-0.000 000 000 742 147 53 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 53(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 53(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 53| = 0.000 000 000 742 147 53


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 53.

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 53 × 2 = 0 + 0.000 000 001 484 295 06;
  • 2) 0.000 000 001 484 295 06 × 2 = 0 + 0.000 000 002 968 590 12;
  • 3) 0.000 000 002 968 590 12 × 2 = 0 + 0.000 000 005 937 180 24;
  • 4) 0.000 000 005 937 180 24 × 2 = 0 + 0.000 000 011 874 360 48;
  • 5) 0.000 000 011 874 360 48 × 2 = 0 + 0.000 000 023 748 720 96;
  • 6) 0.000 000 023 748 720 96 × 2 = 0 + 0.000 000 047 497 441 92;
  • 7) 0.000 000 047 497 441 92 × 2 = 0 + 0.000 000 094 994 883 84;
  • 8) 0.000 000 094 994 883 84 × 2 = 0 + 0.000 000 189 989 767 68;
  • 9) 0.000 000 189 989 767 68 × 2 = 0 + 0.000 000 379 979 535 36;
  • 10) 0.000 000 379 979 535 36 × 2 = 0 + 0.000 000 759 959 070 72;
  • 11) 0.000 000 759 959 070 72 × 2 = 0 + 0.000 001 519 918 141 44;
  • 12) 0.000 001 519 918 141 44 × 2 = 0 + 0.000 003 039 836 282 88;
  • 13) 0.000 003 039 836 282 88 × 2 = 0 + 0.000 006 079 672 565 76;
  • 14) 0.000 006 079 672 565 76 × 2 = 0 + 0.000 012 159 345 131 52;
  • 15) 0.000 012 159 345 131 52 × 2 = 0 + 0.000 024 318 690 263 04;
  • 16) 0.000 024 318 690 263 04 × 2 = 0 + 0.000 048 637 380 526 08;
  • 17) 0.000 048 637 380 526 08 × 2 = 0 + 0.000 097 274 761 052 16;
  • 18) 0.000 097 274 761 052 16 × 2 = 0 + 0.000 194 549 522 104 32;
  • 19) 0.000 194 549 522 104 32 × 2 = 0 + 0.000 389 099 044 208 64;
  • 20) 0.000 389 099 044 208 64 × 2 = 0 + 0.000 778 198 088 417 28;
  • 21) 0.000 778 198 088 417 28 × 2 = 0 + 0.001 556 396 176 834 56;
  • 22) 0.001 556 396 176 834 56 × 2 = 0 + 0.003 112 792 353 669 12;
  • 23) 0.003 112 792 353 669 12 × 2 = 0 + 0.006 225 584 707 338 24;
  • 24) 0.006 225 584 707 338 24 × 2 = 0 + 0.012 451 169 414 676 48;
  • 25) 0.012 451 169 414 676 48 × 2 = 0 + 0.024 902 338 829 352 96;
  • 26) 0.024 902 338 829 352 96 × 2 = 0 + 0.049 804 677 658 705 92;
  • 27) 0.049 804 677 658 705 92 × 2 = 0 + 0.099 609 355 317 411 84;
  • 28) 0.099 609 355 317 411 84 × 2 = 0 + 0.199 218 710 634 823 68;
  • 29) 0.199 218 710 634 823 68 × 2 = 0 + 0.398 437 421 269 647 36;
  • 30) 0.398 437 421 269 647 36 × 2 = 0 + 0.796 874 842 539 294 72;
  • 31) 0.796 874 842 539 294 72 × 2 = 1 + 0.593 749 685 078 589 44;
  • 32) 0.593 749 685 078 589 44 × 2 = 1 + 0.187 499 370 157 178 88;
  • 33) 0.187 499 370 157 178 88 × 2 = 0 + 0.374 998 740 314 357 76;
  • 34) 0.374 998 740 314 357 76 × 2 = 0 + 0.749 997 480 628 715 52;
  • 35) 0.749 997 480 628 715 52 × 2 = 1 + 0.499 994 961 257 431 04;
  • 36) 0.499 994 961 257 431 04 × 2 = 0 + 0.999 989 922 514 862 08;
  • 37) 0.999 989 922 514 862 08 × 2 = 1 + 0.999 979 845 029 724 16;
  • 38) 0.999 979 845 029 724 16 × 2 = 1 + 0.999 959 690 059 448 32;
  • 39) 0.999 959 690 059 448 32 × 2 = 1 + 0.999 919 380 118 896 64;
  • 40) 0.999 919 380 118 896 64 × 2 = 1 + 0.999 838 760 237 793 28;
  • 41) 0.999 838 760 237 793 28 × 2 = 1 + 0.999 677 520 475 586 56;
  • 42) 0.999 677 520 475 586 56 × 2 = 1 + 0.999 355 040 951 173 12;
  • 43) 0.999 355 040 951 173 12 × 2 = 1 + 0.998 710 081 902 346 24;
  • 44) 0.998 710 081 902 346 24 × 2 = 1 + 0.997 420 163 804 692 48;
  • 45) 0.997 420 163 804 692 48 × 2 = 1 + 0.994 840 327 609 384 96;
  • 46) 0.994 840 327 609 384 96 × 2 = 1 + 0.989 680 655 218 769 92;
  • 47) 0.989 680 655 218 769 92 × 2 = 1 + 0.979 361 310 437 539 84;
  • 48) 0.979 361 310 437 539 84 × 2 = 1 + 0.958 722 620 875 079 68;
  • 49) 0.958 722 620 875 079 68 × 2 = 1 + 0.917 445 241 750 159 36;
  • 50) 0.917 445 241 750 159 36 × 2 = 1 + 0.834 890 483 500 318 72;
  • 51) 0.834 890 483 500 318 72 × 2 = 1 + 0.669 780 967 000 637 44;
  • 52) 0.669 780 967 000 637 44 × 2 = 1 + 0.339 561 934 001 274 88;
  • 53) 0.339 561 934 001 274 88 × 2 = 0 + 0.679 123 868 002 549 76;
  • 54) 0.679 123 868 002 549 76 × 2 = 1 + 0.358 247 736 005 099 52;

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 53(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 53(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(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 53(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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 101


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 1101 =

100 1011 1111 1111 1111 1101


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 1101


Decimal number -0.000 000 000 742 147 53 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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