-0.016 738 891 601 562 872 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 872(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
-0.016 738 891 601 562 872(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:

|-0.016 738 891 601 562 872| = 0.016 738 891 601 562 872


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: 0.016 738 891 601 562 872.

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) 0.016 738 891 601 562 872 × 2 = 0 + 0.033 477 783 203 125 744;
  • 2) 0.033 477 783 203 125 744 × 2 = 0 + 0.066 955 566 406 251 488;
  • 3) 0.066 955 566 406 251 488 × 2 = 0 + 0.133 911 132 812 502 976;
  • 4) 0.133 911 132 812 502 976 × 2 = 0 + 0.267 822 265 625 005 952;
  • 5) 0.267 822 265 625 005 952 × 2 = 0 + 0.535 644 531 250 011 904;
  • 6) 0.535 644 531 250 011 904 × 2 = 1 + 0.071 289 062 500 023 808;
  • 7) 0.071 289 062 500 023 808 × 2 = 0 + 0.142 578 125 000 047 616;
  • 8) 0.142 578 125 000 047 616 × 2 = 0 + 0.285 156 250 000 095 232;
  • 9) 0.285 156 250 000 095 232 × 2 = 0 + 0.570 312 500 000 190 464;
  • 10) 0.570 312 500 000 190 464 × 2 = 1 + 0.140 625 000 000 380 928;
  • 11) 0.140 625 000 000 380 928 × 2 = 0 + 0.281 250 000 000 761 856;
  • 12) 0.281 250 000 000 761 856 × 2 = 0 + 0.562 500 000 001 523 712;
  • 13) 0.562 500 000 001 523 712 × 2 = 1 + 0.125 000 000 003 047 424;
  • 14) 0.125 000 000 003 047 424 × 2 = 0 + 0.250 000 000 006 094 848;
  • 15) 0.250 000 000 006 094 848 × 2 = 0 + 0.500 000 000 012 189 696;
  • 16) 0.500 000 000 012 189 696 × 2 = 1 + 0.000 000 000 024 379 392;
  • 17) 0.000 000 000 024 379 392 × 2 = 0 + 0.000 000 000 048 758 784;
  • 18) 0.000 000 000 048 758 784 × 2 = 0 + 0.000 000 000 097 517 568;
  • 19) 0.000 000 000 097 517 568 × 2 = 0 + 0.000 000 000 195 035 136;
  • 20) 0.000 000 000 195 035 136 × 2 = 0 + 0.000 000 000 390 070 272;
  • 21) 0.000 000 000 390 070 272 × 2 = 0 + 0.000 000 000 780 140 544;
  • 22) 0.000 000 000 780 140 544 × 2 = 0 + 0.000 000 001 560 281 088;
  • 23) 0.000 000 001 560 281 088 × 2 = 0 + 0.000 000 003 120 562 176;
  • 24) 0.000 000 003 120 562 176 × 2 = 0 + 0.000 000 006 241 124 352;
  • 25) 0.000 000 006 241 124 352 × 2 = 0 + 0.000 000 012 482 248 704;
  • 26) 0.000 000 012 482 248 704 × 2 = 0 + 0.000 000 024 964 497 408;
  • 27) 0.000 000 024 964 497 408 × 2 = 0 + 0.000 000 049 928 994 816;
  • 28) 0.000 000 049 928 994 816 × 2 = 0 + 0.000 000 099 857 989 632;
  • 29) 0.000 000 099 857 989 632 × 2 = 0 + 0.000 000 199 715 979 264;
  • 30) 0.000 000 199 715 979 264 × 2 = 0 + 0.000 000 399 431 958 528;
  • 31) 0.000 000 399 431 958 528 × 2 = 0 + 0.000 000 798 863 917 056;
  • 32) 0.000 000 798 863 917 056 × 2 = 0 + 0.000 001 597 727 834 112;
  • 33) 0.000 001 597 727 834 112 × 2 = 0 + 0.000 003 195 455 668 224;
  • 34) 0.000 003 195 455 668 224 × 2 = 0 + 0.000 006 390 911 336 448;
  • 35) 0.000 006 390 911 336 448 × 2 = 0 + 0.000 012 781 822 672 896;
  • 36) 0.000 012 781 822 672 896 × 2 = 0 + 0.000 025 563 645 345 792;
  • 37) 0.000 025 563 645 345 792 × 2 = 0 + 0.000 051 127 290 691 584;
  • 38) 0.000 051 127 290 691 584 × 2 = 0 + 0.000 102 254 581 383 168;
  • 39) 0.000 102 254 581 383 168 × 2 = 0 + 0.000 204 509 162 766 336;
  • 40) 0.000 204 509 162 766 336 × 2 = 0 + 0.000 409 018 325 532 672;
  • 41) 0.000 409 018 325 532 672 × 2 = 0 + 0.000 818 036 651 065 344;
  • 42) 0.000 818 036 651 065 344 × 2 = 0 + 0.001 636 073 302 130 688;
  • 43) 0.001 636 073 302 130 688 × 2 = 0 + 0.003 272 146 604 261 376;
  • 44) 0.003 272 146 604 261 376 × 2 = 0 + 0.006 544 293 208 522 752;
  • 45) 0.006 544 293 208 522 752 × 2 = 0 + 0.013 088 586 417 045 504;
  • 46) 0.013 088 586 417 045 504 × 2 = 0 + 0.026 177 172 834 091 008;
  • 47) 0.026 177 172 834 091 008 × 2 = 0 + 0.052 354 345 668 182 016;
  • 48) 0.052 354 345 668 182 016 × 2 = 0 + 0.104 708 691 336 364 032;
  • 49) 0.104 708 691 336 364 032 × 2 = 0 + 0.209 417 382 672 728 064;
  • 50) 0.209 417 382 672 728 064 × 2 = 0 + 0.418 834 765 345 456 128;
  • 51) 0.418 834 765 345 456 128 × 2 = 0 + 0.837 669 530 690 912 256;
  • 52) 0.837 669 530 690 912 256 × 2 = 1 + 0.675 339 061 381 824 512;
  • 53) 0.675 339 061 381 824 512 × 2 = 1 + 0.350 678 122 763 649 024;
  • 54) 0.350 678 122 763 649 024 × 2 = 0 + 0.701 356 245 527 298 048;
  • 55) 0.701 356 245 527 298 048 × 2 = 1 + 0.402 712 491 054 596 096;
  • 56) 0.402 712 491 054 596 096 × 2 = 0 + 0.805 424 982 109 192 192;
  • 57) 0.805 424 982 109 192 192 × 2 = 1 + 0.610 849 964 218 384 384;
  • 58) 0.610 849 964 218 384 384 × 2 = 1 + 0.221 699 928 436 768 768;

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:


0.016 738 891 601 562 872(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 1010 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 872(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 1010 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:


0.016 738 891 601 562 872(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 1010 11(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 1010 11(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0110 1011(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 0100 0000 0000 0000 0000 0000 0000 0000 0000 0110 1011


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 0110 1011 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0110 1011


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 0110 1011


Decimal number -0.016 738 891 601 562 872 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0110 1011

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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