-0.000 000 000 742 147 676 624 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 624 3(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 624 3(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 624 3| = 0.000 000 000 742 147 676 624 3


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 624 3.

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) 0.000 000 000 742 147 676 624 3 × 2 = 0 + 0.000 000 001 484 295 353 248 6;
  • 2) 0.000 000 001 484 295 353 248 6 × 2 = 0 + 0.000 000 002 968 590 706 497 2;
  • 3) 0.000 000 002 968 590 706 497 2 × 2 = 0 + 0.000 000 005 937 181 412 994 4;
  • 4) 0.000 000 005 937 181 412 994 4 × 2 = 0 + 0.000 000 011 874 362 825 988 8;
  • 5) 0.000 000 011 874 362 825 988 8 × 2 = 0 + 0.000 000 023 748 725 651 977 6;
  • 6) 0.000 000 023 748 725 651 977 6 × 2 = 0 + 0.000 000 047 497 451 303 955 2;
  • 7) 0.000 000 047 497 451 303 955 2 × 2 = 0 + 0.000 000 094 994 902 607 910 4;
  • 8) 0.000 000 094 994 902 607 910 4 × 2 = 0 + 0.000 000 189 989 805 215 820 8;
  • 9) 0.000 000 189 989 805 215 820 8 × 2 = 0 + 0.000 000 379 979 610 431 641 6;
  • 10) 0.000 000 379 979 610 431 641 6 × 2 = 0 + 0.000 000 759 959 220 863 283 2;
  • 11) 0.000 000 759 959 220 863 283 2 × 2 = 0 + 0.000 001 519 918 441 726 566 4;
  • 12) 0.000 001 519 918 441 726 566 4 × 2 = 0 + 0.000 003 039 836 883 453 132 8;
  • 13) 0.000 003 039 836 883 453 132 8 × 2 = 0 + 0.000 006 079 673 766 906 265 6;
  • 14) 0.000 006 079 673 766 906 265 6 × 2 = 0 + 0.000 012 159 347 533 812 531 2;
  • 15) 0.000 012 159 347 533 812 531 2 × 2 = 0 + 0.000 024 318 695 067 625 062 4;
  • 16) 0.000 024 318 695 067 625 062 4 × 2 = 0 + 0.000 048 637 390 135 250 124 8;
  • 17) 0.000 048 637 390 135 250 124 8 × 2 = 0 + 0.000 097 274 780 270 500 249 6;
  • 18) 0.000 097 274 780 270 500 249 6 × 2 = 0 + 0.000 194 549 560 541 000 499 2;
  • 19) 0.000 194 549 560 541 000 499 2 × 2 = 0 + 0.000 389 099 121 082 000 998 4;
  • 20) 0.000 389 099 121 082 000 998 4 × 2 = 0 + 0.000 778 198 242 164 001 996 8;
  • 21) 0.000 778 198 242 164 001 996 8 × 2 = 0 + 0.001 556 396 484 328 003 993 6;
  • 22) 0.001 556 396 484 328 003 993 6 × 2 = 0 + 0.003 112 792 968 656 007 987 2;
  • 23) 0.003 112 792 968 656 007 987 2 × 2 = 0 + 0.006 225 585 937 312 015 974 4;
  • 24) 0.006 225 585 937 312 015 974 4 × 2 = 0 + 0.012 451 171 874 624 031 948 8;
  • 25) 0.012 451 171 874 624 031 948 8 × 2 = 0 + 0.024 902 343 749 248 063 897 6;
  • 26) 0.024 902 343 749 248 063 897 6 × 2 = 0 + 0.049 804 687 498 496 127 795 2;
  • 27) 0.049 804 687 498 496 127 795 2 × 2 = 0 + 0.099 609 374 996 992 255 590 4;
  • 28) 0.099 609 374 996 992 255 590 4 × 2 = 0 + 0.199 218 749 993 984 511 180 8;
  • 29) 0.199 218 749 993 984 511 180 8 × 2 = 0 + 0.398 437 499 987 969 022 361 6;
  • 30) 0.398 437 499 987 969 022 361 6 × 2 = 0 + 0.796 874 999 975 938 044 723 2;
  • 31) 0.796 874 999 975 938 044 723 2 × 2 = 1 + 0.593 749 999 951 876 089 446 4;
  • 32) 0.593 749 999 951 876 089 446 4 × 2 = 1 + 0.187 499 999 903 752 178 892 8;
  • 33) 0.187 499 999 903 752 178 892 8 × 2 = 0 + 0.374 999 999 807 504 357 785 6;
  • 34) 0.374 999 999 807 504 357 785 6 × 2 = 0 + 0.749 999 999 615 008 715 571 2;
  • 35) 0.749 999 999 615 008 715 571 2 × 2 = 1 + 0.499 999 999 230 017 431 142 4;
  • 36) 0.499 999 999 230 017 431 142 4 × 2 = 0 + 0.999 999 998 460 034 862 284 8;
  • 37) 0.999 999 998 460 034 862 284 8 × 2 = 1 + 0.999 999 996 920 069 724 569 6;
  • 38) 0.999 999 996 920 069 724 569 6 × 2 = 1 + 0.999 999 993 840 139 449 139 2;
  • 39) 0.999 999 993 840 139 449 139 2 × 2 = 1 + 0.999 999 987 680 278 898 278 4;
  • 40) 0.999 999 987 680 278 898 278 4 × 2 = 1 + 0.999 999 975 360 557 796 556 8;
  • 41) 0.999 999 975 360 557 796 556 8 × 2 = 1 + 0.999 999 950 721 115 593 113 6;
  • 42) 0.999 999 950 721 115 593 113 6 × 2 = 1 + 0.999 999 901 442 231 186 227 2;
  • 43) 0.999 999 901 442 231 186 227 2 × 2 = 1 + 0.999 999 802 884 462 372 454 4;
  • 44) 0.999 999 802 884 462 372 454 4 × 2 = 1 + 0.999 999 605 768 924 744 908 8;
  • 45) 0.999 999 605 768 924 744 908 8 × 2 = 1 + 0.999 999 211 537 849 489 817 6;
  • 46) 0.999 999 211 537 849 489 817 6 × 2 = 1 + 0.999 998 423 075 698 979 635 2;
  • 47) 0.999 998 423 075 698 979 635 2 × 2 = 1 + 0.999 996 846 151 397 959 270 4;
  • 48) 0.999 996 846 151 397 959 270 4 × 2 = 1 + 0.999 993 692 302 795 918 540 8;
  • 49) 0.999 993 692 302 795 918 540 8 × 2 = 1 + 0.999 987 384 605 591 837 081 6;
  • 50) 0.999 987 384 605 591 837 081 6 × 2 = 1 + 0.999 974 769 211 183 674 163 2;
  • 51) 0.999 974 769 211 183 674 163 2 × 2 = 1 + 0.999 949 538 422 367 348 326 4;
  • 52) 0.999 949 538 422 367 348 326 4 × 2 = 1 + 0.999 899 076 844 734 696 652 8;
  • 53) 0.999 899 076 844 734 696 652 8 × 2 = 1 + 0.999 798 153 689 469 393 305 6;
  • 54) 0.999 798 153 689 469 393 305 6 × 2 = 1 + 0.999 596 307 378 938 786 611 2;

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:


0.000 000 000 742 147 676 624 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 624 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 624 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 624 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111