-0.016 738 891 601 562 496 530 555 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 555 41(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
-0.016 738 891 601 562 496 530 555 41(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:

|-0.016 738 891 601 562 496 530 555 41| = 0.016 738 891 601 562 496 530 555 41


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: 0.016 738 891 601 562 496 530 555 41.

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) 0.016 738 891 601 562 496 530 555 41 × 2 = 0 + 0.033 477 783 203 124 993 061 110 82;
  • 2) 0.033 477 783 203 124 993 061 110 82 × 2 = 0 + 0.066 955 566 406 249 986 122 221 64;
  • 3) 0.066 955 566 406 249 986 122 221 64 × 2 = 0 + 0.133 911 132 812 499 972 244 443 28;
  • 4) 0.133 911 132 812 499 972 244 443 28 × 2 = 0 + 0.267 822 265 624 999 944 488 886 56;
  • 5) 0.267 822 265 624 999 944 488 886 56 × 2 = 0 + 0.535 644 531 249 999 888 977 773 12;
  • 6) 0.535 644 531 249 999 888 977 773 12 × 2 = 1 + 0.071 289 062 499 999 777 955 546 24;
  • 7) 0.071 289 062 499 999 777 955 546 24 × 2 = 0 + 0.142 578 124 999 999 555 911 092 48;
  • 8) 0.142 578 124 999 999 555 911 092 48 × 2 = 0 + 0.285 156 249 999 999 111 822 184 96;
  • 9) 0.285 156 249 999 999 111 822 184 96 × 2 = 0 + 0.570 312 499 999 998 223 644 369 92;
  • 10) 0.570 312 499 999 998 223 644 369 92 × 2 = 1 + 0.140 624 999 999 996 447 288 739 84;
  • 11) 0.140 624 999 999 996 447 288 739 84 × 2 = 0 + 0.281 249 999 999 992 894 577 479 68;
  • 12) 0.281 249 999 999 992 894 577 479 68 × 2 = 0 + 0.562 499 999 999 985 789 154 959 36;
  • 13) 0.562 499 999 999 985 789 154 959 36 × 2 = 1 + 0.124 999 999 999 971 578 309 918 72;
  • 14) 0.124 999 999 999 971 578 309 918 72 × 2 = 0 + 0.249 999 999 999 943 156 619 837 44;
  • 15) 0.249 999 999 999 943 156 619 837 44 × 2 = 0 + 0.499 999 999 999 886 313 239 674 88;
  • 16) 0.499 999 999 999 886 313 239 674 88 × 2 = 0 + 0.999 999 999 999 772 626 479 349 76;
  • 17) 0.999 999 999 999 772 626 479 349 76 × 2 = 1 + 0.999 999 999 999 545 252 958 699 52;
  • 18) 0.999 999 999 999 545 252 958 699 52 × 2 = 1 + 0.999 999 999 999 090 505 917 399 04;
  • 19) 0.999 999 999 999 090 505 917 399 04 × 2 = 1 + 0.999 999 999 998 181 011 834 798 08;
  • 20) 0.999 999 999 998 181 011 834 798 08 × 2 = 1 + 0.999 999 999 996 362 023 669 596 16;
  • 21) 0.999 999 999 996 362 023 669 596 16 × 2 = 1 + 0.999 999 999 992 724 047 339 192 32;
  • 22) 0.999 999 999 992 724 047 339 192 32 × 2 = 1 + 0.999 999 999 985 448 094 678 384 64;
  • 23) 0.999 999 999 985 448 094 678 384 64 × 2 = 1 + 0.999 999 999 970 896 189 356 769 28;
  • 24) 0.999 999 999 970 896 189 356 769 28 × 2 = 1 + 0.999 999 999 941 792 378 713 538 56;
  • 25) 0.999 999 999 941 792 378 713 538 56 × 2 = 1 + 0.999 999 999 883 584 757 427 077 12;
  • 26) 0.999 999 999 883 584 757 427 077 12 × 2 = 1 + 0.999 999 999 767 169 514 854 154 24;
  • 27) 0.999 999 999 767 169 514 854 154 24 × 2 = 1 + 0.999 999 999 534 339 029 708 308 48;
  • 28) 0.999 999 999 534 339 029 708 308 48 × 2 = 1 + 0.999 999 999 068 678 059 416 616 96;
  • 29) 0.999 999 999 068 678 059 416 616 96 × 2 = 1 + 0.999 999 998 137 356 118 833 233 92;
  • 30) 0.999 999 998 137 356 118 833 233 92 × 2 = 1 + 0.999 999 996 274 712 237 666 467 84;
  • 31) 0.999 999 996 274 712 237 666 467 84 × 2 = 1 + 0.999 999 992 549 424 475 332 935 68;
  • 32) 0.999 999 992 549 424 475 332 935 68 × 2 = 1 + 0.999 999 985 098 848 950 665 871 36;
  • 33) 0.999 999 985 098 848 950 665 871 36 × 2 = 1 + 0.999 999 970 197 697 901 331 742 72;
  • 34) 0.999 999 970 197 697 901 331 742 72 × 2 = 1 + 0.999 999 940 395 395 802 663 485 44;
  • 35) 0.999 999 940 395 395 802 663 485 44 × 2 = 1 + 0.999 999 880 790 791 605 326 970 88;
  • 36) 0.999 999 880 790 791 605 326 970 88 × 2 = 1 + 0.999 999 761 581 583 210 653 941 76;
  • 37) 0.999 999 761 581 583 210 653 941 76 × 2 = 1 + 0.999 999 523 163 166 421 307 883 52;
  • 38) 0.999 999 523 163 166 421 307 883 52 × 2 = 1 + 0.999 999 046 326 332 842 615 767 04;
  • 39) 0.999 999 046 326 332 842 615 767 04 × 2 = 1 + 0.999 998 092 652 665 685 231 534 08;
  • 40) 0.999 998 092 652 665 685 231 534 08 × 2 = 1 + 0.999 996 185 305 331 370 463 068 16;
  • 41) 0.999 996 185 305 331 370 463 068 16 × 2 = 1 + 0.999 992 370 610 662 740 926 136 32;
  • 42) 0.999 992 370 610 662 740 926 136 32 × 2 = 1 + 0.999 984 741 221 325 481 852 272 64;
  • 43) 0.999 984 741 221 325 481 852 272 64 × 2 = 1 + 0.999 969 482 442 650 963 704 545 28;
  • 44) 0.999 969 482 442 650 963 704 545 28 × 2 = 1 + 0.999 938 964 885 301 927 409 090 56;
  • 45) 0.999 938 964 885 301 927 409 090 56 × 2 = 1 + 0.999 877 929 770 603 854 818 181 12;
  • 46) 0.999 877 929 770 603 854 818 181 12 × 2 = 1 + 0.999 755 859 541 207 709 636 362 24;
  • 47) 0.999 755 859 541 207 709 636 362 24 × 2 = 1 + 0.999 511 719 082 415 419 272 724 48;
  • 48) 0.999 511 719 082 415 419 272 724 48 × 2 = 1 + 0.999 023 438 164 830 838 545 448 96;
  • 49) 0.999 023 438 164 830 838 545 448 96 × 2 = 1 + 0.998 046 876 329 661 677 090 897 92;
  • 50) 0.998 046 876 329 661 677 090 897 92 × 2 = 1 + 0.996 093 752 659 323 354 181 795 84;
  • 51) 0.996 093 752 659 323 354 181 795 84 × 2 = 1 + 0.992 187 505 318 646 708 363 591 68;
  • 52) 0.992 187 505 318 646 708 363 591 68 × 2 = 1 + 0.984 375 010 637 293 416 727 183 36;
  • 53) 0.984 375 010 637 293 416 727 183 36 × 2 = 1 + 0.968 750 021 274 586 833 454 366 72;
  • 54) 0.968 750 021 274 586 833 454 366 72 × 2 = 1 + 0.937 500 042 549 173 666 908 733 44;
  • 55) 0.937 500 042 549 173 666 908 733 44 × 2 = 1 + 0.875 000 085 098 347 333 817 466 88;
  • 56) 0.875 000 085 098 347 333 817 466 88 × 2 = 1 + 0.750 000 170 196 694 667 634 933 76;
  • 57) 0.750 000 170 196 694 667 634 933 76 × 2 = 1 + 0.500 000 340 393 389 335 269 867 52;
  • 58) 0.500 000 340 393 389 335 269 867 52 × 2 = 1 + 0.000 000 680 786 778 670 539 735 04;

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:


0.016 738 891 601 562 496 530 555 41(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 555 41(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:


0.016 738 891 601 562 496 530 555 41(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 -0.016 738 891 601 562 496 530 555 41 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