Base ten decimal number 3.141 592 653 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number 3.141 592 653(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 3. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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:

3(10) =


11(2)

3. Convert to binary (base 2) the fractional part: 0.141 592 653. 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) 0.141 592 653 × 2 = 0 + 0.283 185 306;
  • 2) 0.283 185 306 × 2 = 0 + 0.566 370 612;
  • 3) 0.566 370 612 × 2 = 1 + 0.132 741 224;
  • 4) 0.132 741 224 × 2 = 0 + 0.265 482 448;
  • 5) 0.265 482 448 × 2 = 0 + 0.530 964 896;
  • 6) 0.530 964 896 × 2 = 1 + 0.061 929 792;
  • 7) 0.061 929 792 × 2 = 0 + 0.123 859 584;
  • 8) 0.123 859 584 × 2 = 0 + 0.247 719 168;
  • 9) 0.247 719 168 × 2 = 0 + 0.495 438 336;
  • 10) 0.495 438 336 × 2 = 0 + 0.990 876 672;
  • 11) 0.990 876 672 × 2 = 1 + 0.981 753 344;
  • 12) 0.981 753 344 × 2 = 1 + 0.963 506 688;
  • 13) 0.963 506 688 × 2 = 1 + 0.927 013 376;
  • 14) 0.927 013 376 × 2 = 1 + 0.854 026 752;
  • 15) 0.854 026 752 × 2 = 1 + 0.708 053 504;
  • 16) 0.708 053 504 × 2 = 1 + 0.416 107 008;
  • 17) 0.416 107 008 × 2 = 0 + 0.832 214 016;
  • 18) 0.832 214 016 × 2 = 1 + 0.664 428 032;
  • 19) 0.664 428 032 × 2 = 1 + 0.328 856 064;
  • 20) 0.328 856 064 × 2 = 0 + 0.657 712 128;
  • 21) 0.657 712 128 × 2 = 1 + 0.315 424 256;
  • 22) 0.315 424 256 × 2 = 0 + 0.630 848 512;
  • 23) 0.630 848 512 × 2 = 1 + 0.261 697 024;
  • 24) 0.261 697 024 × 2 = 0 + 0.523 394 048;
  • 25) 0.523 394 048 × 2 = 1 + 0.046 788 096;
  • 26) 0.046 788 096 × 2 = 0 + 0.093 576 192;
  • 27) 0.093 576 192 × 2 = 0 + 0.187 152 384;
  • 28) 0.187 152 384 × 2 = 0 + 0.374 304 768;
  • 29) 0.374 304 768 × 2 = 0 + 0.748 609 536;
  • 30) 0.748 609 536 × 2 = 1 + 0.497 219 072;
  • 31) 0.497 219 072 × 2 = 0 + 0.994 438 144;
  • 32) 0.994 438 144 × 2 = 1 + 0.988 876 288;
  • 33) 0.988 876 288 × 2 = 1 + 0.977 752 576;
  • 34) 0.977 752 576 × 2 = 1 + 0.955 505 152;
  • 35) 0.955 505 152 × 2 = 1 + 0.911 010 304;
  • 36) 0.911 010 304 × 2 = 1 + 0.822 020 608;
  • 37) 0.822 020 608 × 2 = 1 + 0.644 041 216;
  • 38) 0.644 041 216 × 2 = 1 + 0.288 082 432;
  • 39) 0.288 082 432 × 2 = 0 + 0.576 164 864;
  • 40) 0.576 164 864 × 2 = 1 + 0.152 329 728;
  • 41) 0.152 329 728 × 2 = 0 + 0.304 659 456;
  • 42) 0.304 659 456 × 2 = 0 + 0.609 318 912;
  • 43) 0.609 318 912 × 2 = 1 + 0.218 637 824;
  • 44) 0.218 637 824 × 2 = 0 + 0.437 275 648;
  • 45) 0.437 275 648 × 2 = 0 + 0.874 551 296;
  • 46) 0.874 551 296 × 2 = 1 + 0.749 102 592;
  • 47) 0.749 102 592 × 2 = 1 + 0.498 205 184;
  • 48) 0.498 205 184 × 2 = 0 + 0.996 410 368;
  • 49) 0.996 410 368 × 2 = 1 + 0.992 820 736;
  • 50) 0.992 820 736 × 2 = 1 + 0.985 641 472;
  • 51) 0.985 641 472 × 2 = 1 + 0.971 282 944;
  • 52) 0.971 282 944 × 2 = 1 + 0.942 565 888;
  • 53) 0.942 565 888 × 2 = 1 + 0.885 131 776;

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...)

4. 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:

0.141 592 653(10) =


0.0010 0100 0011 1111 0110 1010 1000 0101 1111 1101 0010 0110 1111 1(2)

Positive number before normalization:

3.141 592 653(10) =


11.0010 0100 0011 1111 0110 1010 1000 0101 1111 1101 0010 0110 1111 1(2)

5. Normalize the binary representation of the number, shifting the decimal mark 1 positions to the left so that only one non zero digit remains to the left of it:

3.141 592 653(10) =


11.0010 0100 0011 1111 0110 1010 1000 0101 1111 1101 0010 0110 1111 1(2) =


11.0010 0100 0011 1111 0110 1010 1000 0101 1111 1101 0010 0110 1111 1(2) × 20 =


1.1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111 11(2) × 21

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: 0 (a positive number)


Exponent (unadjusted): 1


Mantissa (not normalized): 1.1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111 11

6. 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 =


1 + 2(11-1) - 1 =


(1 + 1 023)(10) =


1 024(10)


  • division = quotient + remainder;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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) =


1024(10) =


100 0000 0000(2)

7. 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. 1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111 11 =


1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111

Conclusion:

The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
100 0000 0000


Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111

Number 3.141 592 653, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 1

      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

3.141 592 653 = 0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0010 1111 1110 1001 0011 0111 Jun 24 15:28 UTC (GMT)
5.2 = 0 - 100 0000 0001 - 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 Jun 24 15:24 UTC (GMT)
0.185 = 0 - 011 1111 1100 - 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 Jun 24 15:23 UTC (GMT)
9 223 372 036 854 785 024 = 0 - 100 0011 1110 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 Jun 24 15:23 UTC (GMT)
-7.5 = 1 - 100 0000 0001 - 1110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 15:21 UTC (GMT)
-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)
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