-0.016 738 891 601 562 496 530 549 43 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 549 43(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
-0.016 738 891 601 562 496 530 549 43(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:

|-0.016 738 891 601 562 496 530 549 43| = 0.016 738 891 601 562 496 530 549 43


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: 0.016 738 891 601 562 496 530 549 43.

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) 0.016 738 891 601 562 496 530 549 43 × 2 = 0 + 0.033 477 783 203 124 993 061 098 86;
  • 2) 0.033 477 783 203 124 993 061 098 86 × 2 = 0 + 0.066 955 566 406 249 986 122 197 72;
  • 3) 0.066 955 566 406 249 986 122 197 72 × 2 = 0 + 0.133 911 132 812 499 972 244 395 44;
  • 4) 0.133 911 132 812 499 972 244 395 44 × 2 = 0 + 0.267 822 265 624 999 944 488 790 88;
  • 5) 0.267 822 265 624 999 944 488 790 88 × 2 = 0 + 0.535 644 531 249 999 888 977 581 76;
  • 6) 0.535 644 531 249 999 888 977 581 76 × 2 = 1 + 0.071 289 062 499 999 777 955 163 52;
  • 7) 0.071 289 062 499 999 777 955 163 52 × 2 = 0 + 0.142 578 124 999 999 555 910 327 04;
  • 8) 0.142 578 124 999 999 555 910 327 04 × 2 = 0 + 0.285 156 249 999 999 111 820 654 08;
  • 9) 0.285 156 249 999 999 111 820 654 08 × 2 = 0 + 0.570 312 499 999 998 223 641 308 16;
  • 10) 0.570 312 499 999 998 223 641 308 16 × 2 = 1 + 0.140 624 999 999 996 447 282 616 32;
  • 11) 0.140 624 999 999 996 447 282 616 32 × 2 = 0 + 0.281 249 999 999 992 894 565 232 64;
  • 12) 0.281 249 999 999 992 894 565 232 64 × 2 = 0 + 0.562 499 999 999 985 789 130 465 28;
  • 13) 0.562 499 999 999 985 789 130 465 28 × 2 = 1 + 0.124 999 999 999 971 578 260 930 56;
  • 14) 0.124 999 999 999 971 578 260 930 56 × 2 = 0 + 0.249 999 999 999 943 156 521 861 12;
  • 15) 0.249 999 999 999 943 156 521 861 12 × 2 = 0 + 0.499 999 999 999 886 313 043 722 24;
  • 16) 0.499 999 999 999 886 313 043 722 24 × 2 = 0 + 0.999 999 999 999 772 626 087 444 48;
  • 17) 0.999 999 999 999 772 626 087 444 48 × 2 = 1 + 0.999 999 999 999 545 252 174 888 96;
  • 18) 0.999 999 999 999 545 252 174 888 96 × 2 = 1 + 0.999 999 999 999 090 504 349 777 92;
  • 19) 0.999 999 999 999 090 504 349 777 92 × 2 = 1 + 0.999 999 999 998 181 008 699 555 84;
  • 20) 0.999 999 999 998 181 008 699 555 84 × 2 = 1 + 0.999 999 999 996 362 017 399 111 68;
  • 21) 0.999 999 999 996 362 017 399 111 68 × 2 = 1 + 0.999 999 999 992 724 034 798 223 36;
  • 22) 0.999 999 999 992 724 034 798 223 36 × 2 = 1 + 0.999 999 999 985 448 069 596 446 72;
  • 23) 0.999 999 999 985 448 069 596 446 72 × 2 = 1 + 0.999 999 999 970 896 139 192 893 44;
  • 24) 0.999 999 999 970 896 139 192 893 44 × 2 = 1 + 0.999 999 999 941 792 278 385 786 88;
  • 25) 0.999 999 999 941 792 278 385 786 88 × 2 = 1 + 0.999 999 999 883 584 556 771 573 76;
  • 26) 0.999 999 999 883 584 556 771 573 76 × 2 = 1 + 0.999 999 999 767 169 113 543 147 52;
  • 27) 0.999 999 999 767 169 113 543 147 52 × 2 = 1 + 0.999 999 999 534 338 227 086 295 04;
  • 28) 0.999 999 999 534 338 227 086 295 04 × 2 = 1 + 0.999 999 999 068 676 454 172 590 08;
  • 29) 0.999 999 999 068 676 454 172 590 08 × 2 = 1 + 0.999 999 998 137 352 908 345 180 16;
  • 30) 0.999 999 998 137 352 908 345 180 16 × 2 = 1 + 0.999 999 996 274 705 816 690 360 32;
  • 31) 0.999 999 996 274 705 816 690 360 32 × 2 = 1 + 0.999 999 992 549 411 633 380 720 64;
  • 32) 0.999 999 992 549 411 633 380 720 64 × 2 = 1 + 0.999 999 985 098 823 266 761 441 28;
  • 33) 0.999 999 985 098 823 266 761 441 28 × 2 = 1 + 0.999 999 970 197 646 533 522 882 56;
  • 34) 0.999 999 970 197 646 533 522 882 56 × 2 = 1 + 0.999 999 940 395 293 067 045 765 12;
  • 35) 0.999 999 940 395 293 067 045 765 12 × 2 = 1 + 0.999 999 880 790 586 134 091 530 24;
  • 36) 0.999 999 880 790 586 134 091 530 24 × 2 = 1 + 0.999 999 761 581 172 268 183 060 48;
  • 37) 0.999 999 761 581 172 268 183 060 48 × 2 = 1 + 0.999 999 523 162 344 536 366 120 96;
  • 38) 0.999 999 523 162 344 536 366 120 96 × 2 = 1 + 0.999 999 046 324 689 072 732 241 92;
  • 39) 0.999 999 046 324 689 072 732 241 92 × 2 = 1 + 0.999 998 092 649 378 145 464 483 84;
  • 40) 0.999 998 092 649 378 145 464 483 84 × 2 = 1 + 0.999 996 185 298 756 290 928 967 68;
  • 41) 0.999 996 185 298 756 290 928 967 68 × 2 = 1 + 0.999 992 370 597 512 581 857 935 36;
  • 42) 0.999 992 370 597 512 581 857 935 36 × 2 = 1 + 0.999 984 741 195 025 163 715 870 72;
  • 43) 0.999 984 741 195 025 163 715 870 72 × 2 = 1 + 0.999 969 482 390 050 327 431 741 44;
  • 44) 0.999 969 482 390 050 327 431 741 44 × 2 = 1 + 0.999 938 964 780 100 654 863 482 88;
  • 45) 0.999 938 964 780 100 654 863 482 88 × 2 = 1 + 0.999 877 929 560 201 309 726 965 76;
  • 46) 0.999 877 929 560 201 309 726 965 76 × 2 = 1 + 0.999 755 859 120 402 619 453 931 52;
  • 47) 0.999 755 859 120 402 619 453 931 52 × 2 = 1 + 0.999 511 718 240 805 238 907 863 04;
  • 48) 0.999 511 718 240 805 238 907 863 04 × 2 = 1 + 0.999 023 436 481 610 477 815 726 08;
  • 49) 0.999 023 436 481 610 477 815 726 08 × 2 = 1 + 0.998 046 872 963 220 955 631 452 16;
  • 50) 0.998 046 872 963 220 955 631 452 16 × 2 = 1 + 0.996 093 745 926 441 911 262 904 32;
  • 51) 0.996 093 745 926 441 911 262 904 32 × 2 = 1 + 0.992 187 491 852 883 822 525 808 64;
  • 52) 0.992 187 491 852 883 822 525 808 64 × 2 = 1 + 0.984 374 983 705 767 645 051 617 28;
  • 53) 0.984 374 983 705 767 645 051 617 28 × 2 = 1 + 0.968 749 967 411 535 290 103 234 56;
  • 54) 0.968 749 967 411 535 290 103 234 56 × 2 = 1 + 0.937 499 934 823 070 580 206 469 12;
  • 55) 0.937 499 934 823 070 580 206 469 12 × 2 = 1 + 0.874 999 869 646 141 160 412 938 24;
  • 56) 0.874 999 869 646 141 160 412 938 24 × 2 = 1 + 0.749 999 739 292 282 320 825 876 48;
  • 57) 0.749 999 739 292 282 320 825 876 48 × 2 = 1 + 0.499 999 478 584 564 641 651 752 96;
  • 58) 0.499 999 478 584 564 641 651 752 96 × 2 = 0 + 0.999 998 957 169 129 283 303 505 92;

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:


0.016 738 891 601 562 496 530 549 43(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 549 43(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:


0.016 738 891 601 562 496 530 549 43(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 -0.016 738 891 601 562 496 530 549 43 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