-0.000 000 000 742 147 675 99 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 99(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
-0.000 000 000 742 147 675 99(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:

|-0.000 000 000 742 147 675 99| = 0.000 000 000 742 147 675 99


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: 0.000 000 000 742 147 675 99.

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) 0.000 000 000 742 147 675 99 × 2 = 0 + 0.000 000 001 484 295 351 98;
  • 2) 0.000 000 001 484 295 351 98 × 2 = 0 + 0.000 000 002 968 590 703 96;
  • 3) 0.000 000 002 968 590 703 96 × 2 = 0 + 0.000 000 005 937 181 407 92;
  • 4) 0.000 000 005 937 181 407 92 × 2 = 0 + 0.000 000 011 874 362 815 84;
  • 5) 0.000 000 011 874 362 815 84 × 2 = 0 + 0.000 000 023 748 725 631 68;
  • 6) 0.000 000 023 748 725 631 68 × 2 = 0 + 0.000 000 047 497 451 263 36;
  • 7) 0.000 000 047 497 451 263 36 × 2 = 0 + 0.000 000 094 994 902 526 72;
  • 8) 0.000 000 094 994 902 526 72 × 2 = 0 + 0.000 000 189 989 805 053 44;
  • 9) 0.000 000 189 989 805 053 44 × 2 = 0 + 0.000 000 379 979 610 106 88;
  • 10) 0.000 000 379 979 610 106 88 × 2 = 0 + 0.000 000 759 959 220 213 76;
  • 11) 0.000 000 759 959 220 213 76 × 2 = 0 + 0.000 001 519 918 440 427 52;
  • 12) 0.000 001 519 918 440 427 52 × 2 = 0 + 0.000 003 039 836 880 855 04;
  • 13) 0.000 003 039 836 880 855 04 × 2 = 0 + 0.000 006 079 673 761 710 08;
  • 14) 0.000 006 079 673 761 710 08 × 2 = 0 + 0.000 012 159 347 523 420 16;
  • 15) 0.000 012 159 347 523 420 16 × 2 = 0 + 0.000 024 318 695 046 840 32;
  • 16) 0.000 024 318 695 046 840 32 × 2 = 0 + 0.000 048 637 390 093 680 64;
  • 17) 0.000 048 637 390 093 680 64 × 2 = 0 + 0.000 097 274 780 187 361 28;
  • 18) 0.000 097 274 780 187 361 28 × 2 = 0 + 0.000 194 549 560 374 722 56;
  • 19) 0.000 194 549 560 374 722 56 × 2 = 0 + 0.000 389 099 120 749 445 12;
  • 20) 0.000 389 099 120 749 445 12 × 2 = 0 + 0.000 778 198 241 498 890 24;
  • 21) 0.000 778 198 241 498 890 24 × 2 = 0 + 0.001 556 396 482 997 780 48;
  • 22) 0.001 556 396 482 997 780 48 × 2 = 0 + 0.003 112 792 965 995 560 96;
  • 23) 0.003 112 792 965 995 560 96 × 2 = 0 + 0.006 225 585 931 991 121 92;
  • 24) 0.006 225 585 931 991 121 92 × 2 = 0 + 0.012 451 171 863 982 243 84;
  • 25) 0.012 451 171 863 982 243 84 × 2 = 0 + 0.024 902 343 727 964 487 68;
  • 26) 0.024 902 343 727 964 487 68 × 2 = 0 + 0.049 804 687 455 928 975 36;
  • 27) 0.049 804 687 455 928 975 36 × 2 = 0 + 0.099 609 374 911 857 950 72;
  • 28) 0.099 609 374 911 857 950 72 × 2 = 0 + 0.199 218 749 823 715 901 44;
  • 29) 0.199 218 749 823 715 901 44 × 2 = 0 + 0.398 437 499 647 431 802 88;
  • 30) 0.398 437 499 647 431 802 88 × 2 = 0 + 0.796 874 999 294 863 605 76;
  • 31) 0.796 874 999 294 863 605 76 × 2 = 1 + 0.593 749 998 589 727 211 52;
  • 32) 0.593 749 998 589 727 211 52 × 2 = 1 + 0.187 499 997 179 454 423 04;
  • 33) 0.187 499 997 179 454 423 04 × 2 = 0 + 0.374 999 994 358 908 846 08;
  • 34) 0.374 999 994 358 908 846 08 × 2 = 0 + 0.749 999 988 717 817 692 16;
  • 35) 0.749 999 988 717 817 692 16 × 2 = 1 + 0.499 999 977 435 635 384 32;
  • 36) 0.499 999 977 435 635 384 32 × 2 = 0 + 0.999 999 954 871 270 768 64;
  • 37) 0.999 999 954 871 270 768 64 × 2 = 1 + 0.999 999 909 742 541 537 28;
  • 38) 0.999 999 909 742 541 537 28 × 2 = 1 + 0.999 999 819 485 083 074 56;
  • 39) 0.999 999 819 485 083 074 56 × 2 = 1 + 0.999 999 638 970 166 149 12;
  • 40) 0.999 999 638 970 166 149 12 × 2 = 1 + 0.999 999 277 940 332 298 24;
  • 41) 0.999 999 277 940 332 298 24 × 2 = 1 + 0.999 998 555 880 664 596 48;
  • 42) 0.999 998 555 880 664 596 48 × 2 = 1 + 0.999 997 111 761 329 192 96;
  • 43) 0.999 997 111 761 329 192 96 × 2 = 1 + 0.999 994 223 522 658 385 92;
  • 44) 0.999 994 223 522 658 385 92 × 2 = 1 + 0.999 988 447 045 316 771 84;
  • 45) 0.999 988 447 045 316 771 84 × 2 = 1 + 0.999 976 894 090 633 543 68;
  • 46) 0.999 976 894 090 633 543 68 × 2 = 1 + 0.999 953 788 181 267 087 36;
  • 47) 0.999 953 788 181 267 087 36 × 2 = 1 + 0.999 907 576 362 534 174 72;
  • 48) 0.999 907 576 362 534 174 72 × 2 = 1 + 0.999 815 152 725 068 349 44;
  • 49) 0.999 815 152 725 068 349 44 × 2 = 1 + 0.999 630 305 450 136 698 88;
  • 50) 0.999 630 305 450 136 698 88 × 2 = 1 + 0.999 260 610 900 273 397 76;
  • 51) 0.999 260 610 900 273 397 76 × 2 = 1 + 0.998 521 221 800 546 795 52;
  • 52) 0.998 521 221 800 546 795 52 × 2 = 1 + 0.997 042 443 601 093 591 04;
  • 53) 0.997 042 443 601 093 591 04 × 2 = 1 + 0.994 084 887 202 187 182 08;
  • 54) 0.994 084 887 202 187 182 08 × 2 = 1 + 0.988 169 774 404 374 364 16;

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:


0.000 000 000 742 147 675 99(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 99(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:


0.000 000 000 742 147 675 99(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 -0.000 000 000 742 147 675 99 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