7.200 000 000 000 000 177 635 683 722 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 7.200 000 000 000 000 177 635 683 722(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
7.200 000 000 000 000 177 635 683 722(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. First, convert to binary (in base 2) the integer part: 7.
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;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

7(10) =

111(2)


3. Convert to binary (base 2) the fractional part: 0.200 000 000 000 000 177 635 683 722.

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) 0.200 000 000 000 000 177 635 683 722 × 2 = 0 + 0.400 000 000 000 000 355 271 367 444;
  • 2) 0.400 000 000 000 000 355 271 367 444 × 2 = 0 + 0.800 000 000 000 000 710 542 734 888;
  • 3) 0.800 000 000 000 000 710 542 734 888 × 2 = 1 + 0.600 000 000 000 001 421 085 469 776;
  • 4) 0.600 000 000 000 001 421 085 469 776 × 2 = 1 + 0.200 000 000 000 002 842 170 939 552;
  • 5) 0.200 000 000 000 002 842 170 939 552 × 2 = 0 + 0.400 000 000 000 005 684 341 879 104;
  • 6) 0.400 000 000 000 005 684 341 879 104 × 2 = 0 + 0.800 000 000 000 011 368 683 758 208;
  • 7) 0.800 000 000 000 011 368 683 758 208 × 2 = 1 + 0.600 000 000 000 022 737 367 516 416;
  • 8) 0.600 000 000 000 022 737 367 516 416 × 2 = 1 + 0.200 000 000 000 045 474 735 032 832;
  • 9) 0.200 000 000 000 045 474 735 032 832 × 2 = 0 + 0.400 000 000 000 090 949 470 065 664;
  • 10) 0.400 000 000 000 090 949 470 065 664 × 2 = 0 + 0.800 000 000 000 181 898 940 131 328;
  • 11) 0.800 000 000 000 181 898 940 131 328 × 2 = 1 + 0.600 000 000 000 363 797 880 262 656;
  • 12) 0.600 000 000 000 363 797 880 262 656 × 2 = 1 + 0.200 000 000 000 727 595 760 525 312;
  • 13) 0.200 000 000 000 727 595 760 525 312 × 2 = 0 + 0.400 000 000 001 455 191 521 050 624;
  • 14) 0.400 000 000 001 455 191 521 050 624 × 2 = 0 + 0.800 000 000 002 910 383 042 101 248;
  • 15) 0.800 000 000 002 910 383 042 101 248 × 2 = 1 + 0.600 000 000 005 820 766 084 202 496;
  • 16) 0.600 000 000 005 820 766 084 202 496 × 2 = 1 + 0.200 000 000 011 641 532 168 404 992;
  • 17) 0.200 000 000 011 641 532 168 404 992 × 2 = 0 + 0.400 000 000 023 283 064 336 809 984;
  • 18) 0.400 000 000 023 283 064 336 809 984 × 2 = 0 + 0.800 000 000 046 566 128 673 619 968;
  • 19) 0.800 000 000 046 566 128 673 619 968 × 2 = 1 + 0.600 000 000 093 132 257 347 239 936;
  • 20) 0.600 000 000 093 132 257 347 239 936 × 2 = 1 + 0.200 000 000 186 264 514 694 479 872;
  • 21) 0.200 000 000 186 264 514 694 479 872 × 2 = 0 + 0.400 000 000 372 529 029 388 959 744;
  • 22) 0.400 000 000 372 529 029 388 959 744 × 2 = 0 + 0.800 000 000 745 058 058 777 919 488;
  • 23) 0.800 000 000 745 058 058 777 919 488 × 2 = 1 + 0.600 000 001 490 116 117 555 838 976;
  • 24) 0.600 000 001 490 116 117 555 838 976 × 2 = 1 + 0.200 000 002 980 232 235 111 677 952;
  • 25) 0.200 000 002 980 232 235 111 677 952 × 2 = 0 + 0.400 000 005 960 464 470 223 355 904;
  • 26) 0.400 000 005 960 464 470 223 355 904 × 2 = 0 + 0.800 000 011 920 928 940 446 711 808;
  • 27) 0.800 000 011 920 928 940 446 711 808 × 2 = 1 + 0.600 000 023 841 857 880 893 423 616;
  • 28) 0.600 000 023 841 857 880 893 423 616 × 2 = 1 + 0.200 000 047 683 715 761 786 847 232;
  • 29) 0.200 000 047 683 715 761 786 847 232 × 2 = 0 + 0.400 000 095 367 431 523 573 694 464;
  • 30) 0.400 000 095 367 431 523 573 694 464 × 2 = 0 + 0.800 000 190 734 863 047 147 388 928;
  • 31) 0.800 000 190 734 863 047 147 388 928 × 2 = 1 + 0.600 000 381 469 726 094 294 777 856;
  • 32) 0.600 000 381 469 726 094 294 777 856 × 2 = 1 + 0.200 000 762 939 452 188 589 555 712;
  • 33) 0.200 000 762 939 452 188 589 555 712 × 2 = 0 + 0.400 001 525 878 904 377 179 111 424;
  • 34) 0.400 001 525 878 904 377 179 111 424 × 2 = 0 + 0.800 003 051 757 808 754 358 222 848;
  • 35) 0.800 003 051 757 808 754 358 222 848 × 2 = 1 + 0.600 006 103 515 617 508 716 445 696;
  • 36) 0.600 006 103 515 617 508 716 445 696 × 2 = 1 + 0.200 012 207 031 235 017 432 891 392;
  • 37) 0.200 012 207 031 235 017 432 891 392 × 2 = 0 + 0.400 024 414 062 470 034 865 782 784;
  • 38) 0.400 024 414 062 470 034 865 782 784 × 2 = 0 + 0.800 048 828 124 940 069 731 565 568;
  • 39) 0.800 048 828 124 940 069 731 565 568 × 2 = 1 + 0.600 097 656 249 880 139 463 131 136;
  • 40) 0.600 097 656 249 880 139 463 131 136 × 2 = 1 + 0.200 195 312 499 760 278 926 262 272;
  • 41) 0.200 195 312 499 760 278 926 262 272 × 2 = 0 + 0.400 390 624 999 520 557 852 524 544;
  • 42) 0.400 390 624 999 520 557 852 524 544 × 2 = 0 + 0.800 781 249 999 041 115 705 049 088;
  • 43) 0.800 781 249 999 041 115 705 049 088 × 2 = 1 + 0.601 562 499 998 082 231 410 098 176;
  • 44) 0.601 562 499 998 082 231 410 098 176 × 2 = 1 + 0.203 124 999 996 164 462 820 196 352;
  • 45) 0.203 124 999 996 164 462 820 196 352 × 2 = 0 + 0.406 249 999 992 328 925 640 392 704;
  • 46) 0.406 249 999 992 328 925 640 392 704 × 2 = 0 + 0.812 499 999 984 657 851 280 785 408;
  • 47) 0.812 499 999 984 657 851 280 785 408 × 2 = 1 + 0.624 999 999 969 315 702 561 570 816;
  • 48) 0.624 999 999 969 315 702 561 570 816 × 2 = 1 + 0.249 999 999 938 631 405 123 141 632;
  • 49) 0.249 999 999 938 631 405 123 141 632 × 2 = 0 + 0.499 999 999 877 262 810 246 283 264;
  • 50) 0.499 999 999 877 262 810 246 283 264 × 2 = 0 + 0.999 999 999 754 525 620 492 566 528;
  • 51) 0.999 999 999 754 525 620 492 566 528 × 2 = 1 + 0.999 999 999 509 051 240 985 133 056;
  • 52) 0.999 999 999 509 051 240 985 133 056 × 2 = 1 + 0.999 999 999 018 102 481 970 266 112;
  • 53) 0.999 999 999 018 102 481 970 266 112 × 2 = 1 + 0.999 999 998 036 204 963 940 532 224;

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).


4. 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:


0.200 000 000 000 000 177 635 683 722(10) =

0.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1(2)

5. Positive number before normalization:

7.200 000 000 000 000 177 635 683 722(10) =

111.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


7.200 000 000 000 000 177 635 683 722(10) =

111.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1(2) =

111.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1(2) × 20 =

1.1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 111(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(10)


9. 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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1025(10) =

100 0000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 111 =

1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0001

Mantissa (52 bits) =
1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


Decimal number 7.200 000 000 000 000 177 635 683 722 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100

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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