1.703 864 207 537 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.703 864 207 537(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
1.703 864 207 537(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.703 864 207 537.

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) 0.703 864 207 537 × 2 = 1 + 0.407 728 415 074;
  • 2) 0.407 728 415 074 × 2 = 0 + 0.815 456 830 148;
  • 3) 0.815 456 830 148 × 2 = 1 + 0.630 913 660 296;
  • 4) 0.630 913 660 296 × 2 = 1 + 0.261 827 320 592;
  • 5) 0.261 827 320 592 × 2 = 0 + 0.523 654 641 184;
  • 6) 0.523 654 641 184 × 2 = 1 + 0.047 309 282 368;
  • 7) 0.047 309 282 368 × 2 = 0 + 0.094 618 564 736;
  • 8) 0.094 618 564 736 × 2 = 0 + 0.189 237 129 472;
  • 9) 0.189 237 129 472 × 2 = 0 + 0.378 474 258 944;
  • 10) 0.378 474 258 944 × 2 = 0 + 0.756 948 517 888;
  • 11) 0.756 948 517 888 × 2 = 1 + 0.513 897 035 776;
  • 12) 0.513 897 035 776 × 2 = 1 + 0.027 794 071 552;
  • 13) 0.027 794 071 552 × 2 = 0 + 0.055 588 143 104;
  • 14) 0.055 588 143 104 × 2 = 0 + 0.111 176 286 208;
  • 15) 0.111 176 286 208 × 2 = 0 + 0.222 352 572 416;
  • 16) 0.222 352 572 416 × 2 = 0 + 0.444 705 144 832;
  • 17) 0.444 705 144 832 × 2 = 0 + 0.889 410 289 664;
  • 18) 0.889 410 289 664 × 2 = 1 + 0.778 820 579 328;
  • 19) 0.778 820 579 328 × 2 = 1 + 0.557 641 158 656;
  • 20) 0.557 641 158 656 × 2 = 1 + 0.115 282 317 312;
  • 21) 0.115 282 317 312 × 2 = 0 + 0.230 564 634 624;
  • 22) 0.230 564 634 624 × 2 = 0 + 0.461 129 269 248;
  • 23) 0.461 129 269 248 × 2 = 0 + 0.922 258 538 496;
  • 24) 0.922 258 538 496 × 2 = 1 + 0.844 517 076 992;
  • 25) 0.844 517 076 992 × 2 = 1 + 0.689 034 153 984;
  • 26) 0.689 034 153 984 × 2 = 1 + 0.378 068 307 968;
  • 27) 0.378 068 307 968 × 2 = 0 + 0.756 136 615 936;
  • 28) 0.756 136 615 936 × 2 = 1 + 0.512 273 231 872;
  • 29) 0.512 273 231 872 × 2 = 1 + 0.024 546 463 744;
  • 30) 0.024 546 463 744 × 2 = 0 + 0.049 092 927 488;
  • 31) 0.049 092 927 488 × 2 = 0 + 0.098 185 854 976;
  • 32) 0.098 185 854 976 × 2 = 0 + 0.196 371 709 952;
  • 33) 0.196 371 709 952 × 2 = 0 + 0.392 743 419 904;
  • 34) 0.392 743 419 904 × 2 = 0 + 0.785 486 839 808;
  • 35) 0.785 486 839 808 × 2 = 1 + 0.570 973 679 616;
  • 36) 0.570 973 679 616 × 2 = 1 + 0.141 947 359 232;
  • 37) 0.141 947 359 232 × 2 = 0 + 0.283 894 718 464;
  • 38) 0.283 894 718 464 × 2 = 0 + 0.567 789 436 928;
  • 39) 0.567 789 436 928 × 2 = 1 + 0.135 578 873 856;
  • 40) 0.135 578 873 856 × 2 = 0 + 0.271 157 747 712;
  • 41) 0.271 157 747 712 × 2 = 0 + 0.542 315 495 424;
  • 42) 0.542 315 495 424 × 2 = 1 + 0.084 630 990 848;
  • 43) 0.084 630 990 848 × 2 = 0 + 0.169 261 981 696;
  • 44) 0.169 261 981 696 × 2 = 0 + 0.338 523 963 392;
  • 45) 0.338 523 963 392 × 2 = 0 + 0.677 047 926 784;
  • 46) 0.677 047 926 784 × 2 = 1 + 0.354 095 853 568;
  • 47) 0.354 095 853 568 × 2 = 0 + 0.708 191 707 136;
  • 48) 0.708 191 707 136 × 2 = 1 + 0.416 383 414 272;
  • 49) 0.416 383 414 272 × 2 = 0 + 0.832 766 828 544;
  • 50) 0.832 766 828 544 × 2 = 1 + 0.665 533 657 088;
  • 51) 0.665 533 657 088 × 2 = 1 + 0.331 067 314 176;
  • 52) 0.331 067 314 176 × 2 = 0 + 0.662 134 628 352;
  • 53) 0.662 134 628 352 × 2 = 1 + 0.324 269 256 704;

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:


0.703 864 207 537(10) =

0.1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110 1(2)

5. Positive number before normalization:

1.703 864 207 537(10) =

1.1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.703 864 207 537(10) =

1.1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110 1(2) =

1.1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110 1(2) × 20


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): 0

Mantissa (not normalized):
1.1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

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) =

1023(10) =

011 1111 1111(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. 1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110 1 =

1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110


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) =
011 1111 1111

Mantissa (52 bits) =
1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110


Decimal number 1.703 864 207 537 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 0100 0011 0000 0111 0001 1101 1000 0011 0010 0100 0101 0110

» Convert decimal 1.703 864 207 44 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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