-0.000 000 000 742 146 96 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 96(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
-0.000 000 000 742 146 96(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:

|-0.000 000 000 742 146 96| = 0.000 000 000 742 146 96


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: 0.000 000 000 742 146 96.

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) 0.000 000 000 742 146 96 × 2 = 0 + 0.000 000 001 484 293 92;
  • 2) 0.000 000 001 484 293 92 × 2 = 0 + 0.000 000 002 968 587 84;
  • 3) 0.000 000 002 968 587 84 × 2 = 0 + 0.000 000 005 937 175 68;
  • 4) 0.000 000 005 937 175 68 × 2 = 0 + 0.000 000 011 874 351 36;
  • 5) 0.000 000 011 874 351 36 × 2 = 0 + 0.000 000 023 748 702 72;
  • 6) 0.000 000 023 748 702 72 × 2 = 0 + 0.000 000 047 497 405 44;
  • 7) 0.000 000 047 497 405 44 × 2 = 0 + 0.000 000 094 994 810 88;
  • 8) 0.000 000 094 994 810 88 × 2 = 0 + 0.000 000 189 989 621 76;
  • 9) 0.000 000 189 989 621 76 × 2 = 0 + 0.000 000 379 979 243 52;
  • 10) 0.000 000 379 979 243 52 × 2 = 0 + 0.000 000 759 958 487 04;
  • 11) 0.000 000 759 958 487 04 × 2 = 0 + 0.000 001 519 916 974 08;
  • 12) 0.000 001 519 916 974 08 × 2 = 0 + 0.000 003 039 833 948 16;
  • 13) 0.000 003 039 833 948 16 × 2 = 0 + 0.000 006 079 667 896 32;
  • 14) 0.000 006 079 667 896 32 × 2 = 0 + 0.000 012 159 335 792 64;
  • 15) 0.000 012 159 335 792 64 × 2 = 0 + 0.000 024 318 671 585 28;
  • 16) 0.000 024 318 671 585 28 × 2 = 0 + 0.000 048 637 343 170 56;
  • 17) 0.000 048 637 343 170 56 × 2 = 0 + 0.000 097 274 686 341 12;
  • 18) 0.000 097 274 686 341 12 × 2 = 0 + 0.000 194 549 372 682 24;
  • 19) 0.000 194 549 372 682 24 × 2 = 0 + 0.000 389 098 745 364 48;
  • 20) 0.000 389 098 745 364 48 × 2 = 0 + 0.000 778 197 490 728 96;
  • 21) 0.000 778 197 490 728 96 × 2 = 0 + 0.001 556 394 981 457 92;
  • 22) 0.001 556 394 981 457 92 × 2 = 0 + 0.003 112 789 962 915 84;
  • 23) 0.003 112 789 962 915 84 × 2 = 0 + 0.006 225 579 925 831 68;
  • 24) 0.006 225 579 925 831 68 × 2 = 0 + 0.012 451 159 851 663 36;
  • 25) 0.012 451 159 851 663 36 × 2 = 0 + 0.024 902 319 703 326 72;
  • 26) 0.024 902 319 703 326 72 × 2 = 0 + 0.049 804 639 406 653 44;
  • 27) 0.049 804 639 406 653 44 × 2 = 0 + 0.099 609 278 813 306 88;
  • 28) 0.099 609 278 813 306 88 × 2 = 0 + 0.199 218 557 626 613 76;
  • 29) 0.199 218 557 626 613 76 × 2 = 0 + 0.398 437 115 253 227 52;
  • 30) 0.398 437 115 253 227 52 × 2 = 0 + 0.796 874 230 506 455 04;
  • 31) 0.796 874 230 506 455 04 × 2 = 1 + 0.593 748 461 012 910 08;
  • 32) 0.593 748 461 012 910 08 × 2 = 1 + 0.187 496 922 025 820 16;
  • 33) 0.187 496 922 025 820 16 × 2 = 0 + 0.374 993 844 051 640 32;
  • 34) 0.374 993 844 051 640 32 × 2 = 0 + 0.749 987 688 103 280 64;
  • 35) 0.749 987 688 103 280 64 × 2 = 1 + 0.499 975 376 206 561 28;
  • 36) 0.499 975 376 206 561 28 × 2 = 0 + 0.999 950 752 413 122 56;
  • 37) 0.999 950 752 413 122 56 × 2 = 1 + 0.999 901 504 826 245 12;
  • 38) 0.999 901 504 826 245 12 × 2 = 1 + 0.999 803 009 652 490 24;
  • 39) 0.999 803 009 652 490 24 × 2 = 1 + 0.999 606 019 304 980 48;
  • 40) 0.999 606 019 304 980 48 × 2 = 1 + 0.999 212 038 609 960 96;
  • 41) 0.999 212 038 609 960 96 × 2 = 1 + 0.998 424 077 219 921 92;
  • 42) 0.998 424 077 219 921 92 × 2 = 1 + 0.996 848 154 439 843 84;
  • 43) 0.996 848 154 439 843 84 × 2 = 1 + 0.993 696 308 879 687 68;
  • 44) 0.993 696 308 879 687 68 × 2 = 1 + 0.987 392 617 759 375 36;
  • 45) 0.987 392 617 759 375 36 × 2 = 1 + 0.974 785 235 518 750 72;
  • 46) 0.974 785 235 518 750 72 × 2 = 1 + 0.949 570 471 037 501 44;
  • 47) 0.949 570 471 037 501 44 × 2 = 1 + 0.899 140 942 075 002 88;
  • 48) 0.899 140 942 075 002 88 × 2 = 1 + 0.798 281 884 150 005 76;
  • 49) 0.798 281 884 150 005 76 × 2 = 1 + 0.596 563 768 300 011 52;
  • 50) 0.596 563 768 300 011 52 × 2 = 1 + 0.193 127 536 600 023 04;
  • 51) 0.193 127 536 600 023 04 × 2 = 0 + 0.386 255 073 200 046 08;
  • 52) 0.386 255 073 200 046 08 × 2 = 0 + 0.772 510 146 400 092 16;
  • 53) 0.772 510 146 400 092 16 × 2 = 1 + 0.545 020 292 800 184 32;
  • 54) 0.545 020 292 800 184 32 × 2 = 1 + 0.090 040 585 600 368 64;

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:


0.000 000 000 742 146 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 11(2)

6. Positive number before normalization:

0.000 000 000 742 146 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 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:


0.000 000 000 742 146 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 11(2) × 20 =

1.1001 0111 1111 1111 1110 011(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 1110 011


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 0011 =

100 1011 1111 1111 1111 0011


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 0011


Decimal number -0.000 000 000 742 146 96 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 0011

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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