1.745 459 324 169 999 826 274 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 274(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
1.745 459 324 169 999 826 274(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: 0.745 459 324 169 999 826 274.

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) 0.745 459 324 169 999 826 274 × 2 = 1 + 0.490 918 648 339 999 652 548;
  • 2) 0.490 918 648 339 999 652 548 × 2 = 0 + 0.981 837 296 679 999 305 096;
  • 3) 0.981 837 296 679 999 305 096 × 2 = 1 + 0.963 674 593 359 998 610 192;
  • 4) 0.963 674 593 359 998 610 192 × 2 = 1 + 0.927 349 186 719 997 220 384;
  • 5) 0.927 349 186 719 997 220 384 × 2 = 1 + 0.854 698 373 439 994 440 768;
  • 6) 0.854 698 373 439 994 440 768 × 2 = 1 + 0.709 396 746 879 988 881 536;
  • 7) 0.709 396 746 879 988 881 536 × 2 = 1 + 0.418 793 493 759 977 763 072;
  • 8) 0.418 793 493 759 977 763 072 × 2 = 0 + 0.837 586 987 519 955 526 144;
  • 9) 0.837 586 987 519 955 526 144 × 2 = 1 + 0.675 173 975 039 911 052 288;
  • 10) 0.675 173 975 039 911 052 288 × 2 = 1 + 0.350 347 950 079 822 104 576;
  • 11) 0.350 347 950 079 822 104 576 × 2 = 0 + 0.700 695 900 159 644 209 152;
  • 12) 0.700 695 900 159 644 209 152 × 2 = 1 + 0.401 391 800 319 288 418 304;
  • 13) 0.401 391 800 319 288 418 304 × 2 = 0 + 0.802 783 600 638 576 836 608;
  • 14) 0.802 783 600 638 576 836 608 × 2 = 1 + 0.605 567 201 277 153 673 216;
  • 15) 0.605 567 201 277 153 673 216 × 2 = 1 + 0.211 134 402 554 307 346 432;
  • 16) 0.211 134 402 554 307 346 432 × 2 = 0 + 0.422 268 805 108 614 692 864;
  • 17) 0.422 268 805 108 614 692 864 × 2 = 0 + 0.844 537 610 217 229 385 728;
  • 18) 0.844 537 610 217 229 385 728 × 2 = 1 + 0.689 075 220 434 458 771 456;
  • 19) 0.689 075 220 434 458 771 456 × 2 = 1 + 0.378 150 440 868 917 542 912;
  • 20) 0.378 150 440 868 917 542 912 × 2 = 0 + 0.756 300 881 737 835 085 824;
  • 21) 0.756 300 881 737 835 085 824 × 2 = 1 + 0.512 601 763 475 670 171 648;
  • 22) 0.512 601 763 475 670 171 648 × 2 = 1 + 0.025 203 526 951 340 343 296;
  • 23) 0.025 203 526 951 340 343 296 × 2 = 0 + 0.050 407 053 902 680 686 592;
  • 24) 0.050 407 053 902 680 686 592 × 2 = 0 + 0.100 814 107 805 361 373 184;
  • 25) 0.100 814 107 805 361 373 184 × 2 = 0 + 0.201 628 215 610 722 746 368;
  • 26) 0.201 628 215 610 722 746 368 × 2 = 0 + 0.403 256 431 221 445 492 736;
  • 27) 0.403 256 431 221 445 492 736 × 2 = 0 + 0.806 512 862 442 890 985 472;
  • 28) 0.806 512 862 442 890 985 472 × 2 = 1 + 0.613 025 724 885 781 970 944;
  • 29) 0.613 025 724 885 781 970 944 × 2 = 1 + 0.226 051 449 771 563 941 888;
  • 30) 0.226 051 449 771 563 941 888 × 2 = 0 + 0.452 102 899 543 127 883 776;
  • 31) 0.452 102 899 543 127 883 776 × 2 = 0 + 0.904 205 799 086 255 767 552;
  • 32) 0.904 205 799 086 255 767 552 × 2 = 1 + 0.808 411 598 172 511 535 104;
  • 33) 0.808 411 598 172 511 535 104 × 2 = 1 + 0.616 823 196 345 023 070 208;
  • 34) 0.616 823 196 345 023 070 208 × 2 = 1 + 0.233 646 392 690 046 140 416;
  • 35) 0.233 646 392 690 046 140 416 × 2 = 0 + 0.467 292 785 380 092 280 832;
  • 36) 0.467 292 785 380 092 280 832 × 2 = 0 + 0.934 585 570 760 184 561 664;
  • 37) 0.934 585 570 760 184 561 664 × 2 = 1 + 0.869 171 141 520 369 123 328;
  • 38) 0.869 171 141 520 369 123 328 × 2 = 1 + 0.738 342 283 040 738 246 656;
  • 39) 0.738 342 283 040 738 246 656 × 2 = 1 + 0.476 684 566 081 476 493 312;
  • 40) 0.476 684 566 081 476 493 312 × 2 = 0 + 0.953 369 132 162 952 986 624;
  • 41) 0.953 369 132 162 952 986 624 × 2 = 1 + 0.906 738 264 325 905 973 248;
  • 42) 0.906 738 264 325 905 973 248 × 2 = 1 + 0.813 476 528 651 811 946 496;
  • 43) 0.813 476 528 651 811 946 496 × 2 = 1 + 0.626 953 057 303 623 892 992;
  • 44) 0.626 953 057 303 623 892 992 × 2 = 1 + 0.253 906 114 607 247 785 984;
  • 45) 0.253 906 114 607 247 785 984 × 2 = 0 + 0.507 812 229 214 495 571 968;
  • 46) 0.507 812 229 214 495 571 968 × 2 = 1 + 0.015 624 458 428 991 143 936;
  • 47) 0.015 624 458 428 991 143 936 × 2 = 0 + 0.031 248 916 857 982 287 872;
  • 48) 0.031 248 916 857 982 287 872 × 2 = 0 + 0.062 497 833 715 964 575 744;
  • 49) 0.062 497 833 715 964 575 744 × 2 = 0 + 0.124 995 667 431 929 151 488;
  • 50) 0.124 995 667 431 929 151 488 × 2 = 0 + 0.249 991 334 863 858 302 976;
  • 51) 0.249 991 334 863 858 302 976 × 2 = 0 + 0.499 982 669 727 716 605 952;
  • 52) 0.499 982 669 727 716 605 952 × 2 = 0 + 0.999 965 339 455 433 211 904;
  • 53) 0.999 965 339 455 433 211 904 × 2 = 1 + 0.999 930 678 910 866 423 808;

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:


0.745 459 324 169 999 826 274(10) =


0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 274(10) =


1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 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:


1.745 459 324 169 999 826 274(10) =


1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =


1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 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 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 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 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =


1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 274 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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