-480.529 599 999 999 959 436 713 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -480.529 599 999 999 959 436 713(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
-480.529 599 999 999 959 436 713(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:

|-480.529 599 999 999 959 436 713| = 480.529 599 999 999 959 436 713


2. First, convert to binary (in base 2) the integer part: 480.
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;
  • 480 ÷ 2 = 240 + 0;
  • 240 ÷ 2 = 120 + 0;
  • 120 ÷ 2 = 60 + 0;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

480(10) =

1 1110 0000(2)


4. Convert to binary (base 2) the fractional part: 0.529 599 999 999 959 436 713.

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) 0.529 599 999 999 959 436 713 × 2 = 1 + 0.059 199 999 999 918 873 426;
  • 2) 0.059 199 999 999 918 873 426 × 2 = 0 + 0.118 399 999 999 837 746 852;
  • 3) 0.118 399 999 999 837 746 852 × 2 = 0 + 0.236 799 999 999 675 493 704;
  • 4) 0.236 799 999 999 675 493 704 × 2 = 0 + 0.473 599 999 999 350 987 408;
  • 5) 0.473 599 999 999 350 987 408 × 2 = 0 + 0.947 199 999 998 701 974 816;
  • 6) 0.947 199 999 998 701 974 816 × 2 = 1 + 0.894 399 999 997 403 949 632;
  • 7) 0.894 399 999 997 403 949 632 × 2 = 1 + 0.788 799 999 994 807 899 264;
  • 8) 0.788 799 999 994 807 899 264 × 2 = 1 + 0.577 599 999 989 615 798 528;
  • 9) 0.577 599 999 989 615 798 528 × 2 = 1 + 0.155 199 999 979 231 597 056;
  • 10) 0.155 199 999 979 231 597 056 × 2 = 0 + 0.310 399 999 958 463 194 112;
  • 11) 0.310 399 999 958 463 194 112 × 2 = 0 + 0.620 799 999 916 926 388 224;
  • 12) 0.620 799 999 916 926 388 224 × 2 = 1 + 0.241 599 999 833 852 776 448;
  • 13) 0.241 599 999 833 852 776 448 × 2 = 0 + 0.483 199 999 667 705 552 896;
  • 14) 0.483 199 999 667 705 552 896 × 2 = 0 + 0.966 399 999 335 411 105 792;
  • 15) 0.966 399 999 335 411 105 792 × 2 = 1 + 0.932 799 998 670 822 211 584;
  • 16) 0.932 799 998 670 822 211 584 × 2 = 1 + 0.865 599 997 341 644 423 168;
  • 17) 0.865 599 997 341 644 423 168 × 2 = 1 + 0.731 199 994 683 288 846 336;
  • 18) 0.731 199 994 683 288 846 336 × 2 = 1 + 0.462 399 989 366 577 692 672;
  • 19) 0.462 399 989 366 577 692 672 × 2 = 0 + 0.924 799 978 733 155 385 344;
  • 20) 0.924 799 978 733 155 385 344 × 2 = 1 + 0.849 599 957 466 310 770 688;
  • 21) 0.849 599 957 466 310 770 688 × 2 = 1 + 0.699 199 914 932 621 541 376;
  • 22) 0.699 199 914 932 621 541 376 × 2 = 1 + 0.398 399 829 865 243 082 752;
  • 23) 0.398 399 829 865 243 082 752 × 2 = 0 + 0.796 799 659 730 486 165 504;
  • 24) 0.796 799 659 730 486 165 504 × 2 = 1 + 0.593 599 319 460 972 331 008;
  • 25) 0.593 599 319 460 972 331 008 × 2 = 1 + 0.187 198 638 921 944 662 016;
  • 26) 0.187 198 638 921 944 662 016 × 2 = 0 + 0.374 397 277 843 889 324 032;
  • 27) 0.374 397 277 843 889 324 032 × 2 = 0 + 0.748 794 555 687 778 648 064;
  • 28) 0.748 794 555 687 778 648 064 × 2 = 1 + 0.497 589 111 375 557 296 128;
  • 29) 0.497 589 111 375 557 296 128 × 2 = 0 + 0.995 178 222 751 114 592 256;
  • 30) 0.995 178 222 751 114 592 256 × 2 = 1 + 0.990 356 445 502 229 184 512;
  • 31) 0.990 356 445 502 229 184 512 × 2 = 1 + 0.980 712 891 004 458 369 024;
  • 32) 0.980 712 891 004 458 369 024 × 2 = 1 + 0.961 425 782 008 916 738 048;
  • 33) 0.961 425 782 008 916 738 048 × 2 = 1 + 0.922 851 564 017 833 476 096;
  • 34) 0.922 851 564 017 833 476 096 × 2 = 1 + 0.845 703 128 035 666 952 192;
  • 35) 0.845 703 128 035 666 952 192 × 2 = 1 + 0.691 406 256 071 333 904 384;
  • 36) 0.691 406 256 071 333 904 384 × 2 = 1 + 0.382 812 512 142 667 808 768;
  • 37) 0.382 812 512 142 667 808 768 × 2 = 0 + 0.765 625 024 285 335 617 536;
  • 38) 0.765 625 024 285 335 617 536 × 2 = 1 + 0.531 250 048 570 671 235 072;
  • 39) 0.531 250 048 570 671 235 072 × 2 = 1 + 0.062 500 097 141 342 470 144;
  • 40) 0.062 500 097 141 342 470 144 × 2 = 0 + 0.125 000 194 282 684 940 288;
  • 41) 0.125 000 194 282 684 940 288 × 2 = 0 + 0.250 000 388 565 369 880 576;
  • 42) 0.250 000 388 565 369 880 576 × 2 = 0 + 0.500 000 777 130 739 761 152;
  • 43) 0.500 000 777 130 739 761 152 × 2 = 1 + 0.000 001 554 261 479 522 304;
  • 44) 0.000 001 554 261 479 522 304 × 2 = 0 + 0.000 003 108 522 959 044 608;
  • 45) 0.000 003 108 522 959 044 608 × 2 = 0 + 0.000 006 217 045 918 089 216;
  • 46) 0.000 006 217 045 918 089 216 × 2 = 0 + 0.000 012 434 091 836 178 432;
  • 47) 0.000 012 434 091 836 178 432 × 2 = 0 + 0.000 024 868 183 672 356 864;
  • 48) 0.000 024 868 183 672 356 864 × 2 = 0 + 0.000 049 736 367 344 713 728;
  • 49) 0.000 049 736 367 344 713 728 × 2 = 0 + 0.000 099 472 734 689 427 456;
  • 50) 0.000 099 472 734 689 427 456 × 2 = 0 + 0.000 198 945 469 378 854 912;
  • 51) 0.000 198 945 469 378 854 912 × 2 = 0 + 0.000 397 890 938 757 709 824;
  • 52) 0.000 397 890 938 757 709 824 × 2 = 0 + 0.000 795 781 877 515 419 648;
  • 53) 0.000 795 781 877 515 419 648 × 2 = 0 + 0.001 591 563 755 030 839 296;

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:


0.529 599 999 999 959 436 713(10) =

0.1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0000 0000 0(2)

6. Positive number before normalization:

480.529 599 999 999 959 436 713(10) =

1 1110 0000.1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left, so that only one non zero digit remains to the left of it:


480.529 599 999 999 959 436 713(10) =

1 1110 0000.1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0000 0000 0(2) =

1 1110 0000.1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0000 0000 0(2) × 20 =

1.1110 0000 1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0000 0000 0(2) × 28


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): 8

Mantissa (not normalized):
1.1110 0000 1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0000 0000 0


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

8 + 2(11-1) - 1 =

(8 + 1 023)(10) =

1 031(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;
  • 1 031 ÷ 2 = 515 + 1;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1031(10) =

100 0000 0111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1110 0000 1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010 0 0000 0000 =

1110 0000 1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010


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) =
100 0000 0111

Mantissa (52 bits) =
1110 0000 1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010


Decimal number -480.529 599 999 999 959 436 713 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0111 - 1110 0000 1000 0111 1001 0011 1101 1101 1001 0111 1111 0110 0010

» Convert decimal -480.529 599 999 999 959 436 705 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