-0.016 738 891 601 562 496 626 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 626(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
-0.016 738 891 601 562 496 626(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:

|-0.016 738 891 601 562 496 626| = 0.016 738 891 601 562 496 626


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: 0.016 738 891 601 562 496 626.

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) 0.016 738 891 601 562 496 626 × 2 = 0 + 0.033 477 783 203 124 993 252;
  • 2) 0.033 477 783 203 124 993 252 × 2 = 0 + 0.066 955 566 406 249 986 504;
  • 3) 0.066 955 566 406 249 986 504 × 2 = 0 + 0.133 911 132 812 499 973 008;
  • 4) 0.133 911 132 812 499 973 008 × 2 = 0 + 0.267 822 265 624 999 946 016;
  • 5) 0.267 822 265 624 999 946 016 × 2 = 0 + 0.535 644 531 249 999 892 032;
  • 6) 0.535 644 531 249 999 892 032 × 2 = 1 + 0.071 289 062 499 999 784 064;
  • 7) 0.071 289 062 499 999 784 064 × 2 = 0 + 0.142 578 124 999 999 568 128;
  • 8) 0.142 578 124 999 999 568 128 × 2 = 0 + 0.285 156 249 999 999 136 256;
  • 9) 0.285 156 249 999 999 136 256 × 2 = 0 + 0.570 312 499 999 998 272 512;
  • 10) 0.570 312 499 999 998 272 512 × 2 = 1 + 0.140 624 999 999 996 545 024;
  • 11) 0.140 624 999 999 996 545 024 × 2 = 0 + 0.281 249 999 999 993 090 048;
  • 12) 0.281 249 999 999 993 090 048 × 2 = 0 + 0.562 499 999 999 986 180 096;
  • 13) 0.562 499 999 999 986 180 096 × 2 = 1 + 0.124 999 999 999 972 360 192;
  • 14) 0.124 999 999 999 972 360 192 × 2 = 0 + 0.249 999 999 999 944 720 384;
  • 15) 0.249 999 999 999 944 720 384 × 2 = 0 + 0.499 999 999 999 889 440 768;
  • 16) 0.499 999 999 999 889 440 768 × 2 = 0 + 0.999 999 999 999 778 881 536;
  • 17) 0.999 999 999 999 778 881 536 × 2 = 1 + 0.999 999 999 999 557 763 072;
  • 18) 0.999 999 999 999 557 763 072 × 2 = 1 + 0.999 999 999 999 115 526 144;
  • 19) 0.999 999 999 999 115 526 144 × 2 = 1 + 0.999 999 999 998 231 052 288;
  • 20) 0.999 999 999 998 231 052 288 × 2 = 1 + 0.999 999 999 996 462 104 576;
  • 21) 0.999 999 999 996 462 104 576 × 2 = 1 + 0.999 999 999 992 924 209 152;
  • 22) 0.999 999 999 992 924 209 152 × 2 = 1 + 0.999 999 999 985 848 418 304;
  • 23) 0.999 999 999 985 848 418 304 × 2 = 1 + 0.999 999 999 971 696 836 608;
  • 24) 0.999 999 999 971 696 836 608 × 2 = 1 + 0.999 999 999 943 393 673 216;
  • 25) 0.999 999 999 943 393 673 216 × 2 = 1 + 0.999 999 999 886 787 346 432;
  • 26) 0.999 999 999 886 787 346 432 × 2 = 1 + 0.999 999 999 773 574 692 864;
  • 27) 0.999 999 999 773 574 692 864 × 2 = 1 + 0.999 999 999 547 149 385 728;
  • 28) 0.999 999 999 547 149 385 728 × 2 = 1 + 0.999 999 999 094 298 771 456;
  • 29) 0.999 999 999 094 298 771 456 × 2 = 1 + 0.999 999 998 188 597 542 912;
  • 30) 0.999 999 998 188 597 542 912 × 2 = 1 + 0.999 999 996 377 195 085 824;
  • 31) 0.999 999 996 377 195 085 824 × 2 = 1 + 0.999 999 992 754 390 171 648;
  • 32) 0.999 999 992 754 390 171 648 × 2 = 1 + 0.999 999 985 508 780 343 296;
  • 33) 0.999 999 985 508 780 343 296 × 2 = 1 + 0.999 999 971 017 560 686 592;
  • 34) 0.999 999 971 017 560 686 592 × 2 = 1 + 0.999 999 942 035 121 373 184;
  • 35) 0.999 999 942 035 121 373 184 × 2 = 1 + 0.999 999 884 070 242 746 368;
  • 36) 0.999 999 884 070 242 746 368 × 2 = 1 + 0.999 999 768 140 485 492 736;
  • 37) 0.999 999 768 140 485 492 736 × 2 = 1 + 0.999 999 536 280 970 985 472;
  • 38) 0.999 999 536 280 970 985 472 × 2 = 1 + 0.999 999 072 561 941 970 944;
  • 39) 0.999 999 072 561 941 970 944 × 2 = 1 + 0.999 998 145 123 883 941 888;
  • 40) 0.999 998 145 123 883 941 888 × 2 = 1 + 0.999 996 290 247 767 883 776;
  • 41) 0.999 996 290 247 767 883 776 × 2 = 1 + 0.999 992 580 495 535 767 552;
  • 42) 0.999 992 580 495 535 767 552 × 2 = 1 + 0.999 985 160 991 071 535 104;
  • 43) 0.999 985 160 991 071 535 104 × 2 = 1 + 0.999 970 321 982 143 070 208;
  • 44) 0.999 970 321 982 143 070 208 × 2 = 1 + 0.999 940 643 964 286 140 416;
  • 45) 0.999 940 643 964 286 140 416 × 2 = 1 + 0.999 881 287 928 572 280 832;
  • 46) 0.999 881 287 928 572 280 832 × 2 = 1 + 0.999 762 575 857 144 561 664;
  • 47) 0.999 762 575 857 144 561 664 × 2 = 1 + 0.999 525 151 714 289 123 328;
  • 48) 0.999 525 151 714 289 123 328 × 2 = 1 + 0.999 050 303 428 578 246 656;
  • 49) 0.999 050 303 428 578 246 656 × 2 = 1 + 0.998 100 606 857 156 493 312;
  • 50) 0.998 100 606 857 156 493 312 × 2 = 1 + 0.996 201 213 714 312 986 624;
  • 51) 0.996 201 213 714 312 986 624 × 2 = 1 + 0.992 402 427 428 625 973 248;
  • 52) 0.992 402 427 428 625 973 248 × 2 = 1 + 0.984 804 854 857 251 946 496;
  • 53) 0.984 804 854 857 251 946 496 × 2 = 1 + 0.969 609 709 714 503 892 992;
  • 54) 0.969 609 709 714 503 892 992 × 2 = 1 + 0.939 219 419 429 007 785 984;
  • 55) 0.939 219 419 429 007 785 984 × 2 = 1 + 0.878 438 838 858 015 571 968;
  • 56) 0.878 438 838 858 015 571 968 × 2 = 1 + 0.756 877 677 716 031 143 936;
  • 57) 0.756 877 677 716 031 143 936 × 2 = 1 + 0.513 755 355 432 062 287 872;
  • 58) 0.513 755 355 432 062 287 872 × 2 = 1 + 0.027 510 710 864 124 575 744;

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:


0.016 738 891 601 562 496 626(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 626(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:


0.016 738 891 601 562 496 626(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 -0.016 738 891 601 562 496 626 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