-0.000 000 000 742 148 02 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 02(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
-0.000 000 000 742 148 02(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:

|-0.000 000 000 742 148 02| = 0.000 000 000 742 148 02


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: 0.000 000 000 742 148 02.

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) 0.000 000 000 742 148 02 × 2 = 0 + 0.000 000 001 484 296 04;
  • 2) 0.000 000 001 484 296 04 × 2 = 0 + 0.000 000 002 968 592 08;
  • 3) 0.000 000 002 968 592 08 × 2 = 0 + 0.000 000 005 937 184 16;
  • 4) 0.000 000 005 937 184 16 × 2 = 0 + 0.000 000 011 874 368 32;
  • 5) 0.000 000 011 874 368 32 × 2 = 0 + 0.000 000 023 748 736 64;
  • 6) 0.000 000 023 748 736 64 × 2 = 0 + 0.000 000 047 497 473 28;
  • 7) 0.000 000 047 497 473 28 × 2 = 0 + 0.000 000 094 994 946 56;
  • 8) 0.000 000 094 994 946 56 × 2 = 0 + 0.000 000 189 989 893 12;
  • 9) 0.000 000 189 989 893 12 × 2 = 0 + 0.000 000 379 979 786 24;
  • 10) 0.000 000 379 979 786 24 × 2 = 0 + 0.000 000 759 959 572 48;
  • 11) 0.000 000 759 959 572 48 × 2 = 0 + 0.000 001 519 919 144 96;
  • 12) 0.000 001 519 919 144 96 × 2 = 0 + 0.000 003 039 838 289 92;
  • 13) 0.000 003 039 838 289 92 × 2 = 0 + 0.000 006 079 676 579 84;
  • 14) 0.000 006 079 676 579 84 × 2 = 0 + 0.000 012 159 353 159 68;
  • 15) 0.000 012 159 353 159 68 × 2 = 0 + 0.000 024 318 706 319 36;
  • 16) 0.000 024 318 706 319 36 × 2 = 0 + 0.000 048 637 412 638 72;
  • 17) 0.000 048 637 412 638 72 × 2 = 0 + 0.000 097 274 825 277 44;
  • 18) 0.000 097 274 825 277 44 × 2 = 0 + 0.000 194 549 650 554 88;
  • 19) 0.000 194 549 650 554 88 × 2 = 0 + 0.000 389 099 301 109 76;
  • 20) 0.000 389 099 301 109 76 × 2 = 0 + 0.000 778 198 602 219 52;
  • 21) 0.000 778 198 602 219 52 × 2 = 0 + 0.001 556 397 204 439 04;
  • 22) 0.001 556 397 204 439 04 × 2 = 0 + 0.003 112 794 408 878 08;
  • 23) 0.003 112 794 408 878 08 × 2 = 0 + 0.006 225 588 817 756 16;
  • 24) 0.006 225 588 817 756 16 × 2 = 0 + 0.012 451 177 635 512 32;
  • 25) 0.012 451 177 635 512 32 × 2 = 0 + 0.024 902 355 271 024 64;
  • 26) 0.024 902 355 271 024 64 × 2 = 0 + 0.049 804 710 542 049 28;
  • 27) 0.049 804 710 542 049 28 × 2 = 0 + 0.099 609 421 084 098 56;
  • 28) 0.099 609 421 084 098 56 × 2 = 0 + 0.199 218 842 168 197 12;
  • 29) 0.199 218 842 168 197 12 × 2 = 0 + 0.398 437 684 336 394 24;
  • 30) 0.398 437 684 336 394 24 × 2 = 0 + 0.796 875 368 672 788 48;
  • 31) 0.796 875 368 672 788 48 × 2 = 1 + 0.593 750 737 345 576 96;
  • 32) 0.593 750 737 345 576 96 × 2 = 1 + 0.187 501 474 691 153 92;
  • 33) 0.187 501 474 691 153 92 × 2 = 0 + 0.375 002 949 382 307 84;
  • 34) 0.375 002 949 382 307 84 × 2 = 0 + 0.750 005 898 764 615 68;
  • 35) 0.750 005 898 764 615 68 × 2 = 1 + 0.500 011 797 529 231 36;
  • 36) 0.500 011 797 529 231 36 × 2 = 1 + 0.000 023 595 058 462 72;
  • 37) 0.000 023 595 058 462 72 × 2 = 0 + 0.000 047 190 116 925 44;
  • 38) 0.000 047 190 116 925 44 × 2 = 0 + 0.000 094 380 233 850 88;
  • 39) 0.000 094 380 233 850 88 × 2 = 0 + 0.000 188 760 467 701 76;
  • 40) 0.000 188 760 467 701 76 × 2 = 0 + 0.000 377 520 935 403 52;
  • 41) 0.000 377 520 935 403 52 × 2 = 0 + 0.000 755 041 870 807 04;
  • 42) 0.000 755 041 870 807 04 × 2 = 0 + 0.001 510 083 741 614 08;
  • 43) 0.001 510 083 741 614 08 × 2 = 0 + 0.003 020 167 483 228 16;
  • 44) 0.003 020 167 483 228 16 × 2 = 0 + 0.006 040 334 966 456 32;
  • 45) 0.006 040 334 966 456 32 × 2 = 0 + 0.012 080 669 932 912 64;
  • 46) 0.012 080 669 932 912 64 × 2 = 0 + 0.024 161 339 865 825 28;
  • 47) 0.024 161 339 865 825 28 × 2 = 0 + 0.048 322 679 731 650 56;
  • 48) 0.048 322 679 731 650 56 × 2 = 0 + 0.096 645 359 463 301 12;
  • 49) 0.096 645 359 463 301 12 × 2 = 0 + 0.193 290 718 926 602 24;
  • 50) 0.193 290 718 926 602 24 × 2 = 0 + 0.386 581 437 853 204 48;
  • 51) 0.386 581 437 853 204 48 × 2 = 0 + 0.773 162 875 706 408 96;
  • 52) 0.773 162 875 706 408 96 × 2 = 1 + 0.546 325 751 412 817 92;
  • 53) 0.546 325 751 412 817 92 × 2 = 1 + 0.092 651 502 825 635 84;
  • 54) 0.092 651 502 825 635 84 × 2 = 0 + 0.185 303 005 651 271 68;

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:


0.000 000 000 742 148 02(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2)

6. Positive number before normalization:

0.000 000 000 742 148 02(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(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:


0.000 000 000 742 148 02(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 10(2) × 20 =

1.1001 1000 0000 0000 0000 110(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 1000 0000 0000 0000 110


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 1100 0000 0000 0000 0110 =

100 1100 0000 0000 0000 0110


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 1100 0000 0000 0000 0110


Decimal number -0.000 000 000 742 148 02 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0110

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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