Convert -4 459 335 333 705 705 605 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number -4 459 335 333 705 705 605(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-4 459 335 333 705 705 605| = 4 459 335 333 705 705 605

2. 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;
  • 4 459 335 333 705 705 605 ÷ 2 = 2 229 667 666 852 852 802 + 1;
  • 2 229 667 666 852 852 802 ÷ 2 = 1 114 833 833 426 426 401 + 0;
  • 1 114 833 833 426 426 401 ÷ 2 = 557 416 916 713 213 200 + 1;
  • 557 416 916 713 213 200 ÷ 2 = 278 708 458 356 606 600 + 0;
  • 278 708 458 356 606 600 ÷ 2 = 139 354 229 178 303 300 + 0;
  • 139 354 229 178 303 300 ÷ 2 = 69 677 114 589 151 650 + 0;
  • 69 677 114 589 151 650 ÷ 2 = 34 838 557 294 575 825 + 0;
  • 34 838 557 294 575 825 ÷ 2 = 17 419 278 647 287 912 + 1;
  • 17 419 278 647 287 912 ÷ 2 = 8 709 639 323 643 956 + 0;
  • 8 709 639 323 643 956 ÷ 2 = 4 354 819 661 821 978 + 0;
  • 4 354 819 661 821 978 ÷ 2 = 2 177 409 830 910 989 + 0;
  • 2 177 409 830 910 989 ÷ 2 = 1 088 704 915 455 494 + 1;
  • 1 088 704 915 455 494 ÷ 2 = 544 352 457 727 747 + 0;
  • 544 352 457 727 747 ÷ 2 = 272 176 228 863 873 + 1;
  • 272 176 228 863 873 ÷ 2 = 136 088 114 431 936 + 1;
  • 136 088 114 431 936 ÷ 2 = 68 044 057 215 968 + 0;
  • 68 044 057 215 968 ÷ 2 = 34 022 028 607 984 + 0;
  • 34 022 028 607 984 ÷ 2 = 17 011 014 303 992 + 0;
  • 17 011 014 303 992 ÷ 2 = 8 505 507 151 996 + 0;
  • 8 505 507 151 996 ÷ 2 = 4 252 753 575 998 + 0;
  • 4 252 753 575 998 ÷ 2 = 2 126 376 787 999 + 0;
  • 2 126 376 787 999 ÷ 2 = 1 063 188 393 999 + 1;
  • 1 063 188 393 999 ÷ 2 = 531 594 196 999 + 1;
  • 531 594 196 999 ÷ 2 = 265 797 098 499 + 1;
  • 265 797 098 499 ÷ 2 = 132 898 549 249 + 1;
  • 132 898 549 249 ÷ 2 = 66 449 274 624 + 1;
  • 66 449 274 624 ÷ 2 = 33 224 637 312 + 0;
  • 33 224 637 312 ÷ 2 = 16 612 318 656 + 0;
  • 16 612 318 656 ÷ 2 = 8 306 159 328 + 0;
  • 8 306 159 328 ÷ 2 = 4 153 079 664 + 0;
  • 4 153 079 664 ÷ 2 = 2 076 539 832 + 0;
  • 2 076 539 832 ÷ 2 = 1 038 269 916 + 0;
  • 1 038 269 916 ÷ 2 = 519 134 958 + 0;
  • 519 134 958 ÷ 2 = 259 567 479 + 0;
  • 259 567 479 ÷ 2 = 129 783 739 + 1;
  • 129 783 739 ÷ 2 = 64 891 869 + 1;
  • 64 891 869 ÷ 2 = 32 445 934 + 1;
  • 32 445 934 ÷ 2 = 16 222 967 + 0;
  • 16 222 967 ÷ 2 = 8 111 483 + 1;
  • 8 111 483 ÷ 2 = 4 055 741 + 1;
  • 4 055 741 ÷ 2 = 2 027 870 + 1;
  • 2 027 870 ÷ 2 = 1 013 935 + 0;
  • 1 013 935 ÷ 2 = 506 967 + 1;
  • 506 967 ÷ 2 = 253 483 + 1;
  • 253 483 ÷ 2 = 126 741 + 1;
  • 126 741 ÷ 2 = 63 370 + 1;
  • 63 370 ÷ 2 = 31 685 + 0;
  • 31 685 ÷ 2 = 15 842 + 1;
  • 15 842 ÷ 2 = 7 921 + 0;
  • 7 921 ÷ 2 = 3 960 + 1;
  • 3 960 ÷ 2 = 1 980 + 0;
  • 1 980 ÷ 2 = 990 + 0;
  • 990 ÷ 2 = 495 + 0;
  • 495 ÷ 2 = 247 + 1;
  • 247 ÷ 2 = 123 + 1;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 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 positive number.

Take all the remainders starting from the bottom of the list constructed above.

4 459 335 333 705 705 605(10) =


11 1101 1110 0010 1011 1101 1101 1100 0000 0011 1110 0000 0110 1000 1000 0101(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 61 positions to the left so that only one non zero digit remains to the left of it:

4 459 335 333 705 705 605(10) =


11 1101 1110 0010 1011 1101 1101 1100 0000 0011 1110 0000 0110 1000 1000 0101(2) =


11 1101 1110 0010 1011 1101 1101 1100 0000 0011 1110 0000 0110 1000 1000 0101(2) × 20 =


1.1110 1111 0001 0101 1110 1110 1110 0000 0001 1111 0000 0011 0100 0100 0010 1(2) × 261


5. 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): 61


Mantissa (not normalized):
1.1110 1111 0001 0101 1110 1110 1110 0000 0001 1111 0000 0011 0100 0100 0010 1


6. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


61 + 2(11-1) - 1 =


(61 + 1 023)(10) =


1 084(10)


7. 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;
  • 1 084 ÷ 2 = 542 + 0;
  • 542 ÷ 2 = 271 + 0;
  • 271 ÷ 2 = 135 + 1;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

8. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:

Exponent (adjusted) =


1084(10) =


100 0011 1100(2)


9. 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 1111 0001 0101 1110 1110 1110 0000 0001 1111 0000 0011 0100 0 1000 0101 =


1110 1111 0001 0101 1110 1110 1110 0000 0001 1111 0000 0011 0100


10. 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 0011 1100


Mantissa (52 bits) =
1110 1111 0001 0101 1110 1110 1110 0000 0001 1111 0000 0011 0100


Conclusion:

Number -4 459 335 333 705 705 605 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
1 - 100 0011 1100 - 1110 1111 0001 0101 1110 1110 1110 0000 0001 1111 0000 0011 0100

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 1

      49
    • 0

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 1

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 1

      36
    • 1

      35
    • 1

      34
    • 1

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

-4 459 335 333 705 705 606 = ? ... -4 459 335 333 705 705 604 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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