-80.298 375 123 480 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -80.298 375 123 480 7(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
-80.298 375 123 480 7(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:

|-80.298 375 123 480 7| = 80.298 375 123 480 7


2. First, convert to binary (in base 2) the integer part: 80.
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;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

80(10) =

101 0000(2)


4. Convert to binary (base 2) the fractional part: 0.298 375 123 480 7.

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) 0.298 375 123 480 7 × 2 = 0 + 0.596 750 246 961 4;
  • 2) 0.596 750 246 961 4 × 2 = 1 + 0.193 500 493 922 8;
  • 3) 0.193 500 493 922 8 × 2 = 0 + 0.387 000 987 845 6;
  • 4) 0.387 000 987 845 6 × 2 = 0 + 0.774 001 975 691 2;
  • 5) 0.774 001 975 691 2 × 2 = 1 + 0.548 003 951 382 4;
  • 6) 0.548 003 951 382 4 × 2 = 1 + 0.096 007 902 764 8;
  • 7) 0.096 007 902 764 8 × 2 = 0 + 0.192 015 805 529 6;
  • 8) 0.192 015 805 529 6 × 2 = 0 + 0.384 031 611 059 2;
  • 9) 0.384 031 611 059 2 × 2 = 0 + 0.768 063 222 118 4;
  • 10) 0.768 063 222 118 4 × 2 = 1 + 0.536 126 444 236 8;
  • 11) 0.536 126 444 236 8 × 2 = 1 + 0.072 252 888 473 6;
  • 12) 0.072 252 888 473 6 × 2 = 0 + 0.144 505 776 947 2;
  • 13) 0.144 505 776 947 2 × 2 = 0 + 0.289 011 553 894 4;
  • 14) 0.289 011 553 894 4 × 2 = 0 + 0.578 023 107 788 8;
  • 15) 0.578 023 107 788 8 × 2 = 1 + 0.156 046 215 577 6;
  • 16) 0.156 046 215 577 6 × 2 = 0 + 0.312 092 431 155 2;
  • 17) 0.312 092 431 155 2 × 2 = 0 + 0.624 184 862 310 4;
  • 18) 0.624 184 862 310 4 × 2 = 1 + 0.248 369 724 620 8;
  • 19) 0.248 369 724 620 8 × 2 = 0 + 0.496 739 449 241 6;
  • 20) 0.496 739 449 241 6 × 2 = 0 + 0.993 478 898 483 2;
  • 21) 0.993 478 898 483 2 × 2 = 1 + 0.986 957 796 966 4;
  • 22) 0.986 957 796 966 4 × 2 = 1 + 0.973 915 593 932 8;
  • 23) 0.973 915 593 932 8 × 2 = 1 + 0.947 831 187 865 6;
  • 24) 0.947 831 187 865 6 × 2 = 1 + 0.895 662 375 731 2;
  • 25) 0.895 662 375 731 2 × 2 = 1 + 0.791 324 751 462 4;
  • 26) 0.791 324 751 462 4 × 2 = 1 + 0.582 649 502 924 8;
  • 27) 0.582 649 502 924 8 × 2 = 1 + 0.165 299 005 849 6;
  • 28) 0.165 299 005 849 6 × 2 = 0 + 0.330 598 011 699 2;
  • 29) 0.330 598 011 699 2 × 2 = 0 + 0.661 196 023 398 4;
  • 30) 0.661 196 023 398 4 × 2 = 1 + 0.322 392 046 796 8;
  • 31) 0.322 392 046 796 8 × 2 = 0 + 0.644 784 093 593 6;
  • 32) 0.644 784 093 593 6 × 2 = 1 + 0.289 568 187 187 2;
  • 33) 0.289 568 187 187 2 × 2 = 0 + 0.579 136 374 374 4;
  • 34) 0.579 136 374 374 4 × 2 = 1 + 0.158 272 748 748 8;
  • 35) 0.158 272 748 748 8 × 2 = 0 + 0.316 545 497 497 6;
  • 36) 0.316 545 497 497 6 × 2 = 0 + 0.633 090 994 995 2;
  • 37) 0.633 090 994 995 2 × 2 = 1 + 0.266 181 989 990 4;
  • 38) 0.266 181 989 990 4 × 2 = 0 + 0.532 363 979 980 8;
  • 39) 0.532 363 979 980 8 × 2 = 1 + 0.064 727 959 961 6;
  • 40) 0.064 727 959 961 6 × 2 = 0 + 0.129 455 919 923 2;
  • 41) 0.129 455 919 923 2 × 2 = 0 + 0.258 911 839 846 4;
  • 42) 0.258 911 839 846 4 × 2 = 0 + 0.517 823 679 692 8;
  • 43) 0.517 823 679 692 8 × 2 = 1 + 0.035 647 359 385 6;
  • 44) 0.035 647 359 385 6 × 2 = 0 + 0.071 294 718 771 2;
  • 45) 0.071 294 718 771 2 × 2 = 0 + 0.142 589 437 542 4;
  • 46) 0.142 589 437 542 4 × 2 = 0 + 0.285 178 875 084 8;
  • 47) 0.285 178 875 084 8 × 2 = 0 + 0.570 357 750 169 6;
  • 48) 0.570 357 750 169 6 × 2 = 1 + 0.140 715 500 339 2;
  • 49) 0.140 715 500 339 2 × 2 = 0 + 0.281 431 000 678 4;
  • 50) 0.281 431 000 678 4 × 2 = 0 + 0.562 862 001 356 8;
  • 51) 0.562 862 001 356 8 × 2 = 1 + 0.125 724 002 713 6;
  • 52) 0.125 724 002 713 6 × 2 = 0 + 0.251 448 005 427 2;
  • 53) 0.251 448 005 427 2 × 2 = 0 + 0.502 896 010 854 4;

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:


0.298 375 123 480 7(10) =

0.0100 1100 0110 0010 0100 1111 1110 0101 0100 1010 0010 0001 0010 0(2)

6. Positive number before normalization:

80.298 375 123 480 7(10) =

101 0000.0100 1100 0110 0010 0100 1111 1110 0101 0100 1010 0010 0001 0010 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


80.298 375 123 480 7(10) =

101 0000.0100 1100 0110 0010 0100 1111 1110 0101 0100 1010 0010 0001 0010 0(2) =

101 0000.0100 1100 0110 0010 0100 1111 1110 0101 0100 1010 0010 0001 0010 0(2) × 20 =

1.0100 0001 0011 0001 1000 1001 0011 1111 1001 0101 0010 1000 1000 0100 100(2) × 26


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):
1.0100 0001 0011 0001 1000 1001 0011 1111 1001 0101 0010 1000 1000 0100 100


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) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 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;

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) =

1029(10) =

100 0000 0101(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, 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. 0100 0001 0011 0001 1000 1001 0011 1111 1001 0101 0010 1000 1000 010 0100 =

0100 0001 0011 0001 1000 1001 0011 1111 1001 0101 0010 1000 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) =
100 0000 0101

Mantissa (52 bits) =
0100 0001 0011 0001 1000 1001 0011 1111 1001 0101 0010 1000 1000


Decimal number -80.298 375 123 480 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 0100 0001 0011 0001 1000 1001 0011 1111 1001 0101 0010 1000 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