-480.529 599 999 999 959 436 665 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -480.529 599 999 999 959 436 665(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
-480.529 599 999 999 959 436 665(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:

|-480.529 599 999 999 959 436 665| = 480.529 599 999 999 959 436 665


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: 0.529 599 999 999 959 436 665.

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) 0.529 599 999 999 959 436 665 × 2 = 1 + 0.059 199 999 999 918 873 33;
  • 2) 0.059 199 999 999 918 873 33 × 2 = 0 + 0.118 399 999 999 837 746 66;
  • 3) 0.118 399 999 999 837 746 66 × 2 = 0 + 0.236 799 999 999 675 493 32;
  • 4) 0.236 799 999 999 675 493 32 × 2 = 0 + 0.473 599 999 999 350 986 64;
  • 5) 0.473 599 999 999 350 986 64 × 2 = 0 + 0.947 199 999 998 701 973 28;
  • 6) 0.947 199 999 998 701 973 28 × 2 = 1 + 0.894 399 999 997 403 946 56;
  • 7) 0.894 399 999 997 403 946 56 × 2 = 1 + 0.788 799 999 994 807 893 12;
  • 8) 0.788 799 999 994 807 893 12 × 2 = 1 + 0.577 599 999 989 615 786 24;
  • 9) 0.577 599 999 989 615 786 24 × 2 = 1 + 0.155 199 999 979 231 572 48;
  • 10) 0.155 199 999 979 231 572 48 × 2 = 0 + 0.310 399 999 958 463 144 96;
  • 11) 0.310 399 999 958 463 144 96 × 2 = 0 + 0.620 799 999 916 926 289 92;
  • 12) 0.620 799 999 916 926 289 92 × 2 = 1 + 0.241 599 999 833 852 579 84;
  • 13) 0.241 599 999 833 852 579 84 × 2 = 0 + 0.483 199 999 667 705 159 68;
  • 14) 0.483 199 999 667 705 159 68 × 2 = 0 + 0.966 399 999 335 410 319 36;
  • 15) 0.966 399 999 335 410 319 36 × 2 = 1 + 0.932 799 998 670 820 638 72;
  • 16) 0.932 799 998 670 820 638 72 × 2 = 1 + 0.865 599 997 341 641 277 44;
  • 17) 0.865 599 997 341 641 277 44 × 2 = 1 + 0.731 199 994 683 282 554 88;
  • 18) 0.731 199 994 683 282 554 88 × 2 = 1 + 0.462 399 989 366 565 109 76;
  • 19) 0.462 399 989 366 565 109 76 × 2 = 0 + 0.924 799 978 733 130 219 52;
  • 20) 0.924 799 978 733 130 219 52 × 2 = 1 + 0.849 599 957 466 260 439 04;
  • 21) 0.849 599 957 466 260 439 04 × 2 = 1 + 0.699 199 914 932 520 878 08;
  • 22) 0.699 199 914 932 520 878 08 × 2 = 1 + 0.398 399 829 865 041 756 16;
  • 23) 0.398 399 829 865 041 756 16 × 2 = 0 + 0.796 799 659 730 083 512 32;
  • 24) 0.796 799 659 730 083 512 32 × 2 = 1 + 0.593 599 319 460 167 024 64;
  • 25) 0.593 599 319 460 167 024 64 × 2 = 1 + 0.187 198 638 920 334 049 28;
  • 26) 0.187 198 638 920 334 049 28 × 2 = 0 + 0.374 397 277 840 668 098 56;
  • 27) 0.374 397 277 840 668 098 56 × 2 = 0 + 0.748 794 555 681 336 197 12;
  • 28) 0.748 794 555 681 336 197 12 × 2 = 1 + 0.497 589 111 362 672 394 24;
  • 29) 0.497 589 111 362 672 394 24 × 2 = 0 + 0.995 178 222 725 344 788 48;
  • 30) 0.995 178 222 725 344 788 48 × 2 = 1 + 0.990 356 445 450 689 576 96;
  • 31) 0.990 356 445 450 689 576 96 × 2 = 1 + 0.980 712 890 901 379 153 92;
  • 32) 0.980 712 890 901 379 153 92 × 2 = 1 + 0.961 425 781 802 758 307 84;
  • 33) 0.961 425 781 802 758 307 84 × 2 = 1 + 0.922 851 563 605 516 615 68;
  • 34) 0.922 851 563 605 516 615 68 × 2 = 1 + 0.845 703 127 211 033 231 36;
  • 35) 0.845 703 127 211 033 231 36 × 2 = 1 + 0.691 406 254 422 066 462 72;
  • 36) 0.691 406 254 422 066 462 72 × 2 = 1 + 0.382 812 508 844 132 925 44;
  • 37) 0.382 812 508 844 132 925 44 × 2 = 0 + 0.765 625 017 688 265 850 88;
  • 38) 0.765 625 017 688 265 850 88 × 2 = 1 + 0.531 250 035 376 531 701 76;
  • 39) 0.531 250 035 376 531 701 76 × 2 = 1 + 0.062 500 070 753 063 403 52;
  • 40) 0.062 500 070 753 063 403 52 × 2 = 0 + 0.125 000 141 506 126 807 04;
  • 41) 0.125 000 141 506 126 807 04 × 2 = 0 + 0.250 000 283 012 253 614 08;
  • 42) 0.250 000 283 012 253 614 08 × 2 = 0 + 0.500 000 566 024 507 228 16;
  • 43) 0.500 000 566 024 507 228 16 × 2 = 1 + 0.000 001 132 049 014 456 32;
  • 44) 0.000 001 132 049 014 456 32 × 2 = 0 + 0.000 002 264 098 028 912 64;
  • 45) 0.000 002 264 098 028 912 64 × 2 = 0 + 0.000 004 528 196 057 825 28;
  • 46) 0.000 004 528 196 057 825 28 × 2 = 0 + 0.000 009 056 392 115 650 56;
  • 47) 0.000 009 056 392 115 650 56 × 2 = 0 + 0.000 018 112 784 231 301 12;
  • 48) 0.000 018 112 784 231 301 12 × 2 = 0 + 0.000 036 225 568 462 602 24;
  • 49) 0.000 036 225 568 462 602 24 × 2 = 0 + 0.000 072 451 136 925 204 48;
  • 50) 0.000 072 451 136 925 204 48 × 2 = 0 + 0.000 144 902 273 850 408 96;
  • 51) 0.000 144 902 273 850 408 96 × 2 = 0 + 0.000 289 804 547 700 817 92;
  • 52) 0.000 289 804 547 700 817 92 × 2 = 0 + 0.000 579 609 095 401 635 84;
  • 53) 0.000 579 609 095 401 635 84 × 2 = 0 + 0.001 159 218 190 803 271 68;

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:


0.529 599 999 999 959 436 665(10) =

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

6. Positive number before normalization:

480.529 599 999 999 959 436 665(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:


480.529 599 999 999 959 436 665(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 -480.529 599 999 999 959 436 665 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 -480.529 599 999 999 959 436 713 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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