Base ten decimal number -8.038 469 019 917 503 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number -8.038 469 019 917 503(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. We start with the positive version of the number:

|-8.038 469 019 917 503| = 8.038 469 019 917 503

2. First, convert to binary (base 2) the integer part: 8. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 8 ÷ 2 = 4 + 0;
  • 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, by taking all the remainders starting from the bottom of the list constructed above:

8(10) =


1000(2)

4. Convert to binary (base 2) the fractional part: 0.038 469 019 917 503. Multiply it 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) 0.038 469 019 917 503 × 2 = 0 + 0.076 938 039 835 006;
  • 2) 0.076 938 039 835 006 × 2 = 0 + 0.153 876 079 670 012;
  • 3) 0.153 876 079 670 012 × 2 = 0 + 0.307 752 159 340 024;
  • 4) 0.307 752 159 340 024 × 2 = 0 + 0.615 504 318 680 048;
  • 5) 0.615 504 318 680 048 × 2 = 1 + 0.231 008 637 360 096;
  • 6) 0.231 008 637 360 096 × 2 = 0 + 0.462 017 274 720 192;
  • 7) 0.462 017 274 720 192 × 2 = 0 + 0.924 034 549 440 384;
  • 8) 0.924 034 549 440 384 × 2 = 1 + 0.848 069 098 880 768;
  • 9) 0.848 069 098 880 768 × 2 = 1 + 0.696 138 197 761 536;
  • 10) 0.696 138 197 761 536 × 2 = 1 + 0.392 276 395 523 072;
  • 11) 0.392 276 395 523 072 × 2 = 0 + 0.784 552 791 046 144;
  • 12) 0.784 552 791 046 144 × 2 = 1 + 0.569 105 582 092 288;
  • 13) 0.569 105 582 092 288 × 2 = 1 + 0.138 211 164 184 576;
  • 14) 0.138 211 164 184 576 × 2 = 0 + 0.276 422 328 369 152;
  • 15) 0.276 422 328 369 152 × 2 = 0 + 0.552 844 656 738 304;
  • 16) 0.552 844 656 738 304 × 2 = 1 + 0.105 689 313 476 608;
  • 17) 0.105 689 313 476 608 × 2 = 0 + 0.211 378 626 953 216;
  • 18) 0.211 378 626 953 216 × 2 = 0 + 0.422 757 253 906 432;
  • 19) 0.422 757 253 906 432 × 2 = 0 + 0.845 514 507 812 864;
  • 20) 0.845 514 507 812 864 × 2 = 1 + 0.691 029 015 625 728;
  • 21) 0.691 029 015 625 728 × 2 = 1 + 0.382 058 031 251 456;
  • 22) 0.382 058 031 251 456 × 2 = 0 + 0.764 116 062 502 912;
  • 23) 0.764 116 062 502 912 × 2 = 1 + 0.528 232 125 005 824;
  • 24) 0.528 232 125 005 824 × 2 = 1 + 0.056 464 250 011 648;
  • 25) 0.056 464 250 011 648 × 2 = 0 + 0.112 928 500 023 296;
  • 26) 0.112 928 500 023 296 × 2 = 0 + 0.225 857 000 046 592;
  • 27) 0.225 857 000 046 592 × 2 = 0 + 0.451 714 000 093 184;
  • 28) 0.451 714 000 093 184 × 2 = 0 + 0.903 428 000 186 368;
  • 29) 0.903 428 000 186 368 × 2 = 1 + 0.806 856 000 372 736;
  • 30) 0.806 856 000 372 736 × 2 = 1 + 0.613 712 000 745 472;
  • 31) 0.613 712 000 745 472 × 2 = 1 + 0.227 424 001 490 944;
  • 32) 0.227 424 001 490 944 × 2 = 0 + 0.454 848 002 981 888;
  • 33) 0.454 848 002 981 888 × 2 = 0 + 0.909 696 005 963 776;
  • 34) 0.909 696 005 963 776 × 2 = 1 + 0.819 392 011 927 552;
  • 35) 0.819 392 011 927 552 × 2 = 1 + 0.638 784 023 855 104;
  • 36) 0.638 784 023 855 104 × 2 = 1 + 0.277 568 047 710 208;
  • 37) 0.277 568 047 710 208 × 2 = 0 + 0.555 136 095 420 416;
  • 38) 0.555 136 095 420 416 × 2 = 1 + 0.110 272 190 840 832;
  • 39) 0.110 272 190 840 832 × 2 = 0 + 0.220 544 381 681 664;
  • 40) 0.220 544 381 681 664 × 2 = 0 + 0.441 088 763 363 328;
  • 41) 0.441 088 763 363 328 × 2 = 0 + 0.882 177 526 726 656;
  • 42) 0.882 177 526 726 656 × 2 = 1 + 0.764 355 053 453 312;
  • 43) 0.764 355 053 453 312 × 2 = 1 + 0.528 710 106 906 624;
  • 44) 0.528 710 106 906 624 × 2 = 1 + 0.057 420 213 813 248;
  • 45) 0.057 420 213 813 248 × 2 = 0 + 0.114 840 427 626 496;
  • 46) 0.114 840 427 626 496 × 2 = 0 + 0.229 680 855 252 992;
  • 47) 0.229 680 855 252 992 × 2 = 0 + 0.459 361 710 505 984;
  • 48) 0.459 361 710 505 984 × 2 = 0 + 0.918 723 421 011 968;
  • 49) 0.918 723 421 011 968 × 2 = 1 + 0.837 446 842 023 936;
  • 50) 0.837 446 842 023 936 × 2 = 1 + 0.674 893 684 047 872;
  • 51) 0.674 893 684 047 872 × 2 = 1 + 0.349 787 368 095 744;
  • 52) 0.349 787 368 095 744 × 2 = 0 + 0.699 574 736 191 488;
  • 53) 0.699 574 736 191 488 × 2 = 1 + 0.399 149 472 382 976;

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, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.038 469 019 917 503(10) =


0.0000 1001 1101 1001 0001 1011 0000 1110 0111 0100 0111 0000 1110 1(2)

Positive number before normalization:

8.038 469 019 917 503(10) =


1000.0000 1001 1101 1001 0001 1011 0000 1110 0111 0100 0111 0000 1110 1(2)

6. Normalize the binary representation of the number, shifting the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

8.038 469 019 917 503(10) =


1000.0000 1001 1101 1001 0001 1011 0000 1110 0111 0100 0111 0000 1110 1(2) =


1000.0000 1001 1101 1001 0001 1011 0000 1110 0111 0100 0111 0000 1110 1(2) × 20 =


1.0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 0001 1101(2) × 23

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): 1.0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 0001 1101

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:

Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(10)


  • 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;

Exponent (adjusted) =


1026(10) =


100 0000 0010(2)

8. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then 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. 0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 0001 1101 =


0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 0001

Conclusion:

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) =
0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 0001

Number -8.038 469 019 917 503, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


1 - 100 0000 0010 - 0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 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
    • 0

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 0

      42
    • 1

      41
    • 1

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 1

      36
    • 0

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 1

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

Convert decimal numbers from base ten 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 1 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

-8.038 469 019 917 503 = 1 - 100 0000 0010 - 0000 0001 0011 1011 0010 0011 0110 0001 1100 1110 1000 1110 0001 Jun 24 15:20 UTC (GMT)
9.71 = 0 - 100 0000 0010 - 0011 0110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 Jun 24 15:20 UTC (GMT)
5.7 = 0 - 100 0000 0001 - 0110 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 Jun 24 15:20 UTC (GMT)
17 976 931 348 623 157 123 456 789 = 0 - 100 0101 0010 - 1101 1011 1101 1000 0110 1100 1101 0110 0010 0011 1000 1101 1000 Jun 24 15:19 UTC (GMT)
3.141 596 = 0 - 100 0000 0000 - 1001 0010 0001 1111 1101 0001 0101 0110 1001 1111 0100 1001 0000 Jun 24 15:19 UTC (GMT)
1.618 033 988 75 = 0 - 011 1111 1111 - 1001 1110 0011 0111 0111 1001 1011 1001 0111 1111 0110 1000 0001 Jun 24 15:18 UTC (GMT)
406.054 687 5 = 0 - 100 0000 0111 - 1001 0110 0000 1110 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 15:10 UTC (GMT)
0.000 2 = 0 - 011 1111 0010 - 1010 0011 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1100 Jun 24 15:09 UTC (GMT)
1 234.567 = 0 - 100 0000 1001 - 0011 0100 1010 0100 0100 1001 1011 1010 0101 1110 0011 0101 0011 Jun 24 15:08 UTC (GMT)
64.75 = 0 - 100 0000 0101 - 0000 0011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 15:07 UTC (GMT)
-72.938 8 = 1 - 100 0000 0101 - 0010 0011 1100 0001 0101 0100 1100 1001 1000 0101 1111 0000 0110 Jun 24 15:06 UTC (GMT)
16.4 = 0 - 100 0000 0011 - 0000 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 Jun 24 15:04 UTC (GMT)
-160.207 049 = 1 - 100 0000 0110 - 0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100 Jun 24 15:02 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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