-0.000 000 000 742 147 674 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 8(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 8(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 8| = 0.000 000 000 742 147 674 8


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 8.

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 8 × 2 = 0 + 0.000 000 001 484 295 349 6;
  • 2) 0.000 000 001 484 295 349 6 × 2 = 0 + 0.000 000 002 968 590 699 2;
  • 3) 0.000 000 002 968 590 699 2 × 2 = 0 + 0.000 000 005 937 181 398 4;
  • 4) 0.000 000 005 937 181 398 4 × 2 = 0 + 0.000 000 011 874 362 796 8;
  • 5) 0.000 000 011 874 362 796 8 × 2 = 0 + 0.000 000 023 748 725 593 6;
  • 6) 0.000 000 023 748 725 593 6 × 2 = 0 + 0.000 000 047 497 451 187 2;
  • 7) 0.000 000 047 497 451 187 2 × 2 = 0 + 0.000 000 094 994 902 374 4;
  • 8) 0.000 000 094 994 902 374 4 × 2 = 0 + 0.000 000 189 989 804 748 8;
  • 9) 0.000 000 189 989 804 748 8 × 2 = 0 + 0.000 000 379 979 609 497 6;
  • 10) 0.000 000 379 979 609 497 6 × 2 = 0 + 0.000 000 759 959 218 995 2;
  • 11) 0.000 000 759 959 218 995 2 × 2 = 0 + 0.000 001 519 918 437 990 4;
  • 12) 0.000 001 519 918 437 990 4 × 2 = 0 + 0.000 003 039 836 875 980 8;
  • 13) 0.000 003 039 836 875 980 8 × 2 = 0 + 0.000 006 079 673 751 961 6;
  • 14) 0.000 006 079 673 751 961 6 × 2 = 0 + 0.000 012 159 347 503 923 2;
  • 15) 0.000 012 159 347 503 923 2 × 2 = 0 + 0.000 024 318 695 007 846 4;
  • 16) 0.000 024 318 695 007 846 4 × 2 = 0 + 0.000 048 637 390 015 692 8;
  • 17) 0.000 048 637 390 015 692 8 × 2 = 0 + 0.000 097 274 780 031 385 6;
  • 18) 0.000 097 274 780 031 385 6 × 2 = 0 + 0.000 194 549 560 062 771 2;
  • 19) 0.000 194 549 560 062 771 2 × 2 = 0 + 0.000 389 099 120 125 542 4;
  • 20) 0.000 389 099 120 125 542 4 × 2 = 0 + 0.000 778 198 240 251 084 8;
  • 21) 0.000 778 198 240 251 084 8 × 2 = 0 + 0.001 556 396 480 502 169 6;
  • 22) 0.001 556 396 480 502 169 6 × 2 = 0 + 0.003 112 792 961 004 339 2;
  • 23) 0.003 112 792 961 004 339 2 × 2 = 0 + 0.006 225 585 922 008 678 4;
  • 24) 0.006 225 585 922 008 678 4 × 2 = 0 + 0.012 451 171 844 017 356 8;
  • 25) 0.012 451 171 844 017 356 8 × 2 = 0 + 0.024 902 343 688 034 713 6;
  • 26) 0.024 902 343 688 034 713 6 × 2 = 0 + 0.049 804 687 376 069 427 2;
  • 27) 0.049 804 687 376 069 427 2 × 2 = 0 + 0.099 609 374 752 138 854 4;
  • 28) 0.099 609 374 752 138 854 4 × 2 = 0 + 0.199 218 749 504 277 708 8;
  • 29) 0.199 218 749 504 277 708 8 × 2 = 0 + 0.398 437 499 008 555 417 6;
  • 30) 0.398 437 499 008 555 417 6 × 2 = 0 + 0.796 874 998 017 110 835 2;
  • 31) 0.796 874 998 017 110 835 2 × 2 = 1 + 0.593 749 996 034 221 670 4;
  • 32) 0.593 749 996 034 221 670 4 × 2 = 1 + 0.187 499 992 068 443 340 8;
  • 33) 0.187 499 992 068 443 340 8 × 2 = 0 + 0.374 999 984 136 886 681 6;
  • 34) 0.374 999 984 136 886 681 6 × 2 = 0 + 0.749 999 968 273 773 363 2;
  • 35) 0.749 999 968 273 773 363 2 × 2 = 1 + 0.499 999 936 547 546 726 4;
  • 36) 0.499 999 936 547 546 726 4 × 2 = 0 + 0.999 999 873 095 093 452 8;
  • 37) 0.999 999 873 095 093 452 8 × 2 = 1 + 0.999 999 746 190 186 905 6;
  • 38) 0.999 999 746 190 186 905 6 × 2 = 1 + 0.999 999 492 380 373 811 2;
  • 39) 0.999 999 492 380 373 811 2 × 2 = 1 + 0.999 998 984 760 747 622 4;
  • 40) 0.999 998 984 760 747 622 4 × 2 = 1 + 0.999 997 969 521 495 244 8;
  • 41) 0.999 997 969 521 495 244 8 × 2 = 1 + 0.999 995 939 042 990 489 6;
  • 42) 0.999 995 939 042 990 489 6 × 2 = 1 + 0.999 991 878 085 980 979 2;
  • 43) 0.999 991 878 085 980 979 2 × 2 = 1 + 0.999 983 756 171 961 958 4;
  • 44) 0.999 983 756 171 961 958 4 × 2 = 1 + 0.999 967 512 343 923 916 8;
  • 45) 0.999 967 512 343 923 916 8 × 2 = 1 + 0.999 935 024 687 847 833 6;
  • 46) 0.999 935 024 687 847 833 6 × 2 = 1 + 0.999 870 049 375 695 667 2;
  • 47) 0.999 870 049 375 695 667 2 × 2 = 1 + 0.999 740 098 751 391 334 4;
  • 48) 0.999 740 098 751 391 334 4 × 2 = 1 + 0.999 480 197 502 782 668 8;
  • 49) 0.999 480 197 502 782 668 8 × 2 = 1 + 0.998 960 395 005 565 337 6;
  • 50) 0.998 960 395 005 565 337 6 × 2 = 1 + 0.997 920 790 011 130 675 2;
  • 51) 0.997 920 790 011 130 675 2 × 2 = 1 + 0.995 841 580 022 261 350 4;
  • 52) 0.995 841 580 022 261 350 4 × 2 = 1 + 0.991 683 160 044 522 700 8;
  • 53) 0.991 683 160 044 522 700 8 × 2 = 1 + 0.983 366 320 089 045 401 6;
  • 54) 0.983 366 320 089 045 401 6 × 2 = 1 + 0.966 732 640 178 090 803 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:


0.000 000 000 742 147 674 8(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 8(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 8(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 8 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