-0.000 000 000 742 147 674 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 2(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
-0.000 000 000 742 147 674 2(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:

|-0.000 000 000 742 147 674 2| = 0.000 000 000 742 147 674 2


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: 0.000 000 000 742 147 674 2.

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) 0.000 000 000 742 147 674 2 × 2 = 0 + 0.000 000 001 484 295 348 4;
  • 2) 0.000 000 001 484 295 348 4 × 2 = 0 + 0.000 000 002 968 590 696 8;
  • 3) 0.000 000 002 968 590 696 8 × 2 = 0 + 0.000 000 005 937 181 393 6;
  • 4) 0.000 000 005 937 181 393 6 × 2 = 0 + 0.000 000 011 874 362 787 2;
  • 5) 0.000 000 011 874 362 787 2 × 2 = 0 + 0.000 000 023 748 725 574 4;
  • 6) 0.000 000 023 748 725 574 4 × 2 = 0 + 0.000 000 047 497 451 148 8;
  • 7) 0.000 000 047 497 451 148 8 × 2 = 0 + 0.000 000 094 994 902 297 6;
  • 8) 0.000 000 094 994 902 297 6 × 2 = 0 + 0.000 000 189 989 804 595 2;
  • 9) 0.000 000 189 989 804 595 2 × 2 = 0 + 0.000 000 379 979 609 190 4;
  • 10) 0.000 000 379 979 609 190 4 × 2 = 0 + 0.000 000 759 959 218 380 8;
  • 11) 0.000 000 759 959 218 380 8 × 2 = 0 + 0.000 001 519 918 436 761 6;
  • 12) 0.000 001 519 918 436 761 6 × 2 = 0 + 0.000 003 039 836 873 523 2;
  • 13) 0.000 003 039 836 873 523 2 × 2 = 0 + 0.000 006 079 673 747 046 4;
  • 14) 0.000 006 079 673 747 046 4 × 2 = 0 + 0.000 012 159 347 494 092 8;
  • 15) 0.000 012 159 347 494 092 8 × 2 = 0 + 0.000 024 318 694 988 185 6;
  • 16) 0.000 024 318 694 988 185 6 × 2 = 0 + 0.000 048 637 389 976 371 2;
  • 17) 0.000 048 637 389 976 371 2 × 2 = 0 + 0.000 097 274 779 952 742 4;
  • 18) 0.000 097 274 779 952 742 4 × 2 = 0 + 0.000 194 549 559 905 484 8;
  • 19) 0.000 194 549 559 905 484 8 × 2 = 0 + 0.000 389 099 119 810 969 6;
  • 20) 0.000 389 099 119 810 969 6 × 2 = 0 + 0.000 778 198 239 621 939 2;
  • 21) 0.000 778 198 239 621 939 2 × 2 = 0 + 0.001 556 396 479 243 878 4;
  • 22) 0.001 556 396 479 243 878 4 × 2 = 0 + 0.003 112 792 958 487 756 8;
  • 23) 0.003 112 792 958 487 756 8 × 2 = 0 + 0.006 225 585 916 975 513 6;
  • 24) 0.006 225 585 916 975 513 6 × 2 = 0 + 0.012 451 171 833 951 027 2;
  • 25) 0.012 451 171 833 951 027 2 × 2 = 0 + 0.024 902 343 667 902 054 4;
  • 26) 0.024 902 343 667 902 054 4 × 2 = 0 + 0.049 804 687 335 804 108 8;
  • 27) 0.049 804 687 335 804 108 8 × 2 = 0 + 0.099 609 374 671 608 217 6;
  • 28) 0.099 609 374 671 608 217 6 × 2 = 0 + 0.199 218 749 343 216 435 2;
  • 29) 0.199 218 749 343 216 435 2 × 2 = 0 + 0.398 437 498 686 432 870 4;
  • 30) 0.398 437 498 686 432 870 4 × 2 = 0 + 0.796 874 997 372 865 740 8;
  • 31) 0.796 874 997 372 865 740 8 × 2 = 1 + 0.593 749 994 745 731 481 6;
  • 32) 0.593 749 994 745 731 481 6 × 2 = 1 + 0.187 499 989 491 462 963 2;
  • 33) 0.187 499 989 491 462 963 2 × 2 = 0 + 0.374 999 978 982 925 926 4;
  • 34) 0.374 999 978 982 925 926 4 × 2 = 0 + 0.749 999 957 965 851 852 8;
  • 35) 0.749 999 957 965 851 852 8 × 2 = 1 + 0.499 999 915 931 703 705 6;
  • 36) 0.499 999 915 931 703 705 6 × 2 = 0 + 0.999 999 831 863 407 411 2;
  • 37) 0.999 999 831 863 407 411 2 × 2 = 1 + 0.999 999 663 726 814 822 4;
  • 38) 0.999 999 663 726 814 822 4 × 2 = 1 + 0.999 999 327 453 629 644 8;
  • 39) 0.999 999 327 453 629 644 8 × 2 = 1 + 0.999 998 654 907 259 289 6;
  • 40) 0.999 998 654 907 259 289 6 × 2 = 1 + 0.999 997 309 814 518 579 2;
  • 41) 0.999 997 309 814 518 579 2 × 2 = 1 + 0.999 994 619 629 037 158 4;
  • 42) 0.999 994 619 629 037 158 4 × 2 = 1 + 0.999 989 239 258 074 316 8;
  • 43) 0.999 989 239 258 074 316 8 × 2 = 1 + 0.999 978 478 516 148 633 6;
  • 44) 0.999 978 478 516 148 633 6 × 2 = 1 + 0.999 956 957 032 297 267 2;
  • 45) 0.999 956 957 032 297 267 2 × 2 = 1 + 0.999 913 914 064 594 534 4;
  • 46) 0.999 913 914 064 594 534 4 × 2 = 1 + 0.999 827 828 129 189 068 8;
  • 47) 0.999 827 828 129 189 068 8 × 2 = 1 + 0.999 655 656 258 378 137 6;
  • 48) 0.999 655 656 258 378 137 6 × 2 = 1 + 0.999 311 312 516 756 275 2;
  • 49) 0.999 311 312 516 756 275 2 × 2 = 1 + 0.998 622 625 033 512 550 4;
  • 50) 0.998 622 625 033 512 550 4 × 2 = 1 + 0.997 245 250 067 025 100 8;
  • 51) 0.997 245 250 067 025 100 8 × 2 = 1 + 0.994 490 500 134 050 201 6;
  • 52) 0.994 490 500 134 050 201 6 × 2 = 1 + 0.988 981 000 268 100 403 2;
  • 53) 0.988 981 000 268 100 403 2 × 2 = 1 + 0.977 962 000 536 200 806 4;
  • 54) 0.977 962 000 536 200 806 4 × 2 = 1 + 0.955 924 001 072 401 612 8;

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:


0.000 000 000 742 147 674 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 2(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:


0.000 000 000 742 147 674 2(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 -0.000 000 000 742 147 674 2 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