Convert -9.954 034 474 611 633 623 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 -9.954 034 474 611 633 623(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:

|-9.954 034 474 611 633 623| = 9.954 034 474 611 633 623

2. First, convert to the binary (base 2) the integer part: 9.
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;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

9(10) =


1001(2)


4. Convert to the binary (base 2) the fractional part: 0.954 034 474 611 633 623.

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) 0.954 034 474 611 633 623 × 2 = 1 + 0.908 068 949 223 267 246;
  • 2) 0.908 068 949 223 267 246 × 2 = 1 + 0.816 137 898 446 534 492;
  • 3) 0.816 137 898 446 534 492 × 2 = 1 + 0.632 275 796 893 068 984;
  • 4) 0.632 275 796 893 068 984 × 2 = 1 + 0.264 551 593 786 137 968;
  • 5) 0.264 551 593 786 137 968 × 2 = 0 + 0.529 103 187 572 275 936;
  • 6) 0.529 103 187 572 275 936 × 2 = 1 + 0.058 206 375 144 551 872;
  • 7) 0.058 206 375 144 551 872 × 2 = 0 + 0.116 412 750 289 103 744;
  • 8) 0.116 412 750 289 103 744 × 2 = 0 + 0.232 825 500 578 207 488;
  • 9) 0.232 825 500 578 207 488 × 2 = 0 + 0.465 651 001 156 414 976;
  • 10) 0.465 651 001 156 414 976 × 2 = 0 + 0.931 302 002 312 829 952;
  • 11) 0.931 302 002 312 829 952 × 2 = 1 + 0.862 604 004 625 659 904;
  • 12) 0.862 604 004 625 659 904 × 2 = 1 + 0.725 208 009 251 319 808;
  • 13) 0.725 208 009 251 319 808 × 2 = 1 + 0.450 416 018 502 639 616;
  • 14) 0.450 416 018 502 639 616 × 2 = 0 + 0.900 832 037 005 279 232;
  • 15) 0.900 832 037 005 279 232 × 2 = 1 + 0.801 664 074 010 558 464;
  • 16) 0.801 664 074 010 558 464 × 2 = 1 + 0.603 328 148 021 116 928;
  • 17) 0.603 328 148 021 116 928 × 2 = 1 + 0.206 656 296 042 233 856;
  • 18) 0.206 656 296 042 233 856 × 2 = 0 + 0.413 312 592 084 467 712;
  • 19) 0.413 312 592 084 467 712 × 2 = 0 + 0.826 625 184 168 935 424;
  • 20) 0.826 625 184 168 935 424 × 2 = 1 + 0.653 250 368 337 870 848;
  • 21) 0.653 250 368 337 870 848 × 2 = 1 + 0.306 500 736 675 741 696;
  • 22) 0.306 500 736 675 741 696 × 2 = 0 + 0.613 001 473 351 483 392;
  • 23) 0.613 001 473 351 483 392 × 2 = 1 + 0.226 002 946 702 966 784;
  • 24) 0.226 002 946 702 966 784 × 2 = 0 + 0.452 005 893 405 933 568;
  • 25) 0.452 005 893 405 933 568 × 2 = 0 + 0.904 011 786 811 867 136;
  • 26) 0.904 011 786 811 867 136 × 2 = 1 + 0.808 023 573 623 734 272;
  • 27) 0.808 023 573 623 734 272 × 2 = 1 + 0.616 047 147 247 468 544;
  • 28) 0.616 047 147 247 468 544 × 2 = 1 + 0.232 094 294 494 937 088;
  • 29) 0.232 094 294 494 937 088 × 2 = 0 + 0.464 188 588 989 874 176;
  • 30) 0.464 188 588 989 874 176 × 2 = 0 + 0.928 377 177 979 748 352;
  • 31) 0.928 377 177 979 748 352 × 2 = 1 + 0.856 754 355 959 496 704;
  • 32) 0.856 754 355 959 496 704 × 2 = 1 + 0.713 508 711 918 993 408;
  • 33) 0.713 508 711 918 993 408 × 2 = 1 + 0.427 017 423 837 986 816;
  • 34) 0.427 017 423 837 986 816 × 2 = 0 + 0.854 034 847 675 973 632;
  • 35) 0.854 034 847 675 973 632 × 2 = 1 + 0.708 069 695 351 947 264;
  • 36) 0.708 069 695 351 947 264 × 2 = 1 + 0.416 139 390 703 894 528;
  • 37) 0.416 139 390 703 894 528 × 2 = 0 + 0.832 278 781 407 789 056;
  • 38) 0.832 278 781 407 789 056 × 2 = 1 + 0.664 557 562 815 578 112;
  • 39) 0.664 557 562 815 578 112 × 2 = 1 + 0.329 115 125 631 156 224;
  • 40) 0.329 115 125 631 156 224 × 2 = 0 + 0.658 230 251 262 312 448;
  • 41) 0.658 230 251 262 312 448 × 2 = 1 + 0.316 460 502 524 624 896;
  • 42) 0.316 460 502 524 624 896 × 2 = 0 + 0.632 921 005 049 249 792;
  • 43) 0.632 921 005 049 249 792 × 2 = 1 + 0.265 842 010 098 499 584;
  • 44) 0.265 842 010 098 499 584 × 2 = 0 + 0.531 684 020 196 999 168;
  • 45) 0.531 684 020 196 999 168 × 2 = 1 + 0.063 368 040 393 998 336;
  • 46) 0.063 368 040 393 998 336 × 2 = 0 + 0.126 736 080 787 996 672;
  • 47) 0.126 736 080 787 996 672 × 2 = 0 + 0.253 472 161 575 993 344;
  • 48) 0.253 472 161 575 993 344 × 2 = 0 + 0.506 944 323 151 986 688;
  • 49) 0.506 944 323 151 986 688 × 2 = 1 + 0.013 888 646 303 973 376;
  • 50) 0.013 888 646 303 973 376 × 2 = 0 + 0.027 777 292 607 946 752;
  • 51) 0.027 777 292 607 946 752 × 2 = 0 + 0.055 554 585 215 893 504;
  • 52) 0.055 554 585 215 893 504 × 2 = 0 + 0.111 109 170 431 787 008;
  • 53) 0.111 109 170 431 787 008 × 2 = 0 + 0.222 218 340 863 574 016;

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...)


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:

0.954 034 474 611 633 623(10) =


0.1111 0100 0011 1011 1001 1010 0111 0011 1011 0110 1010 1000 1000 0(2)


6. Positive number before normalization:

9.954 034 474 611 633 623(10) =


1001.1111 0100 0011 1011 1001 1010 0111 0011 1011 0110 1010 1000 1000 0(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

9.954 034 474 611 633 623(10) =


1001.1111 0100 0011 1011 1001 1010 0111 0011 1011 0110 1010 1000 1000 0(2) =


1001.1111 0100 0011 1011 1001 1010 0111 0011 1011 0110 1010 1000 1000 0(2) × 20 =


1.0011 1110 1000 0111 0111 0011 0100 1110 0111 0110 1101 0101 0001 0000(2) × 23


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): 3


Mantissa (not normalized):
1.0011 1110 1000 0111 0111 0011 0100 1110 0111 0110 1101 0101 0001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 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) =


1026(10) =


100 0000 0010(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. 0011 1110 1000 0111 0111 0011 0100 1110 0111 0110 1101 0101 0001 0000 =


0011 1110 1000 0111 0111 0011 0100 1110 0111 0110 1101 0101 0001


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 0010


Mantissa (52 bits) =
0011 1110 1000 0111 0111 0011 0100 1110 0111 0110 1101 0101 0001


Conclusion:

Number -9.954 034 474 611 633 623 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
1 - 100 0000 0010 - 0011 1110 1000 0111 0111 0011 0100 1110 0111 0110 1101 0101 0001

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 1

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 0

      35
    • 1

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 1

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

-9.954 034 474 611 633 624 = ? ... -9.954 034 474 611 633 622 = ?


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