-0.000 640 716 485 543 503 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 640 716 485 543 503(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
-0.000 640 716 485 543 503(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:

|-0.000 640 716 485 543 503| = 0.000 640 716 485 543 503


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: 0.000 640 716 485 543 503.

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) 0.000 640 716 485 543 503 × 2 = 0 + 0.001 281 432 971 087 006;
  • 2) 0.001 281 432 971 087 006 × 2 = 0 + 0.002 562 865 942 174 012;
  • 3) 0.002 562 865 942 174 012 × 2 = 0 + 0.005 125 731 884 348 024;
  • 4) 0.005 125 731 884 348 024 × 2 = 0 + 0.010 251 463 768 696 048;
  • 5) 0.010 251 463 768 696 048 × 2 = 0 + 0.020 502 927 537 392 096;
  • 6) 0.020 502 927 537 392 096 × 2 = 0 + 0.041 005 855 074 784 192;
  • 7) 0.041 005 855 074 784 192 × 2 = 0 + 0.082 011 710 149 568 384;
  • 8) 0.082 011 710 149 568 384 × 2 = 0 + 0.164 023 420 299 136 768;
  • 9) 0.164 023 420 299 136 768 × 2 = 0 + 0.328 046 840 598 273 536;
  • 10) 0.328 046 840 598 273 536 × 2 = 0 + 0.656 093 681 196 547 072;
  • 11) 0.656 093 681 196 547 072 × 2 = 1 + 0.312 187 362 393 094 144;
  • 12) 0.312 187 362 393 094 144 × 2 = 0 + 0.624 374 724 786 188 288;
  • 13) 0.624 374 724 786 188 288 × 2 = 1 + 0.248 749 449 572 376 576;
  • 14) 0.248 749 449 572 376 576 × 2 = 0 + 0.497 498 899 144 753 152;
  • 15) 0.497 498 899 144 753 152 × 2 = 0 + 0.994 997 798 289 506 304;
  • 16) 0.994 997 798 289 506 304 × 2 = 1 + 0.989 995 596 579 012 608;
  • 17) 0.989 995 596 579 012 608 × 2 = 1 + 0.979 991 193 158 025 216;
  • 18) 0.979 991 193 158 025 216 × 2 = 1 + 0.959 982 386 316 050 432;
  • 19) 0.959 982 386 316 050 432 × 2 = 1 + 0.919 964 772 632 100 864;
  • 20) 0.919 964 772 632 100 864 × 2 = 1 + 0.839 929 545 264 201 728;
  • 21) 0.839 929 545 264 201 728 × 2 = 1 + 0.679 859 090 528 403 456;
  • 22) 0.679 859 090 528 403 456 × 2 = 1 + 0.359 718 181 056 806 912;
  • 23) 0.359 718 181 056 806 912 × 2 = 0 + 0.719 436 362 113 613 824;
  • 24) 0.719 436 362 113 613 824 × 2 = 1 + 0.438 872 724 227 227 648;
  • 25) 0.438 872 724 227 227 648 × 2 = 0 + 0.877 745 448 454 455 296;
  • 26) 0.877 745 448 454 455 296 × 2 = 1 + 0.755 490 896 908 910 592;
  • 27) 0.755 490 896 908 910 592 × 2 = 1 + 0.510 981 793 817 821 184;
  • 28) 0.510 981 793 817 821 184 × 2 = 1 + 0.021 963 587 635 642 368;
  • 29) 0.021 963 587 635 642 368 × 2 = 0 + 0.043 927 175 271 284 736;
  • 30) 0.043 927 175 271 284 736 × 2 = 0 + 0.087 854 350 542 569 472;
  • 31) 0.087 854 350 542 569 472 × 2 = 0 + 0.175 708 701 085 138 944;
  • 32) 0.175 708 701 085 138 944 × 2 = 0 + 0.351 417 402 170 277 888;
  • 33) 0.351 417 402 170 277 888 × 2 = 0 + 0.702 834 804 340 555 776;
  • 34) 0.702 834 804 340 555 776 × 2 = 1 + 0.405 669 608 681 111 552;
  • 35) 0.405 669 608 681 111 552 × 2 = 0 + 0.811 339 217 362 223 104;
  • 36) 0.811 339 217 362 223 104 × 2 = 1 + 0.622 678 434 724 446 208;
  • 37) 0.622 678 434 724 446 208 × 2 = 1 + 0.245 356 869 448 892 416;
  • 38) 0.245 356 869 448 892 416 × 2 = 0 + 0.490 713 738 897 784 832;
  • 39) 0.490 713 738 897 784 832 × 2 = 0 + 0.981 427 477 795 569 664;
  • 40) 0.981 427 477 795 569 664 × 2 = 1 + 0.962 854 955 591 139 328;
  • 41) 0.962 854 955 591 139 328 × 2 = 1 + 0.925 709 911 182 278 656;
  • 42) 0.925 709 911 182 278 656 × 2 = 1 + 0.851 419 822 364 557 312;
  • 43) 0.851 419 822 364 557 312 × 2 = 1 + 0.702 839 644 729 114 624;
  • 44) 0.702 839 644 729 114 624 × 2 = 1 + 0.405 679 289 458 229 248;
  • 45) 0.405 679 289 458 229 248 × 2 = 0 + 0.811 358 578 916 458 496;
  • 46) 0.811 358 578 916 458 496 × 2 = 1 + 0.622 717 157 832 916 992;
  • 47) 0.622 717 157 832 916 992 × 2 = 1 + 0.245 434 315 665 833 984;
  • 48) 0.245 434 315 665 833 984 × 2 = 0 + 0.490 868 631 331 667 968;
  • 49) 0.490 868 631 331 667 968 × 2 = 0 + 0.981 737 262 663 335 936;
  • 50) 0.981 737 262 663 335 936 × 2 = 1 + 0.963 474 525 326 671 872;
  • 51) 0.963 474 525 326 671 872 × 2 = 1 + 0.926 949 050 653 343 744;
  • 52) 0.926 949 050 653 343 744 × 2 = 1 + 0.853 898 101 306 687 488;
  • 53) 0.853 898 101 306 687 488 × 2 = 1 + 0.707 796 202 613 374 976;
  • 54) 0.707 796 202 613 374 976 × 2 = 1 + 0.415 592 405 226 749 952;
  • 55) 0.415 592 405 226 749 952 × 2 = 0 + 0.831 184 810 453 499 904;
  • 56) 0.831 184 810 453 499 904 × 2 = 1 + 0.662 369 620 906 999 808;
  • 57) 0.662 369 620 906 999 808 × 2 = 1 + 0.324 739 241 813 999 616;
  • 58) 0.324 739 241 813 999 616 × 2 = 0 + 0.649 478 483 627 999 232;
  • 59) 0.649 478 483 627 999 232 × 2 = 1 + 0.298 956 967 255 998 464;
  • 60) 0.298 956 967 255 998 464 × 2 = 0 + 0.597 913 934 511 996 928;
  • 61) 0.597 913 934 511 996 928 × 2 = 1 + 0.195 827 869 023 993 856;
  • 62) 0.195 827 869 023 993 856 × 2 = 0 + 0.391 655 738 047 987 712;
  • 63) 0.391 655 738 047 987 712 × 2 = 0 + 0.783 311 476 095 975 424;

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:


0.000 640 716 485 543 503(10) =

0.0000 0000 0010 1001 1111 1101 0111 0000 0101 1001 1111 0110 0111 1101 1010 100(2)

6. Positive number before normalization:

0.000 640 716 485 543 503(10) =

0.0000 0000 0010 1001 1111 1101 0111 0000 0101 1001 1111 0110 0111 1101 1010 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 640 716 485 543 503(10) =

0.0000 0000 0010 1001 1111 1101 0111 0000 0101 1001 1111 0110 0111 1101 1010 100(2) =

0.0000 0000 0010 1001 1111 1101 0111 0000 0101 1001 1111 0110 0111 1101 1010 100(2) × 20 =

1.0100 1111 1110 1011 1000 0010 1100 1111 1011 0011 1110 1101 0100(2) × 2-11


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): -11

Mantissa (not normalized):
1.0100 1111 1110 1011 1000 0010 1100 1111 1011 0011 1110 1101 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-11 + 2(11-1) - 1 =

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1012(10) =

011 1111 0100(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. 0100 1111 1110 1011 1000 0010 1100 1111 1011 0011 1110 1101 0100 =

0100 1111 1110 1011 1000 0010 1100 1111 1011 0011 1110 1101 0100


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 0100

Mantissa (52 bits) =
0100 1111 1110 1011 1000 0010 1100 1111 1011 0011 1110 1101 0100


Decimal number -0.000 640 716 485 543 503 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0100 - 0100 1111 1110 1011 1000 0010 1100 1111 1011 0011 1110 1101 0100

» Convert decimal -0.000 640 716 485 543 488 to 64 bit double precision IEEE 754 binary floating point representation standard

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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