-0.000 000 000 742 147 676 43 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 43(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 43(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 43| = 0.000 000 000 742 147 676 43


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 43.

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 43 × 2 = 0 + 0.000 000 001 484 295 352 86;
  • 2) 0.000 000 001 484 295 352 86 × 2 = 0 + 0.000 000 002 968 590 705 72;
  • 3) 0.000 000 002 968 590 705 72 × 2 = 0 + 0.000 000 005 937 181 411 44;
  • 4) 0.000 000 005 937 181 411 44 × 2 = 0 + 0.000 000 011 874 362 822 88;
  • 5) 0.000 000 011 874 362 822 88 × 2 = 0 + 0.000 000 023 748 725 645 76;
  • 6) 0.000 000 023 748 725 645 76 × 2 = 0 + 0.000 000 047 497 451 291 52;
  • 7) 0.000 000 047 497 451 291 52 × 2 = 0 + 0.000 000 094 994 902 583 04;
  • 8) 0.000 000 094 994 902 583 04 × 2 = 0 + 0.000 000 189 989 805 166 08;
  • 9) 0.000 000 189 989 805 166 08 × 2 = 0 + 0.000 000 379 979 610 332 16;
  • 10) 0.000 000 379 979 610 332 16 × 2 = 0 + 0.000 000 759 959 220 664 32;
  • 11) 0.000 000 759 959 220 664 32 × 2 = 0 + 0.000 001 519 918 441 328 64;
  • 12) 0.000 001 519 918 441 328 64 × 2 = 0 + 0.000 003 039 836 882 657 28;
  • 13) 0.000 003 039 836 882 657 28 × 2 = 0 + 0.000 006 079 673 765 314 56;
  • 14) 0.000 006 079 673 765 314 56 × 2 = 0 + 0.000 012 159 347 530 629 12;
  • 15) 0.000 012 159 347 530 629 12 × 2 = 0 + 0.000 024 318 695 061 258 24;
  • 16) 0.000 024 318 695 061 258 24 × 2 = 0 + 0.000 048 637 390 122 516 48;
  • 17) 0.000 048 637 390 122 516 48 × 2 = 0 + 0.000 097 274 780 245 032 96;
  • 18) 0.000 097 274 780 245 032 96 × 2 = 0 + 0.000 194 549 560 490 065 92;
  • 19) 0.000 194 549 560 490 065 92 × 2 = 0 + 0.000 389 099 120 980 131 84;
  • 20) 0.000 389 099 120 980 131 84 × 2 = 0 + 0.000 778 198 241 960 263 68;
  • 21) 0.000 778 198 241 960 263 68 × 2 = 0 + 0.001 556 396 483 920 527 36;
  • 22) 0.001 556 396 483 920 527 36 × 2 = 0 + 0.003 112 792 967 841 054 72;
  • 23) 0.003 112 792 967 841 054 72 × 2 = 0 + 0.006 225 585 935 682 109 44;
  • 24) 0.006 225 585 935 682 109 44 × 2 = 0 + 0.012 451 171 871 364 218 88;
  • 25) 0.012 451 171 871 364 218 88 × 2 = 0 + 0.024 902 343 742 728 437 76;
  • 26) 0.024 902 343 742 728 437 76 × 2 = 0 + 0.049 804 687 485 456 875 52;
  • 27) 0.049 804 687 485 456 875 52 × 2 = 0 + 0.099 609 374 970 913 751 04;
  • 28) 0.099 609 374 970 913 751 04 × 2 = 0 + 0.199 218 749 941 827 502 08;
  • 29) 0.199 218 749 941 827 502 08 × 2 = 0 + 0.398 437 499 883 655 004 16;
  • 30) 0.398 437 499 883 655 004 16 × 2 = 0 + 0.796 874 999 767 310 008 32;
  • 31) 0.796 874 999 767 310 008 32 × 2 = 1 + 0.593 749 999 534 620 016 64;
  • 32) 0.593 749 999 534 620 016 64 × 2 = 1 + 0.187 499 999 069 240 033 28;
  • 33) 0.187 499 999 069 240 033 28 × 2 = 0 + 0.374 999 998 138 480 066 56;
  • 34) 0.374 999 998 138 480 066 56 × 2 = 0 + 0.749 999 996 276 960 133 12;
  • 35) 0.749 999 996 276 960 133 12 × 2 = 1 + 0.499 999 992 553 920 266 24;
  • 36) 0.499 999 992 553 920 266 24 × 2 = 0 + 0.999 999 985 107 840 532 48;
  • 37) 0.999 999 985 107 840 532 48 × 2 = 1 + 0.999 999 970 215 681 064 96;
  • 38) 0.999 999 970 215 681 064 96 × 2 = 1 + 0.999 999 940 431 362 129 92;
  • 39) 0.999 999 940 431 362 129 92 × 2 = 1 + 0.999 999 880 862 724 259 84;
  • 40) 0.999 999 880 862 724 259 84 × 2 = 1 + 0.999 999 761 725 448 519 68;
  • 41) 0.999 999 761 725 448 519 68 × 2 = 1 + 0.999 999 523 450 897 039 36;
  • 42) 0.999 999 523 450 897 039 36 × 2 = 1 + 0.999 999 046 901 794 078 72;
  • 43) 0.999 999 046 901 794 078 72 × 2 = 1 + 0.999 998 093 803 588 157 44;
  • 44) 0.999 998 093 803 588 157 44 × 2 = 1 + 0.999 996 187 607 176 314 88;
  • 45) 0.999 996 187 607 176 314 88 × 2 = 1 + 0.999 992 375 214 352 629 76;
  • 46) 0.999 992 375 214 352 629 76 × 2 = 1 + 0.999 984 750 428 705 259 52;
  • 47) 0.999 984 750 428 705 259 52 × 2 = 1 + 0.999 969 500 857 410 519 04;
  • 48) 0.999 969 500 857 410 519 04 × 2 = 1 + 0.999 939 001 714 821 038 08;
  • 49) 0.999 939 001 714 821 038 08 × 2 = 1 + 0.999 878 003 429 642 076 16;
  • 50) 0.999 878 003 429 642 076 16 × 2 = 1 + 0.999 756 006 859 284 152 32;
  • 51) 0.999 756 006 859 284 152 32 × 2 = 1 + 0.999 512 013 718 568 304 64;
  • 52) 0.999 512 013 718 568 304 64 × 2 = 1 + 0.999 024 027 437 136 609 28;
  • 53) 0.999 024 027 437 136 609 28 × 2 = 1 + 0.998 048 054 874 273 218 56;
  • 54) 0.998 048 054 874 273 218 56 × 2 = 1 + 0.996 096 109 748 546 437 12;

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 43(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 43(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 43(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 43 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