-0.016 738 891 601 562 496 525 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 525 6(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
-0.016 738 891 601 562 496 525 6(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:

|-0.016 738 891 601 562 496 525 6| = 0.016 738 891 601 562 496 525 6


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: 0.016 738 891 601 562 496 525 6.

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) 0.016 738 891 601 562 496 525 6 × 2 = 0 + 0.033 477 783 203 124 993 051 2;
  • 2) 0.033 477 783 203 124 993 051 2 × 2 = 0 + 0.066 955 566 406 249 986 102 4;
  • 3) 0.066 955 566 406 249 986 102 4 × 2 = 0 + 0.133 911 132 812 499 972 204 8;
  • 4) 0.133 911 132 812 499 972 204 8 × 2 = 0 + 0.267 822 265 624 999 944 409 6;
  • 5) 0.267 822 265 624 999 944 409 6 × 2 = 0 + 0.535 644 531 249 999 888 819 2;
  • 6) 0.535 644 531 249 999 888 819 2 × 2 = 1 + 0.071 289 062 499 999 777 638 4;
  • 7) 0.071 289 062 499 999 777 638 4 × 2 = 0 + 0.142 578 124 999 999 555 276 8;
  • 8) 0.142 578 124 999 999 555 276 8 × 2 = 0 + 0.285 156 249 999 999 110 553 6;
  • 9) 0.285 156 249 999 999 110 553 6 × 2 = 0 + 0.570 312 499 999 998 221 107 2;
  • 10) 0.570 312 499 999 998 221 107 2 × 2 = 1 + 0.140 624 999 999 996 442 214 4;
  • 11) 0.140 624 999 999 996 442 214 4 × 2 = 0 + 0.281 249 999 999 992 884 428 8;
  • 12) 0.281 249 999 999 992 884 428 8 × 2 = 0 + 0.562 499 999 999 985 768 857 6;
  • 13) 0.562 499 999 999 985 768 857 6 × 2 = 1 + 0.124 999 999 999 971 537 715 2;
  • 14) 0.124 999 999 999 971 537 715 2 × 2 = 0 + 0.249 999 999 999 943 075 430 4;
  • 15) 0.249 999 999 999 943 075 430 4 × 2 = 0 + 0.499 999 999 999 886 150 860 8;
  • 16) 0.499 999 999 999 886 150 860 8 × 2 = 0 + 0.999 999 999 999 772 301 721 6;
  • 17) 0.999 999 999 999 772 301 721 6 × 2 = 1 + 0.999 999 999 999 544 603 443 2;
  • 18) 0.999 999 999 999 544 603 443 2 × 2 = 1 + 0.999 999 999 999 089 206 886 4;
  • 19) 0.999 999 999 999 089 206 886 4 × 2 = 1 + 0.999 999 999 998 178 413 772 8;
  • 20) 0.999 999 999 998 178 413 772 8 × 2 = 1 + 0.999 999 999 996 356 827 545 6;
  • 21) 0.999 999 999 996 356 827 545 6 × 2 = 1 + 0.999 999 999 992 713 655 091 2;
  • 22) 0.999 999 999 992 713 655 091 2 × 2 = 1 + 0.999 999 999 985 427 310 182 4;
  • 23) 0.999 999 999 985 427 310 182 4 × 2 = 1 + 0.999 999 999 970 854 620 364 8;
  • 24) 0.999 999 999 970 854 620 364 8 × 2 = 1 + 0.999 999 999 941 709 240 729 6;
  • 25) 0.999 999 999 941 709 240 729 6 × 2 = 1 + 0.999 999 999 883 418 481 459 2;
  • 26) 0.999 999 999 883 418 481 459 2 × 2 = 1 + 0.999 999 999 766 836 962 918 4;
  • 27) 0.999 999 999 766 836 962 918 4 × 2 = 1 + 0.999 999 999 533 673 925 836 8;
  • 28) 0.999 999 999 533 673 925 836 8 × 2 = 1 + 0.999 999 999 067 347 851 673 6;
  • 29) 0.999 999 999 067 347 851 673 6 × 2 = 1 + 0.999 999 998 134 695 703 347 2;
  • 30) 0.999 999 998 134 695 703 347 2 × 2 = 1 + 0.999 999 996 269 391 406 694 4;
  • 31) 0.999 999 996 269 391 406 694 4 × 2 = 1 + 0.999 999 992 538 782 813 388 8;
  • 32) 0.999 999 992 538 782 813 388 8 × 2 = 1 + 0.999 999 985 077 565 626 777 6;
  • 33) 0.999 999 985 077 565 626 777 6 × 2 = 1 + 0.999 999 970 155 131 253 555 2;
  • 34) 0.999 999 970 155 131 253 555 2 × 2 = 1 + 0.999 999 940 310 262 507 110 4;
  • 35) 0.999 999 940 310 262 507 110 4 × 2 = 1 + 0.999 999 880 620 525 014 220 8;
  • 36) 0.999 999 880 620 525 014 220 8 × 2 = 1 + 0.999 999 761 241 050 028 441 6;
  • 37) 0.999 999 761 241 050 028 441 6 × 2 = 1 + 0.999 999 522 482 100 056 883 2;
  • 38) 0.999 999 522 482 100 056 883 2 × 2 = 1 + 0.999 999 044 964 200 113 766 4;
  • 39) 0.999 999 044 964 200 113 766 4 × 2 = 1 + 0.999 998 089 928 400 227 532 8;
  • 40) 0.999 998 089 928 400 227 532 8 × 2 = 1 + 0.999 996 179 856 800 455 065 6;
  • 41) 0.999 996 179 856 800 455 065 6 × 2 = 1 + 0.999 992 359 713 600 910 131 2;
  • 42) 0.999 992 359 713 600 910 131 2 × 2 = 1 + 0.999 984 719 427 201 820 262 4;
  • 43) 0.999 984 719 427 201 820 262 4 × 2 = 1 + 0.999 969 438 854 403 640 524 8;
  • 44) 0.999 969 438 854 403 640 524 8 × 2 = 1 + 0.999 938 877 708 807 281 049 6;
  • 45) 0.999 938 877 708 807 281 049 6 × 2 = 1 + 0.999 877 755 417 614 562 099 2;
  • 46) 0.999 877 755 417 614 562 099 2 × 2 = 1 + 0.999 755 510 835 229 124 198 4;
  • 47) 0.999 755 510 835 229 124 198 4 × 2 = 1 + 0.999 511 021 670 458 248 396 8;
  • 48) 0.999 511 021 670 458 248 396 8 × 2 = 1 + 0.999 022 043 340 916 496 793 6;
  • 49) 0.999 022 043 340 916 496 793 6 × 2 = 1 + 0.998 044 086 681 832 993 587 2;
  • 50) 0.998 044 086 681 832 993 587 2 × 2 = 1 + 0.996 088 173 363 665 987 174 4;
  • 51) 0.996 088 173 363 665 987 174 4 × 2 = 1 + 0.992 176 346 727 331 974 348 8;
  • 52) 0.992 176 346 727 331 974 348 8 × 2 = 1 + 0.984 352 693 454 663 948 697 6;
  • 53) 0.984 352 693 454 663 948 697 6 × 2 = 1 + 0.968 705 386 909 327 897 395 2;
  • 54) 0.968 705 386 909 327 897 395 2 × 2 = 1 + 0.937 410 773 818 655 794 790 4;
  • 55) 0.937 410 773 818 655 794 790 4 × 2 = 1 + 0.874 821 547 637 311 589 580 8;
  • 56) 0.874 821 547 637 311 589 580 8 × 2 = 1 + 0.749 643 095 274 623 179 161 6;
  • 57) 0.749 643 095 274 623 179 161 6 × 2 = 1 + 0.499 286 190 549 246 358 323 2;
  • 58) 0.499 286 190 549 246 358 323 2 × 2 = 0 + 0.998 572 381 098 492 716 646 4;

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:


0.016 738 891 601 562 496 525 6(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 525 6(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 525 6(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 525 6 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 1110

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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