-0.000 000 000 742 147 673 49 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 49(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
-0.000 000 000 742 147 673 49(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:

|-0.000 000 000 742 147 673 49| = 0.000 000 000 742 147 673 49


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: 0.000 000 000 742 147 673 49.

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) 0.000 000 000 742 147 673 49 × 2 = 0 + 0.000 000 001 484 295 346 98;
  • 2) 0.000 000 001 484 295 346 98 × 2 = 0 + 0.000 000 002 968 590 693 96;
  • 3) 0.000 000 002 968 590 693 96 × 2 = 0 + 0.000 000 005 937 181 387 92;
  • 4) 0.000 000 005 937 181 387 92 × 2 = 0 + 0.000 000 011 874 362 775 84;
  • 5) 0.000 000 011 874 362 775 84 × 2 = 0 + 0.000 000 023 748 725 551 68;
  • 6) 0.000 000 023 748 725 551 68 × 2 = 0 + 0.000 000 047 497 451 103 36;
  • 7) 0.000 000 047 497 451 103 36 × 2 = 0 + 0.000 000 094 994 902 206 72;
  • 8) 0.000 000 094 994 902 206 72 × 2 = 0 + 0.000 000 189 989 804 413 44;
  • 9) 0.000 000 189 989 804 413 44 × 2 = 0 + 0.000 000 379 979 608 826 88;
  • 10) 0.000 000 379 979 608 826 88 × 2 = 0 + 0.000 000 759 959 217 653 76;
  • 11) 0.000 000 759 959 217 653 76 × 2 = 0 + 0.000 001 519 918 435 307 52;
  • 12) 0.000 001 519 918 435 307 52 × 2 = 0 + 0.000 003 039 836 870 615 04;
  • 13) 0.000 003 039 836 870 615 04 × 2 = 0 + 0.000 006 079 673 741 230 08;
  • 14) 0.000 006 079 673 741 230 08 × 2 = 0 + 0.000 012 159 347 482 460 16;
  • 15) 0.000 012 159 347 482 460 16 × 2 = 0 + 0.000 024 318 694 964 920 32;
  • 16) 0.000 024 318 694 964 920 32 × 2 = 0 + 0.000 048 637 389 929 840 64;
  • 17) 0.000 048 637 389 929 840 64 × 2 = 0 + 0.000 097 274 779 859 681 28;
  • 18) 0.000 097 274 779 859 681 28 × 2 = 0 + 0.000 194 549 559 719 362 56;
  • 19) 0.000 194 549 559 719 362 56 × 2 = 0 + 0.000 389 099 119 438 725 12;
  • 20) 0.000 389 099 119 438 725 12 × 2 = 0 + 0.000 778 198 238 877 450 24;
  • 21) 0.000 778 198 238 877 450 24 × 2 = 0 + 0.001 556 396 477 754 900 48;
  • 22) 0.001 556 396 477 754 900 48 × 2 = 0 + 0.003 112 792 955 509 800 96;
  • 23) 0.003 112 792 955 509 800 96 × 2 = 0 + 0.006 225 585 911 019 601 92;
  • 24) 0.006 225 585 911 019 601 92 × 2 = 0 + 0.012 451 171 822 039 203 84;
  • 25) 0.012 451 171 822 039 203 84 × 2 = 0 + 0.024 902 343 644 078 407 68;
  • 26) 0.024 902 343 644 078 407 68 × 2 = 0 + 0.049 804 687 288 156 815 36;
  • 27) 0.049 804 687 288 156 815 36 × 2 = 0 + 0.099 609 374 576 313 630 72;
  • 28) 0.099 609 374 576 313 630 72 × 2 = 0 + 0.199 218 749 152 627 261 44;
  • 29) 0.199 218 749 152 627 261 44 × 2 = 0 + 0.398 437 498 305 254 522 88;
  • 30) 0.398 437 498 305 254 522 88 × 2 = 0 + 0.796 874 996 610 509 045 76;
  • 31) 0.796 874 996 610 509 045 76 × 2 = 1 + 0.593 749 993 221 018 091 52;
  • 32) 0.593 749 993 221 018 091 52 × 2 = 1 + 0.187 499 986 442 036 183 04;
  • 33) 0.187 499 986 442 036 183 04 × 2 = 0 + 0.374 999 972 884 072 366 08;
  • 34) 0.374 999 972 884 072 366 08 × 2 = 0 + 0.749 999 945 768 144 732 16;
  • 35) 0.749 999 945 768 144 732 16 × 2 = 1 + 0.499 999 891 536 289 464 32;
  • 36) 0.499 999 891 536 289 464 32 × 2 = 0 + 0.999 999 783 072 578 928 64;
  • 37) 0.999 999 783 072 578 928 64 × 2 = 1 + 0.999 999 566 145 157 857 28;
  • 38) 0.999 999 566 145 157 857 28 × 2 = 1 + 0.999 999 132 290 315 714 56;
  • 39) 0.999 999 132 290 315 714 56 × 2 = 1 + 0.999 998 264 580 631 429 12;
  • 40) 0.999 998 264 580 631 429 12 × 2 = 1 + 0.999 996 529 161 262 858 24;
  • 41) 0.999 996 529 161 262 858 24 × 2 = 1 + 0.999 993 058 322 525 716 48;
  • 42) 0.999 993 058 322 525 716 48 × 2 = 1 + 0.999 986 116 645 051 432 96;
  • 43) 0.999 986 116 645 051 432 96 × 2 = 1 + 0.999 972 233 290 102 865 92;
  • 44) 0.999 972 233 290 102 865 92 × 2 = 1 + 0.999 944 466 580 205 731 84;
  • 45) 0.999 944 466 580 205 731 84 × 2 = 1 + 0.999 888 933 160 411 463 68;
  • 46) 0.999 888 933 160 411 463 68 × 2 = 1 + 0.999 777 866 320 822 927 36;
  • 47) 0.999 777 866 320 822 927 36 × 2 = 1 + 0.999 555 732 641 645 854 72;
  • 48) 0.999 555 732 641 645 854 72 × 2 = 1 + 0.999 111 465 283 291 709 44;
  • 49) 0.999 111 465 283 291 709 44 × 2 = 1 + 0.998 222 930 566 583 418 88;
  • 50) 0.998 222 930 566 583 418 88 × 2 = 1 + 0.996 445 861 133 166 837 76;
  • 51) 0.996 445 861 133 166 837 76 × 2 = 1 + 0.992 891 722 266 333 675 52;
  • 52) 0.992 891 722 266 333 675 52 × 2 = 1 + 0.985 783 444 532 667 351 04;
  • 53) 0.985 783 444 532 667 351 04 × 2 = 1 + 0.971 566 889 065 334 702 08;
  • 54) 0.971 566 889 065 334 702 08 × 2 = 1 + 0.943 133 778 130 669 404 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:


0.000 000 000 742 147 673 49(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 49(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:


0.000 000 000 742 147 673 49(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 -0.000 000 000 742 147 673 49 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