-0.000 000 000 742 147 675 35 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 35(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 35(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 35| = 0.000 000 000 742 147 675 35


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 35.

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 35 × 2 = 0 + 0.000 000 001 484 295 350 7;
  • 2) 0.000 000 001 484 295 350 7 × 2 = 0 + 0.000 000 002 968 590 701 4;
  • 3) 0.000 000 002 968 590 701 4 × 2 = 0 + 0.000 000 005 937 181 402 8;
  • 4) 0.000 000 005 937 181 402 8 × 2 = 0 + 0.000 000 011 874 362 805 6;
  • 5) 0.000 000 011 874 362 805 6 × 2 = 0 + 0.000 000 023 748 725 611 2;
  • 6) 0.000 000 023 748 725 611 2 × 2 = 0 + 0.000 000 047 497 451 222 4;
  • 7) 0.000 000 047 497 451 222 4 × 2 = 0 + 0.000 000 094 994 902 444 8;
  • 8) 0.000 000 094 994 902 444 8 × 2 = 0 + 0.000 000 189 989 804 889 6;
  • 9) 0.000 000 189 989 804 889 6 × 2 = 0 + 0.000 000 379 979 609 779 2;
  • 10) 0.000 000 379 979 609 779 2 × 2 = 0 + 0.000 000 759 959 219 558 4;
  • 11) 0.000 000 759 959 219 558 4 × 2 = 0 + 0.000 001 519 918 439 116 8;
  • 12) 0.000 001 519 918 439 116 8 × 2 = 0 + 0.000 003 039 836 878 233 6;
  • 13) 0.000 003 039 836 878 233 6 × 2 = 0 + 0.000 006 079 673 756 467 2;
  • 14) 0.000 006 079 673 756 467 2 × 2 = 0 + 0.000 012 159 347 512 934 4;
  • 15) 0.000 012 159 347 512 934 4 × 2 = 0 + 0.000 024 318 695 025 868 8;
  • 16) 0.000 024 318 695 025 868 8 × 2 = 0 + 0.000 048 637 390 051 737 6;
  • 17) 0.000 048 637 390 051 737 6 × 2 = 0 + 0.000 097 274 780 103 475 2;
  • 18) 0.000 097 274 780 103 475 2 × 2 = 0 + 0.000 194 549 560 206 950 4;
  • 19) 0.000 194 549 560 206 950 4 × 2 = 0 + 0.000 389 099 120 413 900 8;
  • 20) 0.000 389 099 120 413 900 8 × 2 = 0 + 0.000 778 198 240 827 801 6;
  • 21) 0.000 778 198 240 827 801 6 × 2 = 0 + 0.001 556 396 481 655 603 2;
  • 22) 0.001 556 396 481 655 603 2 × 2 = 0 + 0.003 112 792 963 311 206 4;
  • 23) 0.003 112 792 963 311 206 4 × 2 = 0 + 0.006 225 585 926 622 412 8;
  • 24) 0.006 225 585 926 622 412 8 × 2 = 0 + 0.012 451 171 853 244 825 6;
  • 25) 0.012 451 171 853 244 825 6 × 2 = 0 + 0.024 902 343 706 489 651 2;
  • 26) 0.024 902 343 706 489 651 2 × 2 = 0 + 0.049 804 687 412 979 302 4;
  • 27) 0.049 804 687 412 979 302 4 × 2 = 0 + 0.099 609 374 825 958 604 8;
  • 28) 0.099 609 374 825 958 604 8 × 2 = 0 + 0.199 218 749 651 917 209 6;
  • 29) 0.199 218 749 651 917 209 6 × 2 = 0 + 0.398 437 499 303 834 419 2;
  • 30) 0.398 437 499 303 834 419 2 × 2 = 0 + 0.796 874 998 607 668 838 4;
  • 31) 0.796 874 998 607 668 838 4 × 2 = 1 + 0.593 749 997 215 337 676 8;
  • 32) 0.593 749 997 215 337 676 8 × 2 = 1 + 0.187 499 994 430 675 353 6;
  • 33) 0.187 499 994 430 675 353 6 × 2 = 0 + 0.374 999 988 861 350 707 2;
  • 34) 0.374 999 988 861 350 707 2 × 2 = 0 + 0.749 999 977 722 701 414 4;
  • 35) 0.749 999 977 722 701 414 4 × 2 = 1 + 0.499 999 955 445 402 828 8;
  • 36) 0.499 999 955 445 402 828 8 × 2 = 0 + 0.999 999 910 890 805 657 6;
  • 37) 0.999 999 910 890 805 657 6 × 2 = 1 + 0.999 999 821 781 611 315 2;
  • 38) 0.999 999 821 781 611 315 2 × 2 = 1 + 0.999 999 643 563 222 630 4;
  • 39) 0.999 999 643 563 222 630 4 × 2 = 1 + 0.999 999 287 126 445 260 8;
  • 40) 0.999 999 287 126 445 260 8 × 2 = 1 + 0.999 998 574 252 890 521 6;
  • 41) 0.999 998 574 252 890 521 6 × 2 = 1 + 0.999 997 148 505 781 043 2;
  • 42) 0.999 997 148 505 781 043 2 × 2 = 1 + 0.999 994 297 011 562 086 4;
  • 43) 0.999 994 297 011 562 086 4 × 2 = 1 + 0.999 988 594 023 124 172 8;
  • 44) 0.999 988 594 023 124 172 8 × 2 = 1 + 0.999 977 188 046 248 345 6;
  • 45) 0.999 977 188 046 248 345 6 × 2 = 1 + 0.999 954 376 092 496 691 2;
  • 46) 0.999 954 376 092 496 691 2 × 2 = 1 + 0.999 908 752 184 993 382 4;
  • 47) 0.999 908 752 184 993 382 4 × 2 = 1 + 0.999 817 504 369 986 764 8;
  • 48) 0.999 817 504 369 986 764 8 × 2 = 1 + 0.999 635 008 739 973 529 6;
  • 49) 0.999 635 008 739 973 529 6 × 2 = 1 + 0.999 270 017 479 947 059 2;
  • 50) 0.999 270 017 479 947 059 2 × 2 = 1 + 0.998 540 034 959 894 118 4;
  • 51) 0.998 540 034 959 894 118 4 × 2 = 1 + 0.997 080 069 919 788 236 8;
  • 52) 0.997 080 069 919 788 236 8 × 2 = 1 + 0.994 160 139 839 576 473 6;
  • 53) 0.994 160 139 839 576 473 6 × 2 = 1 + 0.988 320 279 679 152 947 2;
  • 54) 0.988 320 279 679 152 947 2 × 2 = 1 + 0.976 640 559 358 305 894 4;

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 35(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 35(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 35(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 35 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