-0.000 000 000 001 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 001(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
-0.000 000 000 001(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:

|-0.000 000 000 001| = 0.000 000 000 001


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

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 001.

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) 0.000 000 000 001 × 2 = 0 + 0.000 000 000 002;
  • 2) 0.000 000 000 002 × 2 = 0 + 0.000 000 000 004;
  • 3) 0.000 000 000 004 × 2 = 0 + 0.000 000 000 008;
  • 4) 0.000 000 000 008 × 2 = 0 + 0.000 000 000 016;
  • 5) 0.000 000 000 016 × 2 = 0 + 0.000 000 000 032;
  • 6) 0.000 000 000 032 × 2 = 0 + 0.000 000 000 064;
  • 7) 0.000 000 000 064 × 2 = 0 + 0.000 000 000 128;
  • 8) 0.000 000 000 128 × 2 = 0 + 0.000 000 000 256;
  • 9) 0.000 000 000 256 × 2 = 0 + 0.000 000 000 512;
  • 10) 0.000 000 000 512 × 2 = 0 + 0.000 000 001 024;
  • 11) 0.000 000 001 024 × 2 = 0 + 0.000 000 002 048;
  • 12) 0.000 000 002 048 × 2 = 0 + 0.000 000 004 096;
  • 13) 0.000 000 004 096 × 2 = 0 + 0.000 000 008 192;
  • 14) 0.000 000 008 192 × 2 = 0 + 0.000 000 016 384;
  • 15) 0.000 000 016 384 × 2 = 0 + 0.000 000 032 768;
  • 16) 0.000 000 032 768 × 2 = 0 + 0.000 000 065 536;
  • 17) 0.000 000 065 536 × 2 = 0 + 0.000 000 131 072;
  • 18) 0.000 000 131 072 × 2 = 0 + 0.000 000 262 144;
  • 19) 0.000 000 262 144 × 2 = 0 + 0.000 000 524 288;
  • 20) 0.000 000 524 288 × 2 = 0 + 0.000 001 048 576;
  • 21) 0.000 001 048 576 × 2 = 0 + 0.000 002 097 152;
  • 22) 0.000 002 097 152 × 2 = 0 + 0.000 004 194 304;
  • 23) 0.000 004 194 304 × 2 = 0 + 0.000 008 388 608;
  • 24) 0.000 008 388 608 × 2 = 0 + 0.000 016 777 216;
  • 25) 0.000 016 777 216 × 2 = 0 + 0.000 033 554 432;
  • 26) 0.000 033 554 432 × 2 = 0 + 0.000 067 108 864;
  • 27) 0.000 067 108 864 × 2 = 0 + 0.000 134 217 728;
  • 28) 0.000 134 217 728 × 2 = 0 + 0.000 268 435 456;
  • 29) 0.000 268 435 456 × 2 = 0 + 0.000 536 870 912;
  • 30) 0.000 536 870 912 × 2 = 0 + 0.001 073 741 824;
  • 31) 0.001 073 741 824 × 2 = 0 + 0.002 147 483 648;
  • 32) 0.002 147 483 648 × 2 = 0 + 0.004 294 967 296;
  • 33) 0.004 294 967 296 × 2 = 0 + 0.008 589 934 592;
  • 34) 0.008 589 934 592 × 2 = 0 + 0.017 179 869 184;
  • 35) 0.017 179 869 184 × 2 = 0 + 0.034 359 738 368;
  • 36) 0.034 359 738 368 × 2 = 0 + 0.068 719 476 736;
  • 37) 0.068 719 476 736 × 2 = 0 + 0.137 438 953 472;
  • 38) 0.137 438 953 472 × 2 = 0 + 0.274 877 906 944;
  • 39) 0.274 877 906 944 × 2 = 0 + 0.549 755 813 888;
  • 40) 0.549 755 813 888 × 2 = 1 + 0.099 511 627 776;
  • 41) 0.099 511 627 776 × 2 = 0 + 0.199 023 255 552;
  • 42) 0.199 023 255 552 × 2 = 0 + 0.398 046 511 104;
  • 43) 0.398 046 511 104 × 2 = 0 + 0.796 093 022 208;
  • 44) 0.796 093 022 208 × 2 = 1 + 0.592 186 044 416;
  • 45) 0.592 186 044 416 × 2 = 1 + 0.184 372 088 832;
  • 46) 0.184 372 088 832 × 2 = 0 + 0.368 744 177 664;
  • 47) 0.368 744 177 664 × 2 = 0 + 0.737 488 355 328;
  • 48) 0.737 488 355 328 × 2 = 1 + 0.474 976 710 656;
  • 49) 0.474 976 710 656 × 2 = 0 + 0.949 953 421 312;
  • 50) 0.949 953 421 312 × 2 = 1 + 0.899 906 842 624;
  • 51) 0.899 906 842 624 × 2 = 1 + 0.799 813 685 248;
  • 52) 0.799 813 685 248 × 2 = 1 + 0.599 627 370 496;
  • 53) 0.599 627 370 496 × 2 = 1 + 0.199 254 740 992;
  • 54) 0.199 254 740 992 × 2 = 0 + 0.398 509 481 984;
  • 55) 0.398 509 481 984 × 2 = 0 + 0.797 018 963 968;
  • 56) 0.797 018 963 968 × 2 = 1 + 0.594 037 927 936;
  • 57) 0.594 037 927 936 × 2 = 1 + 0.188 075 855 872;
  • 58) 0.188 075 855 872 × 2 = 0 + 0.376 151 711 744;
  • 59) 0.376 151 711 744 × 2 = 0 + 0.752 303 423 488;
  • 60) 0.752 303 423 488 × 2 = 1 + 0.504 606 846 976;
  • 61) 0.504 606 846 976 × 2 = 1 + 0.009 213 693 952;
  • 62) 0.009 213 693 952 × 2 = 0 + 0.018 427 387 904;
  • 63) 0.018 427 387 904 × 2 = 0 + 0.036 854 775 808;
  • 64) 0.036 854 775 808 × 2 = 0 + 0.073 709 551 616;
  • 65) 0.073 709 551 616 × 2 = 0 + 0.147 419 103 232;
  • 66) 0.147 419 103 232 × 2 = 0 + 0.294 838 206 464;
  • 67) 0.294 838 206 464 × 2 = 0 + 0.589 676 412 928;
  • 68) 0.589 676 412 928 × 2 = 1 + 0.179 352 825 856;
  • 69) 0.179 352 825 856 × 2 = 0 + 0.358 705 651 712;
  • 70) 0.358 705 651 712 × 2 = 0 + 0.717 411 303 424;
  • 71) 0.717 411 303 424 × 2 = 1 + 0.434 822 606 848;
  • 72) 0.434 822 606 848 × 2 = 0 + 0.869 645 213 696;
  • 73) 0.869 645 213 696 × 2 = 1 + 0.739 290 427 392;
  • 74) 0.739 290 427 392 × 2 = 1 + 0.478 580 854 784;
  • 75) 0.478 580 854 784 × 2 = 0 + 0.957 161 709 568;
  • 76) 0.957 161 709 568 × 2 = 1 + 0.914 323 419 136;
  • 77) 0.914 323 419 136 × 2 = 1 + 0.828 646 838 272;
  • 78) 0.828 646 838 272 × 2 = 1 + 0.657 293 676 544;
  • 79) 0.657 293 676 544 × 2 = 1 + 0.314 587 353 088;
  • 80) 0.314 587 353 088 × 2 = 0 + 0.629 174 706 176;
  • 81) 0.629 174 706 176 × 2 = 1 + 0.258 349 412 352;
  • 82) 0.258 349 412 352 × 2 = 0 + 0.516 698 824 704;
  • 83) 0.516 698 824 704 × 2 = 1 + 0.033 397 649 408;
  • 84) 0.033 397 649 408 × 2 = 0 + 0.066 795 298 816;
  • 85) 0.066 795 298 816 × 2 = 0 + 0.133 590 597 632;
  • 86) 0.133 590 597 632 × 2 = 0 + 0.267 181 195 264;
  • 87) 0.267 181 195 264 × 2 = 0 + 0.534 362 390 528;
  • 88) 0.534 362 390 528 × 2 = 1 + 0.068 724 781 056;
  • 89) 0.068 724 781 056 × 2 = 0 + 0.137 449 562 112;
  • 90) 0.137 449 562 112 × 2 = 0 + 0.274 899 124 224;
  • 91) 0.274 899 124 224 × 2 = 0 + 0.549 798 248 448;
  • 92) 0.549 798 248 448 × 2 = 1 + 0.099 596 496 896;

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:


0.000 000 000 001(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001(2)

6. Positive number before normalization:

0.000 000 000 001(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 40 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 001(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001(2) × 20 =

1.0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001(2) × 2-40


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): -40

Mantissa (not normalized):
1.0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-40 + 2(11-1) - 1 =

(-40 + 1 023)(10) =

983(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;
  • 983 ÷ 2 = 491 + 1;
  • 491 ÷ 2 = 245 + 1;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

983(10) =

011 1101 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001 =

0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 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) =
011 1101 0111

Mantissa (52 bits) =
0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001


Decimal number -0.000 000 000 001 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0111 - 0001 1001 0111 1001 1001 1000 0001 0010 1101 1110 1010 0001 0001

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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