-0.016 738 891 601 562 496 530 567 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 567(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
-0.016 738 891 601 562 496 530 567(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:

|-0.016 738 891 601 562 496 530 567| = 0.016 738 891 601 562 496 530 567


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: 0.016 738 891 601 562 496 530 567.

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) 0.016 738 891 601 562 496 530 567 × 2 = 0 + 0.033 477 783 203 124 993 061 134;
  • 2) 0.033 477 783 203 124 993 061 134 × 2 = 0 + 0.066 955 566 406 249 986 122 268;
  • 3) 0.066 955 566 406 249 986 122 268 × 2 = 0 + 0.133 911 132 812 499 972 244 536;
  • 4) 0.133 911 132 812 499 972 244 536 × 2 = 0 + 0.267 822 265 624 999 944 489 072;
  • 5) 0.267 822 265 624 999 944 489 072 × 2 = 0 + 0.535 644 531 249 999 888 978 144;
  • 6) 0.535 644 531 249 999 888 978 144 × 2 = 1 + 0.071 289 062 499 999 777 956 288;
  • 7) 0.071 289 062 499 999 777 956 288 × 2 = 0 + 0.142 578 124 999 999 555 912 576;
  • 8) 0.142 578 124 999 999 555 912 576 × 2 = 0 + 0.285 156 249 999 999 111 825 152;
  • 9) 0.285 156 249 999 999 111 825 152 × 2 = 0 + 0.570 312 499 999 998 223 650 304;
  • 10) 0.570 312 499 999 998 223 650 304 × 2 = 1 + 0.140 624 999 999 996 447 300 608;
  • 11) 0.140 624 999 999 996 447 300 608 × 2 = 0 + 0.281 249 999 999 992 894 601 216;
  • 12) 0.281 249 999 999 992 894 601 216 × 2 = 0 + 0.562 499 999 999 985 789 202 432;
  • 13) 0.562 499 999 999 985 789 202 432 × 2 = 1 + 0.124 999 999 999 971 578 404 864;
  • 14) 0.124 999 999 999 971 578 404 864 × 2 = 0 + 0.249 999 999 999 943 156 809 728;
  • 15) 0.249 999 999 999 943 156 809 728 × 2 = 0 + 0.499 999 999 999 886 313 619 456;
  • 16) 0.499 999 999 999 886 313 619 456 × 2 = 0 + 0.999 999 999 999 772 627 238 912;
  • 17) 0.999 999 999 999 772 627 238 912 × 2 = 1 + 0.999 999 999 999 545 254 477 824;
  • 18) 0.999 999 999 999 545 254 477 824 × 2 = 1 + 0.999 999 999 999 090 508 955 648;
  • 19) 0.999 999 999 999 090 508 955 648 × 2 = 1 + 0.999 999 999 998 181 017 911 296;
  • 20) 0.999 999 999 998 181 017 911 296 × 2 = 1 + 0.999 999 999 996 362 035 822 592;
  • 21) 0.999 999 999 996 362 035 822 592 × 2 = 1 + 0.999 999 999 992 724 071 645 184;
  • 22) 0.999 999 999 992 724 071 645 184 × 2 = 1 + 0.999 999 999 985 448 143 290 368;
  • 23) 0.999 999 999 985 448 143 290 368 × 2 = 1 + 0.999 999 999 970 896 286 580 736;
  • 24) 0.999 999 999 970 896 286 580 736 × 2 = 1 + 0.999 999 999 941 792 573 161 472;
  • 25) 0.999 999 999 941 792 573 161 472 × 2 = 1 + 0.999 999 999 883 585 146 322 944;
  • 26) 0.999 999 999 883 585 146 322 944 × 2 = 1 + 0.999 999 999 767 170 292 645 888;
  • 27) 0.999 999 999 767 170 292 645 888 × 2 = 1 + 0.999 999 999 534 340 585 291 776;
  • 28) 0.999 999 999 534 340 585 291 776 × 2 = 1 + 0.999 999 999 068 681 170 583 552;
  • 29) 0.999 999 999 068 681 170 583 552 × 2 = 1 + 0.999 999 998 137 362 341 167 104;
  • 30) 0.999 999 998 137 362 341 167 104 × 2 = 1 + 0.999 999 996 274 724 682 334 208;
  • 31) 0.999 999 996 274 724 682 334 208 × 2 = 1 + 0.999 999 992 549 449 364 668 416;
  • 32) 0.999 999 992 549 449 364 668 416 × 2 = 1 + 0.999 999 985 098 898 729 336 832;
  • 33) 0.999 999 985 098 898 729 336 832 × 2 = 1 + 0.999 999 970 197 797 458 673 664;
  • 34) 0.999 999 970 197 797 458 673 664 × 2 = 1 + 0.999 999 940 395 594 917 347 328;
  • 35) 0.999 999 940 395 594 917 347 328 × 2 = 1 + 0.999 999 880 791 189 834 694 656;
  • 36) 0.999 999 880 791 189 834 694 656 × 2 = 1 + 0.999 999 761 582 379 669 389 312;
  • 37) 0.999 999 761 582 379 669 389 312 × 2 = 1 + 0.999 999 523 164 759 338 778 624;
  • 38) 0.999 999 523 164 759 338 778 624 × 2 = 1 + 0.999 999 046 329 518 677 557 248;
  • 39) 0.999 999 046 329 518 677 557 248 × 2 = 1 + 0.999 998 092 659 037 355 114 496;
  • 40) 0.999 998 092 659 037 355 114 496 × 2 = 1 + 0.999 996 185 318 074 710 228 992;
  • 41) 0.999 996 185 318 074 710 228 992 × 2 = 1 + 0.999 992 370 636 149 420 457 984;
  • 42) 0.999 992 370 636 149 420 457 984 × 2 = 1 + 0.999 984 741 272 298 840 915 968;
  • 43) 0.999 984 741 272 298 840 915 968 × 2 = 1 + 0.999 969 482 544 597 681 831 936;
  • 44) 0.999 969 482 544 597 681 831 936 × 2 = 1 + 0.999 938 965 089 195 363 663 872;
  • 45) 0.999 938 965 089 195 363 663 872 × 2 = 1 + 0.999 877 930 178 390 727 327 744;
  • 46) 0.999 877 930 178 390 727 327 744 × 2 = 1 + 0.999 755 860 356 781 454 655 488;
  • 47) 0.999 755 860 356 781 454 655 488 × 2 = 1 + 0.999 511 720 713 562 909 310 976;
  • 48) 0.999 511 720 713 562 909 310 976 × 2 = 1 + 0.999 023 441 427 125 818 621 952;
  • 49) 0.999 023 441 427 125 818 621 952 × 2 = 1 + 0.998 046 882 854 251 637 243 904;
  • 50) 0.998 046 882 854 251 637 243 904 × 2 = 1 + 0.996 093 765 708 503 274 487 808;
  • 51) 0.996 093 765 708 503 274 487 808 × 2 = 1 + 0.992 187 531 417 006 548 975 616;
  • 52) 0.992 187 531 417 006 548 975 616 × 2 = 1 + 0.984 375 062 834 013 097 951 232;
  • 53) 0.984 375 062 834 013 097 951 232 × 2 = 1 + 0.968 750 125 668 026 195 902 464;
  • 54) 0.968 750 125 668 026 195 902 464 × 2 = 1 + 0.937 500 251 336 052 391 804 928;
  • 55) 0.937 500 251 336 052 391 804 928 × 2 = 1 + 0.875 000 502 672 104 783 609 856;
  • 56) 0.875 000 502 672 104 783 609 856 × 2 = 1 + 0.750 001 005 344 209 567 219 712;
  • 57) 0.750 001 005 344 209 567 219 712 × 2 = 1 + 0.500 002 010 688 419 134 439 424;
  • 58) 0.500 002 010 688 419 134 439 424 × 2 = 1 + 0.000 004 021 376 838 268 878 848;

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:


0.016 738 891 601 562 496 530 567(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 567(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:


0.016 738 891 601 562 496 530 567(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 -0.016 738 891 601 562 496 530 567 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