-0.016 738 891 601 562 496 817 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 817(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
-0.016 738 891 601 562 496 817(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:

|-0.016 738 891 601 562 496 817| = 0.016 738 891 601 562 496 817


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: 0.016 738 891 601 562 496 817.

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) 0.016 738 891 601 562 496 817 × 2 = 0 + 0.033 477 783 203 124 993 634;
  • 2) 0.033 477 783 203 124 993 634 × 2 = 0 + 0.066 955 566 406 249 987 268;
  • 3) 0.066 955 566 406 249 987 268 × 2 = 0 + 0.133 911 132 812 499 974 536;
  • 4) 0.133 911 132 812 499 974 536 × 2 = 0 + 0.267 822 265 624 999 949 072;
  • 5) 0.267 822 265 624 999 949 072 × 2 = 0 + 0.535 644 531 249 999 898 144;
  • 6) 0.535 644 531 249 999 898 144 × 2 = 1 + 0.071 289 062 499 999 796 288;
  • 7) 0.071 289 062 499 999 796 288 × 2 = 0 + 0.142 578 124 999 999 592 576;
  • 8) 0.142 578 124 999 999 592 576 × 2 = 0 + 0.285 156 249 999 999 185 152;
  • 9) 0.285 156 249 999 999 185 152 × 2 = 0 + 0.570 312 499 999 998 370 304;
  • 10) 0.570 312 499 999 998 370 304 × 2 = 1 + 0.140 624 999 999 996 740 608;
  • 11) 0.140 624 999 999 996 740 608 × 2 = 0 + 0.281 249 999 999 993 481 216;
  • 12) 0.281 249 999 999 993 481 216 × 2 = 0 + 0.562 499 999 999 986 962 432;
  • 13) 0.562 499 999 999 986 962 432 × 2 = 1 + 0.124 999 999 999 973 924 864;
  • 14) 0.124 999 999 999 973 924 864 × 2 = 0 + 0.249 999 999 999 947 849 728;
  • 15) 0.249 999 999 999 947 849 728 × 2 = 0 + 0.499 999 999 999 895 699 456;
  • 16) 0.499 999 999 999 895 699 456 × 2 = 0 + 0.999 999 999 999 791 398 912;
  • 17) 0.999 999 999 999 791 398 912 × 2 = 1 + 0.999 999 999 999 582 797 824;
  • 18) 0.999 999 999 999 582 797 824 × 2 = 1 + 0.999 999 999 999 165 595 648;
  • 19) 0.999 999 999 999 165 595 648 × 2 = 1 + 0.999 999 999 998 331 191 296;
  • 20) 0.999 999 999 998 331 191 296 × 2 = 1 + 0.999 999 999 996 662 382 592;
  • 21) 0.999 999 999 996 662 382 592 × 2 = 1 + 0.999 999 999 993 324 765 184;
  • 22) 0.999 999 999 993 324 765 184 × 2 = 1 + 0.999 999 999 986 649 530 368;
  • 23) 0.999 999 999 986 649 530 368 × 2 = 1 + 0.999 999 999 973 299 060 736;
  • 24) 0.999 999 999 973 299 060 736 × 2 = 1 + 0.999 999 999 946 598 121 472;
  • 25) 0.999 999 999 946 598 121 472 × 2 = 1 + 0.999 999 999 893 196 242 944;
  • 26) 0.999 999 999 893 196 242 944 × 2 = 1 + 0.999 999 999 786 392 485 888;
  • 27) 0.999 999 999 786 392 485 888 × 2 = 1 + 0.999 999 999 572 784 971 776;
  • 28) 0.999 999 999 572 784 971 776 × 2 = 1 + 0.999 999 999 145 569 943 552;
  • 29) 0.999 999 999 145 569 943 552 × 2 = 1 + 0.999 999 998 291 139 887 104;
  • 30) 0.999 999 998 291 139 887 104 × 2 = 1 + 0.999 999 996 582 279 774 208;
  • 31) 0.999 999 996 582 279 774 208 × 2 = 1 + 0.999 999 993 164 559 548 416;
  • 32) 0.999 999 993 164 559 548 416 × 2 = 1 + 0.999 999 986 329 119 096 832;
  • 33) 0.999 999 986 329 119 096 832 × 2 = 1 + 0.999 999 972 658 238 193 664;
  • 34) 0.999 999 972 658 238 193 664 × 2 = 1 + 0.999 999 945 316 476 387 328;
  • 35) 0.999 999 945 316 476 387 328 × 2 = 1 + 0.999 999 890 632 952 774 656;
  • 36) 0.999 999 890 632 952 774 656 × 2 = 1 + 0.999 999 781 265 905 549 312;
  • 37) 0.999 999 781 265 905 549 312 × 2 = 1 + 0.999 999 562 531 811 098 624;
  • 38) 0.999 999 562 531 811 098 624 × 2 = 1 + 0.999 999 125 063 622 197 248;
  • 39) 0.999 999 125 063 622 197 248 × 2 = 1 + 0.999 998 250 127 244 394 496;
  • 40) 0.999 998 250 127 244 394 496 × 2 = 1 + 0.999 996 500 254 488 788 992;
  • 41) 0.999 996 500 254 488 788 992 × 2 = 1 + 0.999 993 000 508 977 577 984;
  • 42) 0.999 993 000 508 977 577 984 × 2 = 1 + 0.999 986 001 017 955 155 968;
  • 43) 0.999 986 001 017 955 155 968 × 2 = 1 + 0.999 972 002 035 910 311 936;
  • 44) 0.999 972 002 035 910 311 936 × 2 = 1 + 0.999 944 004 071 820 623 872;
  • 45) 0.999 944 004 071 820 623 872 × 2 = 1 + 0.999 888 008 143 641 247 744;
  • 46) 0.999 888 008 143 641 247 744 × 2 = 1 + 0.999 776 016 287 282 495 488;
  • 47) 0.999 776 016 287 282 495 488 × 2 = 1 + 0.999 552 032 574 564 990 976;
  • 48) 0.999 552 032 574 564 990 976 × 2 = 1 + 0.999 104 065 149 129 981 952;
  • 49) 0.999 104 065 149 129 981 952 × 2 = 1 + 0.998 208 130 298 259 963 904;
  • 50) 0.998 208 130 298 259 963 904 × 2 = 1 + 0.996 416 260 596 519 927 808;
  • 51) 0.996 416 260 596 519 927 808 × 2 = 1 + 0.992 832 521 193 039 855 616;
  • 52) 0.992 832 521 193 039 855 616 × 2 = 1 + 0.985 665 042 386 079 711 232;
  • 53) 0.985 665 042 386 079 711 232 × 2 = 1 + 0.971 330 084 772 159 422 464;
  • 54) 0.971 330 084 772 159 422 464 × 2 = 1 + 0.942 660 169 544 318 844 928;
  • 55) 0.942 660 169 544 318 844 928 × 2 = 1 + 0.885 320 339 088 637 689 856;
  • 56) 0.885 320 339 088 637 689 856 × 2 = 1 + 0.770 640 678 177 275 379 712;
  • 57) 0.770 640 678 177 275 379 712 × 2 = 1 + 0.541 281 356 354 550 759 424;
  • 58) 0.541 281 356 354 550 759 424 × 2 = 1 + 0.082 562 712 709 101 518 848;

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:


0.016 738 891 601 562 496 817(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 817(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 817(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1017(10) =

011 1111 1001(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 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 817 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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