-0.016 738 891 601 562 807 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 807(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
-0.016 738 891 601 562 807(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:

|-0.016 738 891 601 562 807| = 0.016 738 891 601 562 807


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: 0.016 738 891 601 562 807.

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) 0.016 738 891 601 562 807 × 2 = 0 + 0.033 477 783 203 125 614;
  • 2) 0.033 477 783 203 125 614 × 2 = 0 + 0.066 955 566 406 251 228;
  • 3) 0.066 955 566 406 251 228 × 2 = 0 + 0.133 911 132 812 502 456;
  • 4) 0.133 911 132 812 502 456 × 2 = 0 + 0.267 822 265 625 004 912;
  • 5) 0.267 822 265 625 004 912 × 2 = 0 + 0.535 644 531 250 009 824;
  • 6) 0.535 644 531 250 009 824 × 2 = 1 + 0.071 289 062 500 019 648;
  • 7) 0.071 289 062 500 019 648 × 2 = 0 + 0.142 578 125 000 039 296;
  • 8) 0.142 578 125 000 039 296 × 2 = 0 + 0.285 156 250 000 078 592;
  • 9) 0.285 156 250 000 078 592 × 2 = 0 + 0.570 312 500 000 157 184;
  • 10) 0.570 312 500 000 157 184 × 2 = 1 + 0.140 625 000 000 314 368;
  • 11) 0.140 625 000 000 314 368 × 2 = 0 + 0.281 250 000 000 628 736;
  • 12) 0.281 250 000 000 628 736 × 2 = 0 + 0.562 500 000 001 257 472;
  • 13) 0.562 500 000 001 257 472 × 2 = 1 + 0.125 000 000 002 514 944;
  • 14) 0.125 000 000 002 514 944 × 2 = 0 + 0.250 000 000 005 029 888;
  • 15) 0.250 000 000 005 029 888 × 2 = 0 + 0.500 000 000 010 059 776;
  • 16) 0.500 000 000 010 059 776 × 2 = 1 + 0.000 000 000 020 119 552;
  • 17) 0.000 000 000 020 119 552 × 2 = 0 + 0.000 000 000 040 239 104;
  • 18) 0.000 000 000 040 239 104 × 2 = 0 + 0.000 000 000 080 478 208;
  • 19) 0.000 000 000 080 478 208 × 2 = 0 + 0.000 000 000 160 956 416;
  • 20) 0.000 000 000 160 956 416 × 2 = 0 + 0.000 000 000 321 912 832;
  • 21) 0.000 000 000 321 912 832 × 2 = 0 + 0.000 000 000 643 825 664;
  • 22) 0.000 000 000 643 825 664 × 2 = 0 + 0.000 000 001 287 651 328;
  • 23) 0.000 000 001 287 651 328 × 2 = 0 + 0.000 000 002 575 302 656;
  • 24) 0.000 000 002 575 302 656 × 2 = 0 + 0.000 000 005 150 605 312;
  • 25) 0.000 000 005 150 605 312 × 2 = 0 + 0.000 000 010 301 210 624;
  • 26) 0.000 000 010 301 210 624 × 2 = 0 + 0.000 000 020 602 421 248;
  • 27) 0.000 000 020 602 421 248 × 2 = 0 + 0.000 000 041 204 842 496;
  • 28) 0.000 000 041 204 842 496 × 2 = 0 + 0.000 000 082 409 684 992;
  • 29) 0.000 000 082 409 684 992 × 2 = 0 + 0.000 000 164 819 369 984;
  • 30) 0.000 000 164 819 369 984 × 2 = 0 + 0.000 000 329 638 739 968;
  • 31) 0.000 000 329 638 739 968 × 2 = 0 + 0.000 000 659 277 479 936;
  • 32) 0.000 000 659 277 479 936 × 2 = 0 + 0.000 001 318 554 959 872;
  • 33) 0.000 001 318 554 959 872 × 2 = 0 + 0.000 002 637 109 919 744;
  • 34) 0.000 002 637 109 919 744 × 2 = 0 + 0.000 005 274 219 839 488;
  • 35) 0.000 005 274 219 839 488 × 2 = 0 + 0.000 010 548 439 678 976;
  • 36) 0.000 010 548 439 678 976 × 2 = 0 + 0.000 021 096 879 357 952;
  • 37) 0.000 021 096 879 357 952 × 2 = 0 + 0.000 042 193 758 715 904;
  • 38) 0.000 042 193 758 715 904 × 2 = 0 + 0.000 084 387 517 431 808;
  • 39) 0.000 084 387 517 431 808 × 2 = 0 + 0.000 168 775 034 863 616;
  • 40) 0.000 168 775 034 863 616 × 2 = 0 + 0.000 337 550 069 727 232;
  • 41) 0.000 337 550 069 727 232 × 2 = 0 + 0.000 675 100 139 454 464;
  • 42) 0.000 675 100 139 454 464 × 2 = 0 + 0.001 350 200 278 908 928;
  • 43) 0.001 350 200 278 908 928 × 2 = 0 + 0.002 700 400 557 817 856;
  • 44) 0.002 700 400 557 817 856 × 2 = 0 + 0.005 400 801 115 635 712;
  • 45) 0.005 400 801 115 635 712 × 2 = 0 + 0.010 801 602 231 271 424;
  • 46) 0.010 801 602 231 271 424 × 2 = 0 + 0.021 603 204 462 542 848;
  • 47) 0.021 603 204 462 542 848 × 2 = 0 + 0.043 206 408 925 085 696;
  • 48) 0.043 206 408 925 085 696 × 2 = 0 + 0.086 412 817 850 171 392;
  • 49) 0.086 412 817 850 171 392 × 2 = 0 + 0.172 825 635 700 342 784;
  • 50) 0.172 825 635 700 342 784 × 2 = 0 + 0.345 651 271 400 685 568;
  • 51) 0.345 651 271 400 685 568 × 2 = 0 + 0.691 302 542 801 371 136;
  • 52) 0.691 302 542 801 371 136 × 2 = 1 + 0.382 605 085 602 742 272;
  • 53) 0.382 605 085 602 742 272 × 2 = 0 + 0.765 210 171 205 484 544;
  • 54) 0.765 210 171 205 484 544 × 2 = 1 + 0.530 420 342 410 969 088;
  • 55) 0.530 420 342 410 969 088 × 2 = 1 + 0.060 840 684 821 938 176;
  • 56) 0.060 840 684 821 938 176 × 2 = 0 + 0.121 681 369 643 876 352;
  • 57) 0.121 681 369 643 876 352 × 2 = 0 + 0.243 362 739 287 752 704;
  • 58) 0.243 362 739 287 752 704 × 2 = 0 + 0.486 725 478 575 505 408;

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:


0.016 738 891 601 562 807(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0110 00(2)

6. Positive number before normalization:

0.016 738 891 601 562 807(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0110 00(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:


0.016 738 891 601 562 807(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0110 00(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0110 00(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 1000(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 0101 1000


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 0101 1000 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 1000


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 0101 1000


Decimal number -0.016 738 891 601 562 807 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 0101 1000

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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