17.035 500 000 000 136 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 500 000 000 136(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
17.035 500 000 000 136(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: 17.
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;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

17(10) =

1 0001(2)


3. Convert to binary (base 2) the fractional part: 0.035 500 000 000 136.

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) 0.035 500 000 000 136 × 2 = 0 + 0.071 000 000 000 272;
  • 2) 0.071 000 000 000 272 × 2 = 0 + 0.142 000 000 000 544;
  • 3) 0.142 000 000 000 544 × 2 = 0 + 0.284 000 000 001 088;
  • 4) 0.284 000 000 001 088 × 2 = 0 + 0.568 000 000 002 176;
  • 5) 0.568 000 000 002 176 × 2 = 1 + 0.136 000 000 004 352;
  • 6) 0.136 000 000 004 352 × 2 = 0 + 0.272 000 000 008 704;
  • 7) 0.272 000 000 008 704 × 2 = 0 + 0.544 000 000 017 408;
  • 8) 0.544 000 000 017 408 × 2 = 1 + 0.088 000 000 034 816;
  • 9) 0.088 000 000 034 816 × 2 = 0 + 0.176 000 000 069 632;
  • 10) 0.176 000 000 069 632 × 2 = 0 + 0.352 000 000 139 264;
  • 11) 0.352 000 000 139 264 × 2 = 0 + 0.704 000 000 278 528;
  • 12) 0.704 000 000 278 528 × 2 = 1 + 0.408 000 000 557 056;
  • 13) 0.408 000 000 557 056 × 2 = 0 + 0.816 000 001 114 112;
  • 14) 0.816 000 001 114 112 × 2 = 1 + 0.632 000 002 228 224;
  • 15) 0.632 000 002 228 224 × 2 = 1 + 0.264 000 004 456 448;
  • 16) 0.264 000 004 456 448 × 2 = 0 + 0.528 000 008 912 896;
  • 17) 0.528 000 008 912 896 × 2 = 1 + 0.056 000 017 825 792;
  • 18) 0.056 000 017 825 792 × 2 = 0 + 0.112 000 035 651 584;
  • 19) 0.112 000 035 651 584 × 2 = 0 + 0.224 000 071 303 168;
  • 20) 0.224 000 071 303 168 × 2 = 0 + 0.448 000 142 606 336;
  • 21) 0.448 000 142 606 336 × 2 = 0 + 0.896 000 285 212 672;
  • 22) 0.896 000 285 212 672 × 2 = 1 + 0.792 000 570 425 344;
  • 23) 0.792 000 570 425 344 × 2 = 1 + 0.584 001 140 850 688;
  • 24) 0.584 001 140 850 688 × 2 = 1 + 0.168 002 281 701 376;
  • 25) 0.168 002 281 701 376 × 2 = 0 + 0.336 004 563 402 752;
  • 26) 0.336 004 563 402 752 × 2 = 0 + 0.672 009 126 805 504;
  • 27) 0.672 009 126 805 504 × 2 = 1 + 0.344 018 253 611 008;
  • 28) 0.344 018 253 611 008 × 2 = 0 + 0.688 036 507 222 016;
  • 29) 0.688 036 507 222 016 × 2 = 1 + 0.376 073 014 444 032;
  • 30) 0.376 073 014 444 032 × 2 = 0 + 0.752 146 028 888 064;
  • 31) 0.752 146 028 888 064 × 2 = 1 + 0.504 292 057 776 128;
  • 32) 0.504 292 057 776 128 × 2 = 1 + 0.008 584 115 552 256;
  • 33) 0.008 584 115 552 256 × 2 = 0 + 0.017 168 231 104 512;
  • 34) 0.017 168 231 104 512 × 2 = 0 + 0.034 336 462 209 024;
  • 35) 0.034 336 462 209 024 × 2 = 0 + 0.068 672 924 418 048;
  • 36) 0.068 672 924 418 048 × 2 = 0 + 0.137 345 848 836 096;
  • 37) 0.137 345 848 836 096 × 2 = 0 + 0.274 691 697 672 192;
  • 38) 0.274 691 697 672 192 × 2 = 0 + 0.549 383 395 344 384;
  • 39) 0.549 383 395 344 384 × 2 = 1 + 0.098 766 790 688 768;
  • 40) 0.098 766 790 688 768 × 2 = 0 + 0.197 533 581 377 536;
  • 41) 0.197 533 581 377 536 × 2 = 0 + 0.395 067 162 755 072;
  • 42) 0.395 067 162 755 072 × 2 = 0 + 0.790 134 325 510 144;
  • 43) 0.790 134 325 510 144 × 2 = 1 + 0.580 268 651 020 288;
  • 44) 0.580 268 651 020 288 × 2 = 1 + 0.160 537 302 040 576;
  • 45) 0.160 537 302 040 576 × 2 = 0 + 0.321 074 604 081 152;
  • 46) 0.321 074 604 081 152 × 2 = 0 + 0.642 149 208 162 304;
  • 47) 0.642 149 208 162 304 × 2 = 1 + 0.284 298 416 324 608;
  • 48) 0.284 298 416 324 608 × 2 = 0 + 0.568 596 832 649 216;
  • 49) 0.568 596 832 649 216 × 2 = 1 + 0.137 193 665 298 432;
  • 50) 0.137 193 665 298 432 × 2 = 0 + 0.274 387 330 596 864;
  • 51) 0.274 387 330 596 864 × 2 = 0 + 0.548 774 661 193 728;
  • 52) 0.548 774 661 193 728 × 2 = 1 + 0.097 549 322 387 456;
  • 53) 0.097 549 322 387 456 × 2 = 0 + 0.195 098 644 774 912;

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:


0.035 500 000 000 136(10) =

0.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1001 0(2)

5. Positive number before normalization:

17.035 500 000 000 136(10) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


17.035 500 000 000 136(10) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1001 0(2) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1001 0(2) × 20 =

1.0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1001 0(2) × 24


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): 4

Mantissa (not normalized):
1.0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1001 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 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) =

1027(10) =

100 0000 0011(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. 0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010 1 0010 =

0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010


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 0011

Mantissa (52 bits) =
0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010


Decimal number 17.035 500 000 000 136 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0011 0010

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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