-0.000 000 000 742 149 12 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 12(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
-0.000 000 000 742 149 12(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:

|-0.000 000 000 742 149 12| = 0.000 000 000 742 149 12


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: 0.000 000 000 742 149 12.

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) 0.000 000 000 742 149 12 × 2 = 0 + 0.000 000 001 484 298 24;
  • 2) 0.000 000 001 484 298 24 × 2 = 0 + 0.000 000 002 968 596 48;
  • 3) 0.000 000 002 968 596 48 × 2 = 0 + 0.000 000 005 937 192 96;
  • 4) 0.000 000 005 937 192 96 × 2 = 0 + 0.000 000 011 874 385 92;
  • 5) 0.000 000 011 874 385 92 × 2 = 0 + 0.000 000 023 748 771 84;
  • 6) 0.000 000 023 748 771 84 × 2 = 0 + 0.000 000 047 497 543 68;
  • 7) 0.000 000 047 497 543 68 × 2 = 0 + 0.000 000 094 995 087 36;
  • 8) 0.000 000 094 995 087 36 × 2 = 0 + 0.000 000 189 990 174 72;
  • 9) 0.000 000 189 990 174 72 × 2 = 0 + 0.000 000 379 980 349 44;
  • 10) 0.000 000 379 980 349 44 × 2 = 0 + 0.000 000 759 960 698 88;
  • 11) 0.000 000 759 960 698 88 × 2 = 0 + 0.000 001 519 921 397 76;
  • 12) 0.000 001 519 921 397 76 × 2 = 0 + 0.000 003 039 842 795 52;
  • 13) 0.000 003 039 842 795 52 × 2 = 0 + 0.000 006 079 685 591 04;
  • 14) 0.000 006 079 685 591 04 × 2 = 0 + 0.000 012 159 371 182 08;
  • 15) 0.000 012 159 371 182 08 × 2 = 0 + 0.000 024 318 742 364 16;
  • 16) 0.000 024 318 742 364 16 × 2 = 0 + 0.000 048 637 484 728 32;
  • 17) 0.000 048 637 484 728 32 × 2 = 0 + 0.000 097 274 969 456 64;
  • 18) 0.000 097 274 969 456 64 × 2 = 0 + 0.000 194 549 938 913 28;
  • 19) 0.000 194 549 938 913 28 × 2 = 0 + 0.000 389 099 877 826 56;
  • 20) 0.000 389 099 877 826 56 × 2 = 0 + 0.000 778 199 755 653 12;
  • 21) 0.000 778 199 755 653 12 × 2 = 0 + 0.001 556 399 511 306 24;
  • 22) 0.001 556 399 511 306 24 × 2 = 0 + 0.003 112 799 022 612 48;
  • 23) 0.003 112 799 022 612 48 × 2 = 0 + 0.006 225 598 045 224 96;
  • 24) 0.006 225 598 045 224 96 × 2 = 0 + 0.012 451 196 090 449 92;
  • 25) 0.012 451 196 090 449 92 × 2 = 0 + 0.024 902 392 180 899 84;
  • 26) 0.024 902 392 180 899 84 × 2 = 0 + 0.049 804 784 361 799 68;
  • 27) 0.049 804 784 361 799 68 × 2 = 0 + 0.099 609 568 723 599 36;
  • 28) 0.099 609 568 723 599 36 × 2 = 0 + 0.199 219 137 447 198 72;
  • 29) 0.199 219 137 447 198 72 × 2 = 0 + 0.398 438 274 894 397 44;
  • 30) 0.398 438 274 894 397 44 × 2 = 0 + 0.796 876 549 788 794 88;
  • 31) 0.796 876 549 788 794 88 × 2 = 1 + 0.593 753 099 577 589 76;
  • 32) 0.593 753 099 577 589 76 × 2 = 1 + 0.187 506 199 155 179 52;
  • 33) 0.187 506 199 155 179 52 × 2 = 0 + 0.375 012 398 310 359 04;
  • 34) 0.375 012 398 310 359 04 × 2 = 0 + 0.750 024 796 620 718 08;
  • 35) 0.750 024 796 620 718 08 × 2 = 1 + 0.500 049 593 241 436 16;
  • 36) 0.500 049 593 241 436 16 × 2 = 1 + 0.000 099 186 482 872 32;
  • 37) 0.000 099 186 482 872 32 × 2 = 0 + 0.000 198 372 965 744 64;
  • 38) 0.000 198 372 965 744 64 × 2 = 0 + 0.000 396 745 931 489 28;
  • 39) 0.000 396 745 931 489 28 × 2 = 0 + 0.000 793 491 862 978 56;
  • 40) 0.000 793 491 862 978 56 × 2 = 0 + 0.001 586 983 725 957 12;
  • 41) 0.001 586 983 725 957 12 × 2 = 0 + 0.003 173 967 451 914 24;
  • 42) 0.003 173 967 451 914 24 × 2 = 0 + 0.006 347 934 903 828 48;
  • 43) 0.006 347 934 903 828 48 × 2 = 0 + 0.012 695 869 807 656 96;
  • 44) 0.012 695 869 807 656 96 × 2 = 0 + 0.025 391 739 615 313 92;
  • 45) 0.025 391 739 615 313 92 × 2 = 0 + 0.050 783 479 230 627 84;
  • 46) 0.050 783 479 230 627 84 × 2 = 0 + 0.101 566 958 461 255 68;
  • 47) 0.101 566 958 461 255 68 × 2 = 0 + 0.203 133 916 922 511 36;
  • 48) 0.203 133 916 922 511 36 × 2 = 0 + 0.406 267 833 845 022 72;
  • 49) 0.406 267 833 845 022 72 × 2 = 0 + 0.812 535 667 690 045 44;
  • 50) 0.812 535 667 690 045 44 × 2 = 1 + 0.625 071 335 380 090 88;
  • 51) 0.625 071 335 380 090 88 × 2 = 1 + 0.250 142 670 760 181 76;
  • 52) 0.250 142 670 760 181 76 × 2 = 0 + 0.500 285 341 520 363 52;
  • 53) 0.500 285 341 520 363 52 × 2 = 1 + 0.000 570 683 040 727 04;
  • 54) 0.000 570 683 040 727 04 × 2 = 0 + 0.001 141 366 081 454 08;

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:


0.000 000 000 742 149 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(2)

6. Positive number before normalization:

0.000 000 000 742 149 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(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:


0.000 000 000 742 149 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(2) × 20 =

1.1001 1000 0000 0000 0011 010(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 1000 0000 0000 0011 010


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 1100 0000 0000 0001 1010 =

100 1100 0000 0000 0001 1010


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 1100 0000 0000 0001 1010


Decimal number -0.000 000 000 742 149 12 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0001 1010

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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