-0.000 000 000 742 147 675 22 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 22(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 22(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 22| = 0.000 000 000 742 147 675 22


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 22.

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 22 × 2 = 0 + 0.000 000 001 484 295 350 44;
  • 2) 0.000 000 001 484 295 350 44 × 2 = 0 + 0.000 000 002 968 590 700 88;
  • 3) 0.000 000 002 968 590 700 88 × 2 = 0 + 0.000 000 005 937 181 401 76;
  • 4) 0.000 000 005 937 181 401 76 × 2 = 0 + 0.000 000 011 874 362 803 52;
  • 5) 0.000 000 011 874 362 803 52 × 2 = 0 + 0.000 000 023 748 725 607 04;
  • 6) 0.000 000 023 748 725 607 04 × 2 = 0 + 0.000 000 047 497 451 214 08;
  • 7) 0.000 000 047 497 451 214 08 × 2 = 0 + 0.000 000 094 994 902 428 16;
  • 8) 0.000 000 094 994 902 428 16 × 2 = 0 + 0.000 000 189 989 804 856 32;
  • 9) 0.000 000 189 989 804 856 32 × 2 = 0 + 0.000 000 379 979 609 712 64;
  • 10) 0.000 000 379 979 609 712 64 × 2 = 0 + 0.000 000 759 959 219 425 28;
  • 11) 0.000 000 759 959 219 425 28 × 2 = 0 + 0.000 001 519 918 438 850 56;
  • 12) 0.000 001 519 918 438 850 56 × 2 = 0 + 0.000 003 039 836 877 701 12;
  • 13) 0.000 003 039 836 877 701 12 × 2 = 0 + 0.000 006 079 673 755 402 24;
  • 14) 0.000 006 079 673 755 402 24 × 2 = 0 + 0.000 012 159 347 510 804 48;
  • 15) 0.000 012 159 347 510 804 48 × 2 = 0 + 0.000 024 318 695 021 608 96;
  • 16) 0.000 024 318 695 021 608 96 × 2 = 0 + 0.000 048 637 390 043 217 92;
  • 17) 0.000 048 637 390 043 217 92 × 2 = 0 + 0.000 097 274 780 086 435 84;
  • 18) 0.000 097 274 780 086 435 84 × 2 = 0 + 0.000 194 549 560 172 871 68;
  • 19) 0.000 194 549 560 172 871 68 × 2 = 0 + 0.000 389 099 120 345 743 36;
  • 20) 0.000 389 099 120 345 743 36 × 2 = 0 + 0.000 778 198 240 691 486 72;
  • 21) 0.000 778 198 240 691 486 72 × 2 = 0 + 0.001 556 396 481 382 973 44;
  • 22) 0.001 556 396 481 382 973 44 × 2 = 0 + 0.003 112 792 962 765 946 88;
  • 23) 0.003 112 792 962 765 946 88 × 2 = 0 + 0.006 225 585 925 531 893 76;
  • 24) 0.006 225 585 925 531 893 76 × 2 = 0 + 0.012 451 171 851 063 787 52;
  • 25) 0.012 451 171 851 063 787 52 × 2 = 0 + 0.024 902 343 702 127 575 04;
  • 26) 0.024 902 343 702 127 575 04 × 2 = 0 + 0.049 804 687 404 255 150 08;
  • 27) 0.049 804 687 404 255 150 08 × 2 = 0 + 0.099 609 374 808 510 300 16;
  • 28) 0.099 609 374 808 510 300 16 × 2 = 0 + 0.199 218 749 617 020 600 32;
  • 29) 0.199 218 749 617 020 600 32 × 2 = 0 + 0.398 437 499 234 041 200 64;
  • 30) 0.398 437 499 234 041 200 64 × 2 = 0 + 0.796 874 998 468 082 401 28;
  • 31) 0.796 874 998 468 082 401 28 × 2 = 1 + 0.593 749 996 936 164 802 56;
  • 32) 0.593 749 996 936 164 802 56 × 2 = 1 + 0.187 499 993 872 329 605 12;
  • 33) 0.187 499 993 872 329 605 12 × 2 = 0 + 0.374 999 987 744 659 210 24;
  • 34) 0.374 999 987 744 659 210 24 × 2 = 0 + 0.749 999 975 489 318 420 48;
  • 35) 0.749 999 975 489 318 420 48 × 2 = 1 + 0.499 999 950 978 636 840 96;
  • 36) 0.499 999 950 978 636 840 96 × 2 = 0 + 0.999 999 901 957 273 681 92;
  • 37) 0.999 999 901 957 273 681 92 × 2 = 1 + 0.999 999 803 914 547 363 84;
  • 38) 0.999 999 803 914 547 363 84 × 2 = 1 + 0.999 999 607 829 094 727 68;
  • 39) 0.999 999 607 829 094 727 68 × 2 = 1 + 0.999 999 215 658 189 455 36;
  • 40) 0.999 999 215 658 189 455 36 × 2 = 1 + 0.999 998 431 316 378 910 72;
  • 41) 0.999 998 431 316 378 910 72 × 2 = 1 + 0.999 996 862 632 757 821 44;
  • 42) 0.999 996 862 632 757 821 44 × 2 = 1 + 0.999 993 725 265 515 642 88;
  • 43) 0.999 993 725 265 515 642 88 × 2 = 1 + 0.999 987 450 531 031 285 76;
  • 44) 0.999 987 450 531 031 285 76 × 2 = 1 + 0.999 974 901 062 062 571 52;
  • 45) 0.999 974 901 062 062 571 52 × 2 = 1 + 0.999 949 802 124 125 143 04;
  • 46) 0.999 949 802 124 125 143 04 × 2 = 1 + 0.999 899 604 248 250 286 08;
  • 47) 0.999 899 604 248 250 286 08 × 2 = 1 + 0.999 799 208 496 500 572 16;
  • 48) 0.999 799 208 496 500 572 16 × 2 = 1 + 0.999 598 416 993 001 144 32;
  • 49) 0.999 598 416 993 001 144 32 × 2 = 1 + 0.999 196 833 986 002 288 64;
  • 50) 0.999 196 833 986 002 288 64 × 2 = 1 + 0.998 393 667 972 004 577 28;
  • 51) 0.998 393 667 972 004 577 28 × 2 = 1 + 0.996 787 335 944 009 154 56;
  • 52) 0.996 787 335 944 009 154 56 × 2 = 1 + 0.993 574 671 888 018 309 12;
  • 53) 0.993 574 671 888 018 309 12 × 2 = 1 + 0.987 149 343 776 036 618 24;
  • 54) 0.987 149 343 776 036 618 24 × 2 = 1 + 0.974 298 687 552 073 236 48;

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 22(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 22(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 22(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 22 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