-0.000 000 000 742 147 674 78 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 78(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 78(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 78| = 0.000 000 000 742 147 674 78


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 78.

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 78 × 2 = 0 + 0.000 000 001 484 295 349 56;
  • 2) 0.000 000 001 484 295 349 56 × 2 = 0 + 0.000 000 002 968 590 699 12;
  • 3) 0.000 000 002 968 590 699 12 × 2 = 0 + 0.000 000 005 937 181 398 24;
  • 4) 0.000 000 005 937 181 398 24 × 2 = 0 + 0.000 000 011 874 362 796 48;
  • 5) 0.000 000 011 874 362 796 48 × 2 = 0 + 0.000 000 023 748 725 592 96;
  • 6) 0.000 000 023 748 725 592 96 × 2 = 0 + 0.000 000 047 497 451 185 92;
  • 7) 0.000 000 047 497 451 185 92 × 2 = 0 + 0.000 000 094 994 902 371 84;
  • 8) 0.000 000 094 994 902 371 84 × 2 = 0 + 0.000 000 189 989 804 743 68;
  • 9) 0.000 000 189 989 804 743 68 × 2 = 0 + 0.000 000 379 979 609 487 36;
  • 10) 0.000 000 379 979 609 487 36 × 2 = 0 + 0.000 000 759 959 218 974 72;
  • 11) 0.000 000 759 959 218 974 72 × 2 = 0 + 0.000 001 519 918 437 949 44;
  • 12) 0.000 001 519 918 437 949 44 × 2 = 0 + 0.000 003 039 836 875 898 88;
  • 13) 0.000 003 039 836 875 898 88 × 2 = 0 + 0.000 006 079 673 751 797 76;
  • 14) 0.000 006 079 673 751 797 76 × 2 = 0 + 0.000 012 159 347 503 595 52;
  • 15) 0.000 012 159 347 503 595 52 × 2 = 0 + 0.000 024 318 695 007 191 04;
  • 16) 0.000 024 318 695 007 191 04 × 2 = 0 + 0.000 048 637 390 014 382 08;
  • 17) 0.000 048 637 390 014 382 08 × 2 = 0 + 0.000 097 274 780 028 764 16;
  • 18) 0.000 097 274 780 028 764 16 × 2 = 0 + 0.000 194 549 560 057 528 32;
  • 19) 0.000 194 549 560 057 528 32 × 2 = 0 + 0.000 389 099 120 115 056 64;
  • 20) 0.000 389 099 120 115 056 64 × 2 = 0 + 0.000 778 198 240 230 113 28;
  • 21) 0.000 778 198 240 230 113 28 × 2 = 0 + 0.001 556 396 480 460 226 56;
  • 22) 0.001 556 396 480 460 226 56 × 2 = 0 + 0.003 112 792 960 920 453 12;
  • 23) 0.003 112 792 960 920 453 12 × 2 = 0 + 0.006 225 585 921 840 906 24;
  • 24) 0.006 225 585 921 840 906 24 × 2 = 0 + 0.012 451 171 843 681 812 48;
  • 25) 0.012 451 171 843 681 812 48 × 2 = 0 + 0.024 902 343 687 363 624 96;
  • 26) 0.024 902 343 687 363 624 96 × 2 = 0 + 0.049 804 687 374 727 249 92;
  • 27) 0.049 804 687 374 727 249 92 × 2 = 0 + 0.099 609 374 749 454 499 84;
  • 28) 0.099 609 374 749 454 499 84 × 2 = 0 + 0.199 218 749 498 908 999 68;
  • 29) 0.199 218 749 498 908 999 68 × 2 = 0 + 0.398 437 498 997 817 999 36;
  • 30) 0.398 437 498 997 817 999 36 × 2 = 0 + 0.796 874 997 995 635 998 72;
  • 31) 0.796 874 997 995 635 998 72 × 2 = 1 + 0.593 749 995 991 271 997 44;
  • 32) 0.593 749 995 991 271 997 44 × 2 = 1 + 0.187 499 991 982 543 994 88;
  • 33) 0.187 499 991 982 543 994 88 × 2 = 0 + 0.374 999 983 965 087 989 76;
  • 34) 0.374 999 983 965 087 989 76 × 2 = 0 + 0.749 999 967 930 175 979 52;
  • 35) 0.749 999 967 930 175 979 52 × 2 = 1 + 0.499 999 935 860 351 959 04;
  • 36) 0.499 999 935 860 351 959 04 × 2 = 0 + 0.999 999 871 720 703 918 08;
  • 37) 0.999 999 871 720 703 918 08 × 2 = 1 + 0.999 999 743 441 407 836 16;
  • 38) 0.999 999 743 441 407 836 16 × 2 = 1 + 0.999 999 486 882 815 672 32;
  • 39) 0.999 999 486 882 815 672 32 × 2 = 1 + 0.999 998 973 765 631 344 64;
  • 40) 0.999 998 973 765 631 344 64 × 2 = 1 + 0.999 997 947 531 262 689 28;
  • 41) 0.999 997 947 531 262 689 28 × 2 = 1 + 0.999 995 895 062 525 378 56;
  • 42) 0.999 995 895 062 525 378 56 × 2 = 1 + 0.999 991 790 125 050 757 12;
  • 43) 0.999 991 790 125 050 757 12 × 2 = 1 + 0.999 983 580 250 101 514 24;
  • 44) 0.999 983 580 250 101 514 24 × 2 = 1 + 0.999 967 160 500 203 028 48;
  • 45) 0.999 967 160 500 203 028 48 × 2 = 1 + 0.999 934 321 000 406 056 96;
  • 46) 0.999 934 321 000 406 056 96 × 2 = 1 + 0.999 868 642 000 812 113 92;
  • 47) 0.999 868 642 000 812 113 92 × 2 = 1 + 0.999 737 284 001 624 227 84;
  • 48) 0.999 737 284 001 624 227 84 × 2 = 1 + 0.999 474 568 003 248 455 68;
  • 49) 0.999 474 568 003 248 455 68 × 2 = 1 + 0.998 949 136 006 496 911 36;
  • 50) 0.998 949 136 006 496 911 36 × 2 = 1 + 0.997 898 272 012 993 822 72;
  • 51) 0.997 898 272 012 993 822 72 × 2 = 1 + 0.995 796 544 025 987 645 44;
  • 52) 0.995 796 544 025 987 645 44 × 2 = 1 + 0.991 593 088 051 975 290 88;
  • 53) 0.991 593 088 051 975 290 88 × 2 = 1 + 0.983 186 176 103 950 581 76;
  • 54) 0.983 186 176 103 950 581 76 × 2 = 1 + 0.966 372 352 207 901 163 52;

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 78(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 78(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 78(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 78 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