-0.000 000 000 742 147 675 74 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 74(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 74(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 74| = 0.000 000 000 742 147 675 74


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 74.

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 74 × 2 = 0 + 0.000 000 001 484 295 351 48;
  • 2) 0.000 000 001 484 295 351 48 × 2 = 0 + 0.000 000 002 968 590 702 96;
  • 3) 0.000 000 002 968 590 702 96 × 2 = 0 + 0.000 000 005 937 181 405 92;
  • 4) 0.000 000 005 937 181 405 92 × 2 = 0 + 0.000 000 011 874 362 811 84;
  • 5) 0.000 000 011 874 362 811 84 × 2 = 0 + 0.000 000 023 748 725 623 68;
  • 6) 0.000 000 023 748 725 623 68 × 2 = 0 + 0.000 000 047 497 451 247 36;
  • 7) 0.000 000 047 497 451 247 36 × 2 = 0 + 0.000 000 094 994 902 494 72;
  • 8) 0.000 000 094 994 902 494 72 × 2 = 0 + 0.000 000 189 989 804 989 44;
  • 9) 0.000 000 189 989 804 989 44 × 2 = 0 + 0.000 000 379 979 609 978 88;
  • 10) 0.000 000 379 979 609 978 88 × 2 = 0 + 0.000 000 759 959 219 957 76;
  • 11) 0.000 000 759 959 219 957 76 × 2 = 0 + 0.000 001 519 918 439 915 52;
  • 12) 0.000 001 519 918 439 915 52 × 2 = 0 + 0.000 003 039 836 879 831 04;
  • 13) 0.000 003 039 836 879 831 04 × 2 = 0 + 0.000 006 079 673 759 662 08;
  • 14) 0.000 006 079 673 759 662 08 × 2 = 0 + 0.000 012 159 347 519 324 16;
  • 15) 0.000 012 159 347 519 324 16 × 2 = 0 + 0.000 024 318 695 038 648 32;
  • 16) 0.000 024 318 695 038 648 32 × 2 = 0 + 0.000 048 637 390 077 296 64;
  • 17) 0.000 048 637 390 077 296 64 × 2 = 0 + 0.000 097 274 780 154 593 28;
  • 18) 0.000 097 274 780 154 593 28 × 2 = 0 + 0.000 194 549 560 309 186 56;
  • 19) 0.000 194 549 560 309 186 56 × 2 = 0 + 0.000 389 099 120 618 373 12;
  • 20) 0.000 389 099 120 618 373 12 × 2 = 0 + 0.000 778 198 241 236 746 24;
  • 21) 0.000 778 198 241 236 746 24 × 2 = 0 + 0.001 556 396 482 473 492 48;
  • 22) 0.001 556 396 482 473 492 48 × 2 = 0 + 0.003 112 792 964 946 984 96;
  • 23) 0.003 112 792 964 946 984 96 × 2 = 0 + 0.006 225 585 929 893 969 92;
  • 24) 0.006 225 585 929 893 969 92 × 2 = 0 + 0.012 451 171 859 787 939 84;
  • 25) 0.012 451 171 859 787 939 84 × 2 = 0 + 0.024 902 343 719 575 879 68;
  • 26) 0.024 902 343 719 575 879 68 × 2 = 0 + 0.049 804 687 439 151 759 36;
  • 27) 0.049 804 687 439 151 759 36 × 2 = 0 + 0.099 609 374 878 303 518 72;
  • 28) 0.099 609 374 878 303 518 72 × 2 = 0 + 0.199 218 749 756 607 037 44;
  • 29) 0.199 218 749 756 607 037 44 × 2 = 0 + 0.398 437 499 513 214 074 88;
  • 30) 0.398 437 499 513 214 074 88 × 2 = 0 + 0.796 874 999 026 428 149 76;
  • 31) 0.796 874 999 026 428 149 76 × 2 = 1 + 0.593 749 998 052 856 299 52;
  • 32) 0.593 749 998 052 856 299 52 × 2 = 1 + 0.187 499 996 105 712 599 04;
  • 33) 0.187 499 996 105 712 599 04 × 2 = 0 + 0.374 999 992 211 425 198 08;
  • 34) 0.374 999 992 211 425 198 08 × 2 = 0 + 0.749 999 984 422 850 396 16;
  • 35) 0.749 999 984 422 850 396 16 × 2 = 1 + 0.499 999 968 845 700 792 32;
  • 36) 0.499 999 968 845 700 792 32 × 2 = 0 + 0.999 999 937 691 401 584 64;
  • 37) 0.999 999 937 691 401 584 64 × 2 = 1 + 0.999 999 875 382 803 169 28;
  • 38) 0.999 999 875 382 803 169 28 × 2 = 1 + 0.999 999 750 765 606 338 56;
  • 39) 0.999 999 750 765 606 338 56 × 2 = 1 + 0.999 999 501 531 212 677 12;
  • 40) 0.999 999 501 531 212 677 12 × 2 = 1 + 0.999 999 003 062 425 354 24;
  • 41) 0.999 999 003 062 425 354 24 × 2 = 1 + 0.999 998 006 124 850 708 48;
  • 42) 0.999 998 006 124 850 708 48 × 2 = 1 + 0.999 996 012 249 701 416 96;
  • 43) 0.999 996 012 249 701 416 96 × 2 = 1 + 0.999 992 024 499 402 833 92;
  • 44) 0.999 992 024 499 402 833 92 × 2 = 1 + 0.999 984 048 998 805 667 84;
  • 45) 0.999 984 048 998 805 667 84 × 2 = 1 + 0.999 968 097 997 611 335 68;
  • 46) 0.999 968 097 997 611 335 68 × 2 = 1 + 0.999 936 195 995 222 671 36;
  • 47) 0.999 936 195 995 222 671 36 × 2 = 1 + 0.999 872 391 990 445 342 72;
  • 48) 0.999 872 391 990 445 342 72 × 2 = 1 + 0.999 744 783 980 890 685 44;
  • 49) 0.999 744 783 980 890 685 44 × 2 = 1 + 0.999 489 567 961 781 370 88;
  • 50) 0.999 489 567 961 781 370 88 × 2 = 1 + 0.998 979 135 923 562 741 76;
  • 51) 0.998 979 135 923 562 741 76 × 2 = 1 + 0.997 958 271 847 125 483 52;
  • 52) 0.997 958 271 847 125 483 52 × 2 = 1 + 0.995 916 543 694 250 967 04;
  • 53) 0.995 916 543 694 250 967 04 × 2 = 1 + 0.991 833 087 388 501 934 08;
  • 54) 0.991 833 087 388 501 934 08 × 2 = 1 + 0.983 666 174 777 003 868 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 74(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 74(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 74(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 74 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