-0.000 000 000 742 147 674 95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 95(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
-0.000 000 000 742 147 674 95(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:

|-0.000 000 000 742 147 674 95| = 0.000 000 000 742 147 674 95


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: 0.000 000 000 742 147 674 95.

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) 0.000 000 000 742 147 674 95 × 2 = 0 + 0.000 000 001 484 295 349 9;
  • 2) 0.000 000 001 484 295 349 9 × 2 = 0 + 0.000 000 002 968 590 699 8;
  • 3) 0.000 000 002 968 590 699 8 × 2 = 0 + 0.000 000 005 937 181 399 6;
  • 4) 0.000 000 005 937 181 399 6 × 2 = 0 + 0.000 000 011 874 362 799 2;
  • 5) 0.000 000 011 874 362 799 2 × 2 = 0 + 0.000 000 023 748 725 598 4;
  • 6) 0.000 000 023 748 725 598 4 × 2 = 0 + 0.000 000 047 497 451 196 8;
  • 7) 0.000 000 047 497 451 196 8 × 2 = 0 + 0.000 000 094 994 902 393 6;
  • 8) 0.000 000 094 994 902 393 6 × 2 = 0 + 0.000 000 189 989 804 787 2;
  • 9) 0.000 000 189 989 804 787 2 × 2 = 0 + 0.000 000 379 979 609 574 4;
  • 10) 0.000 000 379 979 609 574 4 × 2 = 0 + 0.000 000 759 959 219 148 8;
  • 11) 0.000 000 759 959 219 148 8 × 2 = 0 + 0.000 001 519 918 438 297 6;
  • 12) 0.000 001 519 918 438 297 6 × 2 = 0 + 0.000 003 039 836 876 595 2;
  • 13) 0.000 003 039 836 876 595 2 × 2 = 0 + 0.000 006 079 673 753 190 4;
  • 14) 0.000 006 079 673 753 190 4 × 2 = 0 + 0.000 012 159 347 506 380 8;
  • 15) 0.000 012 159 347 506 380 8 × 2 = 0 + 0.000 024 318 695 012 761 6;
  • 16) 0.000 024 318 695 012 761 6 × 2 = 0 + 0.000 048 637 390 025 523 2;
  • 17) 0.000 048 637 390 025 523 2 × 2 = 0 + 0.000 097 274 780 051 046 4;
  • 18) 0.000 097 274 780 051 046 4 × 2 = 0 + 0.000 194 549 560 102 092 8;
  • 19) 0.000 194 549 560 102 092 8 × 2 = 0 + 0.000 389 099 120 204 185 6;
  • 20) 0.000 389 099 120 204 185 6 × 2 = 0 + 0.000 778 198 240 408 371 2;
  • 21) 0.000 778 198 240 408 371 2 × 2 = 0 + 0.001 556 396 480 816 742 4;
  • 22) 0.001 556 396 480 816 742 4 × 2 = 0 + 0.003 112 792 961 633 484 8;
  • 23) 0.003 112 792 961 633 484 8 × 2 = 0 + 0.006 225 585 923 266 969 6;
  • 24) 0.006 225 585 923 266 969 6 × 2 = 0 + 0.012 451 171 846 533 939 2;
  • 25) 0.012 451 171 846 533 939 2 × 2 = 0 + 0.024 902 343 693 067 878 4;
  • 26) 0.024 902 343 693 067 878 4 × 2 = 0 + 0.049 804 687 386 135 756 8;
  • 27) 0.049 804 687 386 135 756 8 × 2 = 0 + 0.099 609 374 772 271 513 6;
  • 28) 0.099 609 374 772 271 513 6 × 2 = 0 + 0.199 218 749 544 543 027 2;
  • 29) 0.199 218 749 544 543 027 2 × 2 = 0 + 0.398 437 499 089 086 054 4;
  • 30) 0.398 437 499 089 086 054 4 × 2 = 0 + 0.796 874 998 178 172 108 8;
  • 31) 0.796 874 998 178 172 108 8 × 2 = 1 + 0.593 749 996 356 344 217 6;
  • 32) 0.593 749 996 356 344 217 6 × 2 = 1 + 0.187 499 992 712 688 435 2;
  • 33) 0.187 499 992 712 688 435 2 × 2 = 0 + 0.374 999 985 425 376 870 4;
  • 34) 0.374 999 985 425 376 870 4 × 2 = 0 + 0.749 999 970 850 753 740 8;
  • 35) 0.749 999 970 850 753 740 8 × 2 = 1 + 0.499 999 941 701 507 481 6;
  • 36) 0.499 999 941 701 507 481 6 × 2 = 0 + 0.999 999 883 403 014 963 2;
  • 37) 0.999 999 883 403 014 963 2 × 2 = 1 + 0.999 999 766 806 029 926 4;
  • 38) 0.999 999 766 806 029 926 4 × 2 = 1 + 0.999 999 533 612 059 852 8;
  • 39) 0.999 999 533 612 059 852 8 × 2 = 1 + 0.999 999 067 224 119 705 6;
  • 40) 0.999 999 067 224 119 705 6 × 2 = 1 + 0.999 998 134 448 239 411 2;
  • 41) 0.999 998 134 448 239 411 2 × 2 = 1 + 0.999 996 268 896 478 822 4;
  • 42) 0.999 996 268 896 478 822 4 × 2 = 1 + 0.999 992 537 792 957 644 8;
  • 43) 0.999 992 537 792 957 644 8 × 2 = 1 + 0.999 985 075 585 915 289 6;
  • 44) 0.999 985 075 585 915 289 6 × 2 = 1 + 0.999 970 151 171 830 579 2;
  • 45) 0.999 970 151 171 830 579 2 × 2 = 1 + 0.999 940 302 343 661 158 4;
  • 46) 0.999 940 302 343 661 158 4 × 2 = 1 + 0.999 880 604 687 322 316 8;
  • 47) 0.999 880 604 687 322 316 8 × 2 = 1 + 0.999 761 209 374 644 633 6;
  • 48) 0.999 761 209 374 644 633 6 × 2 = 1 + 0.999 522 418 749 289 267 2;
  • 49) 0.999 522 418 749 289 267 2 × 2 = 1 + 0.999 044 837 498 578 534 4;
  • 50) 0.999 044 837 498 578 534 4 × 2 = 1 + 0.998 089 674 997 157 068 8;
  • 51) 0.998 089 674 997 157 068 8 × 2 = 1 + 0.996 179 349 994 314 137 6;
  • 52) 0.996 179 349 994 314 137 6 × 2 = 1 + 0.992 358 699 988 628 275 2;
  • 53) 0.992 358 699 988 628 275 2 × 2 = 1 + 0.984 717 399 977 256 550 4;
  • 54) 0.984 717 399 977 256 550 4 × 2 = 1 + 0.969 434 799 954 513 100 8;

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:


0.000 000 000 742 147 674 95(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 95(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:


0.000 000 000 742 147 674 95(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 -0.000 000 000 742 147 674 95 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