-0.000 000 000 742 147 676 640 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 640 8(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
-0.000 000 000 742 147 676 640 8(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:

|-0.000 000 000 742 147 676 640 8| = 0.000 000 000 742 147 676 640 8


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: 0.000 000 000 742 147 676 640 8.

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) 0.000 000 000 742 147 676 640 8 × 2 = 0 + 0.000 000 001 484 295 353 281 6;
  • 2) 0.000 000 001 484 295 353 281 6 × 2 = 0 + 0.000 000 002 968 590 706 563 2;
  • 3) 0.000 000 002 968 590 706 563 2 × 2 = 0 + 0.000 000 005 937 181 413 126 4;
  • 4) 0.000 000 005 937 181 413 126 4 × 2 = 0 + 0.000 000 011 874 362 826 252 8;
  • 5) 0.000 000 011 874 362 826 252 8 × 2 = 0 + 0.000 000 023 748 725 652 505 6;
  • 6) 0.000 000 023 748 725 652 505 6 × 2 = 0 + 0.000 000 047 497 451 305 011 2;
  • 7) 0.000 000 047 497 451 305 011 2 × 2 = 0 + 0.000 000 094 994 902 610 022 4;
  • 8) 0.000 000 094 994 902 610 022 4 × 2 = 0 + 0.000 000 189 989 805 220 044 8;
  • 9) 0.000 000 189 989 805 220 044 8 × 2 = 0 + 0.000 000 379 979 610 440 089 6;
  • 10) 0.000 000 379 979 610 440 089 6 × 2 = 0 + 0.000 000 759 959 220 880 179 2;
  • 11) 0.000 000 759 959 220 880 179 2 × 2 = 0 + 0.000 001 519 918 441 760 358 4;
  • 12) 0.000 001 519 918 441 760 358 4 × 2 = 0 + 0.000 003 039 836 883 520 716 8;
  • 13) 0.000 003 039 836 883 520 716 8 × 2 = 0 + 0.000 006 079 673 767 041 433 6;
  • 14) 0.000 006 079 673 767 041 433 6 × 2 = 0 + 0.000 012 159 347 534 082 867 2;
  • 15) 0.000 012 159 347 534 082 867 2 × 2 = 0 + 0.000 024 318 695 068 165 734 4;
  • 16) 0.000 024 318 695 068 165 734 4 × 2 = 0 + 0.000 048 637 390 136 331 468 8;
  • 17) 0.000 048 637 390 136 331 468 8 × 2 = 0 + 0.000 097 274 780 272 662 937 6;
  • 18) 0.000 097 274 780 272 662 937 6 × 2 = 0 + 0.000 194 549 560 545 325 875 2;
  • 19) 0.000 194 549 560 545 325 875 2 × 2 = 0 + 0.000 389 099 121 090 651 750 4;
  • 20) 0.000 389 099 121 090 651 750 4 × 2 = 0 + 0.000 778 198 242 181 303 500 8;
  • 21) 0.000 778 198 242 181 303 500 8 × 2 = 0 + 0.001 556 396 484 362 607 001 6;
  • 22) 0.001 556 396 484 362 607 001 6 × 2 = 0 + 0.003 112 792 968 725 214 003 2;
  • 23) 0.003 112 792 968 725 214 003 2 × 2 = 0 + 0.006 225 585 937 450 428 006 4;
  • 24) 0.006 225 585 937 450 428 006 4 × 2 = 0 + 0.012 451 171 874 900 856 012 8;
  • 25) 0.012 451 171 874 900 856 012 8 × 2 = 0 + 0.024 902 343 749 801 712 025 6;
  • 26) 0.024 902 343 749 801 712 025 6 × 2 = 0 + 0.049 804 687 499 603 424 051 2;
  • 27) 0.049 804 687 499 603 424 051 2 × 2 = 0 + 0.099 609 374 999 206 848 102 4;
  • 28) 0.099 609 374 999 206 848 102 4 × 2 = 0 + 0.199 218 749 998 413 696 204 8;
  • 29) 0.199 218 749 998 413 696 204 8 × 2 = 0 + 0.398 437 499 996 827 392 409 6;
  • 30) 0.398 437 499 996 827 392 409 6 × 2 = 0 + 0.796 874 999 993 654 784 819 2;
  • 31) 0.796 874 999 993 654 784 819 2 × 2 = 1 + 0.593 749 999 987 309 569 638 4;
  • 32) 0.593 749 999 987 309 569 638 4 × 2 = 1 + 0.187 499 999 974 619 139 276 8;
  • 33) 0.187 499 999 974 619 139 276 8 × 2 = 0 + 0.374 999 999 949 238 278 553 6;
  • 34) 0.374 999 999 949 238 278 553 6 × 2 = 0 + 0.749 999 999 898 476 557 107 2;
  • 35) 0.749 999 999 898 476 557 107 2 × 2 = 1 + 0.499 999 999 796 953 114 214 4;
  • 36) 0.499 999 999 796 953 114 214 4 × 2 = 0 + 0.999 999 999 593 906 228 428 8;
  • 37) 0.999 999 999 593 906 228 428 8 × 2 = 1 + 0.999 999 999 187 812 456 857 6;
  • 38) 0.999 999 999 187 812 456 857 6 × 2 = 1 + 0.999 999 998 375 624 913 715 2;
  • 39) 0.999 999 998 375 624 913 715 2 × 2 = 1 + 0.999 999 996 751 249 827 430 4;
  • 40) 0.999 999 996 751 249 827 430 4 × 2 = 1 + 0.999 999 993 502 499 654 860 8;
  • 41) 0.999 999 993 502 499 654 860 8 × 2 = 1 + 0.999 999 987 004 999 309 721 6;
  • 42) 0.999 999 987 004 999 309 721 6 × 2 = 1 + 0.999 999 974 009 998 619 443 2;
  • 43) 0.999 999 974 009 998 619 443 2 × 2 = 1 + 0.999 999 948 019 997 238 886 4;
  • 44) 0.999 999 948 019 997 238 886 4 × 2 = 1 + 0.999 999 896 039 994 477 772 8;
  • 45) 0.999 999 896 039 994 477 772 8 × 2 = 1 + 0.999 999 792 079 988 955 545 6;
  • 46) 0.999 999 792 079 988 955 545 6 × 2 = 1 + 0.999 999 584 159 977 911 091 2;
  • 47) 0.999 999 584 159 977 911 091 2 × 2 = 1 + 0.999 999 168 319 955 822 182 4;
  • 48) 0.999 999 168 319 955 822 182 4 × 2 = 1 + 0.999 998 336 639 911 644 364 8;
  • 49) 0.999 998 336 639 911 644 364 8 × 2 = 1 + 0.999 996 673 279 823 288 729 6;
  • 50) 0.999 996 673 279 823 288 729 6 × 2 = 1 + 0.999 993 346 559 646 577 459 2;
  • 51) 0.999 993 346 559 646 577 459 2 × 2 = 1 + 0.999 986 693 119 293 154 918 4;
  • 52) 0.999 986 693 119 293 154 918 4 × 2 = 1 + 0.999 973 386 238 586 309 836 8;
  • 53) 0.999 973 386 238 586 309 836 8 × 2 = 1 + 0.999 946 772 477 172 619 673 6;
  • 54) 0.999 946 772 477 172 619 673 6 × 2 = 1 + 0.999 893 544 954 345 239 347 2;

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:


0.000 000 000 742 147 676 640 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 640 8(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:


0.000 000 000 742 147 676 640 8(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 -0.000 000 000 742 147 676 640 8 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