-0.000 000 000 060 834 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 060 834(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
-0.000 000 000 060 834(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:

|-0.000 000 000 060 834| = 0.000 000 000 060 834


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: 0.000 000 000 060 834.

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) 0.000 000 000 060 834 × 2 = 0 + 0.000 000 000 121 668;
  • 2) 0.000 000 000 121 668 × 2 = 0 + 0.000 000 000 243 336;
  • 3) 0.000 000 000 243 336 × 2 = 0 + 0.000 000 000 486 672;
  • 4) 0.000 000 000 486 672 × 2 = 0 + 0.000 000 000 973 344;
  • 5) 0.000 000 000 973 344 × 2 = 0 + 0.000 000 001 946 688;
  • 6) 0.000 000 001 946 688 × 2 = 0 + 0.000 000 003 893 376;
  • 7) 0.000 000 003 893 376 × 2 = 0 + 0.000 000 007 786 752;
  • 8) 0.000 000 007 786 752 × 2 = 0 + 0.000 000 015 573 504;
  • 9) 0.000 000 015 573 504 × 2 = 0 + 0.000 000 031 147 008;
  • 10) 0.000 000 031 147 008 × 2 = 0 + 0.000 000 062 294 016;
  • 11) 0.000 000 062 294 016 × 2 = 0 + 0.000 000 124 588 032;
  • 12) 0.000 000 124 588 032 × 2 = 0 + 0.000 000 249 176 064;
  • 13) 0.000 000 249 176 064 × 2 = 0 + 0.000 000 498 352 128;
  • 14) 0.000 000 498 352 128 × 2 = 0 + 0.000 000 996 704 256;
  • 15) 0.000 000 996 704 256 × 2 = 0 + 0.000 001 993 408 512;
  • 16) 0.000 001 993 408 512 × 2 = 0 + 0.000 003 986 817 024;
  • 17) 0.000 003 986 817 024 × 2 = 0 + 0.000 007 973 634 048;
  • 18) 0.000 007 973 634 048 × 2 = 0 + 0.000 015 947 268 096;
  • 19) 0.000 015 947 268 096 × 2 = 0 + 0.000 031 894 536 192;
  • 20) 0.000 031 894 536 192 × 2 = 0 + 0.000 063 789 072 384;
  • 21) 0.000 063 789 072 384 × 2 = 0 + 0.000 127 578 144 768;
  • 22) 0.000 127 578 144 768 × 2 = 0 + 0.000 255 156 289 536;
  • 23) 0.000 255 156 289 536 × 2 = 0 + 0.000 510 312 579 072;
  • 24) 0.000 510 312 579 072 × 2 = 0 + 0.001 020 625 158 144;
  • 25) 0.001 020 625 158 144 × 2 = 0 + 0.002 041 250 316 288;
  • 26) 0.002 041 250 316 288 × 2 = 0 + 0.004 082 500 632 576;
  • 27) 0.004 082 500 632 576 × 2 = 0 + 0.008 165 001 265 152;
  • 28) 0.008 165 001 265 152 × 2 = 0 + 0.016 330 002 530 304;
  • 29) 0.016 330 002 530 304 × 2 = 0 + 0.032 660 005 060 608;
  • 30) 0.032 660 005 060 608 × 2 = 0 + 0.065 320 010 121 216;
  • 31) 0.065 320 010 121 216 × 2 = 0 + 0.130 640 020 242 432;
  • 32) 0.130 640 020 242 432 × 2 = 0 + 0.261 280 040 484 864;
  • 33) 0.261 280 040 484 864 × 2 = 0 + 0.522 560 080 969 728;
  • 34) 0.522 560 080 969 728 × 2 = 1 + 0.045 120 161 939 456;
  • 35) 0.045 120 161 939 456 × 2 = 0 + 0.090 240 323 878 912;
  • 36) 0.090 240 323 878 912 × 2 = 0 + 0.180 480 647 757 824;
  • 37) 0.180 480 647 757 824 × 2 = 0 + 0.360 961 295 515 648;
  • 38) 0.360 961 295 515 648 × 2 = 0 + 0.721 922 591 031 296;
  • 39) 0.721 922 591 031 296 × 2 = 1 + 0.443 845 182 062 592;
  • 40) 0.443 845 182 062 592 × 2 = 0 + 0.887 690 364 125 184;
  • 41) 0.887 690 364 125 184 × 2 = 1 + 0.775 380 728 250 368;
  • 42) 0.775 380 728 250 368 × 2 = 1 + 0.550 761 456 500 736;
  • 43) 0.550 761 456 500 736 × 2 = 1 + 0.101 522 913 001 472;
  • 44) 0.101 522 913 001 472 × 2 = 0 + 0.203 045 826 002 944;
  • 45) 0.203 045 826 002 944 × 2 = 0 + 0.406 091 652 005 888;
  • 46) 0.406 091 652 005 888 × 2 = 0 + 0.812 183 304 011 776;
  • 47) 0.812 183 304 011 776 × 2 = 1 + 0.624 366 608 023 552;
  • 48) 0.624 366 608 023 552 × 2 = 1 + 0.248 733 216 047 104;
  • 49) 0.248 733 216 047 104 × 2 = 0 + 0.497 466 432 094 208;
  • 50) 0.497 466 432 094 208 × 2 = 0 + 0.994 932 864 188 416;
  • 51) 0.994 932 864 188 416 × 2 = 1 + 0.989 865 728 376 832;
  • 52) 0.989 865 728 376 832 × 2 = 1 + 0.979 731 456 753 664;
  • 53) 0.979 731 456 753 664 × 2 = 1 + 0.959 462 913 507 328;
  • 54) 0.959 462 913 507 328 × 2 = 1 + 0.918 925 827 014 656;
  • 55) 0.918 925 827 014 656 × 2 = 1 + 0.837 851 654 029 312;
  • 56) 0.837 851 654 029 312 × 2 = 1 + 0.675 703 308 058 624;
  • 57) 0.675 703 308 058 624 × 2 = 1 + 0.351 406 616 117 248;
  • 58) 0.351 406 616 117 248 × 2 = 0 + 0.702 813 232 234 496;
  • 59) 0.702 813 232 234 496 × 2 = 1 + 0.405 626 464 468 992;
  • 60) 0.405 626 464 468 992 × 2 = 0 + 0.811 252 928 937 984;
  • 61) 0.811 252 928 937 984 × 2 = 1 + 0.622 505 857 875 968;
  • 62) 0.622 505 857 875 968 × 2 = 1 + 0.245 011 715 751 936;
  • 63) 0.245 011 715 751 936 × 2 = 0 + 0.490 023 431 503 872;
  • 64) 0.490 023 431 503 872 × 2 = 0 + 0.980 046 863 007 744;
  • 65) 0.980 046 863 007 744 × 2 = 1 + 0.960 093 726 015 488;
  • 66) 0.960 093 726 015 488 × 2 = 1 + 0.920 187 452 030 976;
  • 67) 0.920 187 452 030 976 × 2 = 1 + 0.840 374 904 061 952;
  • 68) 0.840 374 904 061 952 × 2 = 1 + 0.680 749 808 123 904;
  • 69) 0.680 749 808 123 904 × 2 = 1 + 0.361 499 616 247 808;
  • 70) 0.361 499 616 247 808 × 2 = 0 + 0.722 999 232 495 616;
  • 71) 0.722 999 232 495 616 × 2 = 1 + 0.445 998 464 991 232;
  • 72) 0.445 998 464 991 232 × 2 = 0 + 0.891 996 929 982 464;
  • 73) 0.891 996 929 982 464 × 2 = 1 + 0.783 993 859 964 928;
  • 74) 0.783 993 859 964 928 × 2 = 1 + 0.567 987 719 929 856;
  • 75) 0.567 987 719 929 856 × 2 = 1 + 0.135 975 439 859 712;
  • 76) 0.135 975 439 859 712 × 2 = 0 + 0.271 950 879 719 424;
  • 77) 0.271 950 879 719 424 × 2 = 0 + 0.543 901 759 438 848;
  • 78) 0.543 901 759 438 848 × 2 = 1 + 0.087 803 518 877 696;
  • 79) 0.087 803 518 877 696 × 2 = 0 + 0.175 607 037 755 392;
  • 80) 0.175 607 037 755 392 × 2 = 0 + 0.351 214 075 510 784;
  • 81) 0.351 214 075 510 784 × 2 = 0 + 0.702 428 151 021 568;
  • 82) 0.702 428 151 021 568 × 2 = 1 + 0.404 856 302 043 136;
  • 83) 0.404 856 302 043 136 × 2 = 0 + 0.809 712 604 086 272;
  • 84) 0.809 712 604 086 272 × 2 = 1 + 0.619 425 208 172 544;
  • 85) 0.619 425 208 172 544 × 2 = 1 + 0.238 850 416 345 088;
  • 86) 0.238 850 416 345 088 × 2 = 0 + 0.477 700 832 690 176;

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:


0.000 000 000 060 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0100 0010 1110 0011 0011 1111 1010 1100 1111 1010 1110 0100 0101 10(2)

6. Positive number before normalization:

0.000 000 000 060 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0100 0010 1110 0011 0011 1111 1010 1100 1111 1010 1110 0100 0101 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 34 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 060 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0100 0010 1110 0011 0011 1111 1010 1100 1111 1010 1110 0100 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0100 0010 1110 0011 0011 1111 1010 1100 1111 1010 1110 0100 0101 10(2) × 20 =

1.0000 1011 1000 1100 1111 1110 1011 0011 1110 1011 1001 0001 0110(2) × 2-34


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): -34

Mantissa (not normalized):
1.0000 1011 1000 1100 1111 1110 1011 0011 1110 1011 1001 0001 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-34 + 2(11-1) - 1 =

(-34 + 1 023)(10) =

989(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;
  • 989 ÷ 2 = 494 + 1;
  • 494 ÷ 2 = 247 + 0;
  • 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;

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) =

989(10) =

011 1101 1101(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. 0000 1011 1000 1100 1111 1110 1011 0011 1110 1011 1001 0001 0110 =

0000 1011 1000 1100 1111 1110 1011 0011 1110 1011 1001 0001 0110


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 1101

Mantissa (52 bits) =
0000 1011 1000 1100 1111 1110 1011 0011 1110 1011 1001 0001 0110


Decimal number -0.000 000 000 060 834 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1101 - 0000 1011 1000 1100 1111 1110 1011 0011 1110 1011 1001 0001 0110

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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