 742 323 2 × 2 = 1 + 0.999 999 996 274 704 925 484 646 4;
  • 31) 0.999 999 996 274 704 925 484 646 4 × 2 = 1 + 0.999 999 992 549 409 850 969 292 8;
  • 32) 0.999 999 992 549 409 850 969 292 8 × 2 = 1 + 0.999 999 985 098 819 701 938 585 6;
  • 33) 0.999 999 985 098 819 701 938 585 6 × 2 = 1 + 0.999 999 970 197 639 403 877 171 2;
  • 34) 0.999 999 970 197 639 403 877 171 2 × 2 = 1 + 0.999 999 940 395 278 807 754 342 4;
  • 35) 0.999 999 940 395 278 807 754 342 4 × 2 = 1 + 0.999 999 880 790 557 615 508 684 8;
  • 36) 0.999 999 880 790 557 615 508 684 8 × 2 = 1 + 0.999 999 761 581 115 231 017 369 6;
  • 37) 0.999 999 761 581 115 231 017 369 6 × 2 = 1 + 0.999 999 523 162 230 462 034 739 2;
  • 38) 0.999 999 523 162 230 462 034 739 2 × 2 = 1 + 0.999 999 046 324 460 924 069 478 4;
  • 39) 0.999 999 046 324 460 924 069 478 4 × 2 = 1 + 0.999 998 092 648 921 848 138 956 8;
  • 40) 0.999 998 092 648 921 848 138 956 8 × 2 = 1 + 0.999 996 185 297 843 696 277 913 6;
  • 41) 0.999 996 185 297 843 696 277 913 6 × 2 = 1 + 0.999 992 370 595 687 392 555 827 2;
  • 42) 0.999 992 370 595 687 392 555 827 2 × 2 = 1 + 0.999 984 741 191 374 785 111 654 4;
  • 43) 0.999 984 741 191 374 785 111 654 4 × 2 = 1 + 0.999 969 482 382 749 570 223 308 8;
  • 44) 0.999 969 482 382 749 570 223 308 8 × 2 = 1 + 0.999 938 964 765 499 140 446 617 6;
  • 45) 0.999 938 964 765 499 140 446 617 6 × 2 = 1 + 0.999 877 929 530 998 280 893 235 2;
  • 46) 0.999 877 929 530 998 280 893 235 2 × 2 = 1 + 0.999 755 859 061 996 561 786 470 4;
  • 47) 0.999 755 859 061 996 561 786 470 4 × 2 = 1 + 0.999 511 718 123 993 123 572 940 8;
  • 48) 0.999 511 718 123 993 123 572 940 8 × 2 = 1 + 0.999 023 436 247 986 247 145 881 6;
  • 49) 0.999 023 436 247 986 247 145 881 6 × 2 = 1 + 0.998 046 872 495 972 494 291 763 2;
  • 50) 0.998 046 872 495 972 494 291 763 2 × 2 = 1 + 0.996 093 744 991 944 988 583 526 4;
  • 51) 0.996 093 744 991 944 988 583 526 4 × 2 = 1 + 0.992 187 489 983 889 977 167 052 8;
  • 52) 0.992 187 489 983 889 977 167 052 8 × 2 = 1 + 0.984 374 979 967 779 954 334 105 6;
  • 53) 0.984 374 979 967 779 954 334 105 6 × 2 = 1 + 0.968 749 959 935 559 908 668 211 2;
  • 54) 0.968 749 959 935 559 908 668 211 2 × 2 = 1 + 0.937 499 919 871 119 817 336 422 4;
  • 55) 0.937 499 919 871 119 817 336 422 4 × 2 = 1 + 0.874 999 839 742 239 634 672 844 8;
  • 56) 0.874 999 839 742 239 634 672 844 8 × 2 = 1 + 0.749 999 679 484 479 269 345 689 6;
  • 57) 0.749 999 679 484 479 269 345 689 6 × 2 = 1 + 0.499 999 358 968 958 538 691 379 2;
  • 58) 0.499 999 358 968 958 538 691 379 2 × 2 = 0 + 0.999 998 717 937 917 077 382 758 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:


0.016 738 891 601 562 496 530 548 6(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 548 6(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:


0.016 738 891 601 562 496 530 548 6(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 -0.016 738 891 601 562 496 530 548 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 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