-0.000 000 000 742 147 676 647 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 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
-0.000 000 000 742 147 676 647 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:

|-0.000 000 000 742 147 676 647 8| = 0.000 000 000 742 147 676 647 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: 0.000 000 000 742 147 676 647 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) 0.000 000 000 742 147 676 647 8 × 2 = 0 + 0.000 000 001 484 295 353 295 6;
  • 2) 0.000 000 001 484 295 353 295 6 × 2 = 0 + 0.000 000 002 968 590 706 591 2;
  • 3) 0.000 000 002 968 590 706 591 2 × 2 = 0 + 0.000 000 005 937 181 413 182 4;
  • 4) 0.000 000 005 937 181 413 182 4 × 2 = 0 + 0.000 000 011 874 362 826 364 8;
  • 5) 0.000 000 011 874 362 826 364 8 × 2 = 0 + 0.000 000 023 748 725 652 729 6;
  • 6) 0.000 000 023 748 725 652 729 6 × 2 = 0 + 0.000 000 047 497 451 305 459 2;
  • 7) 0.000 000 047 497 451 305 459 2 × 2 = 0 + 0.000 000 094 994 902 610 918 4;
  • 8) 0.000 000 094 994 902 610 918 4 × 2 = 0 + 0.000 000 189 989 805 221 836 8;
  • 9) 0.000 000 189 989 805 221 836 8 × 2 = 0 + 0.000 000 379 979 610 443 673 6;
  • 10) 0.000 000 379 979 610 443 673 6 × 2 = 0 + 0.000 000 759 959 220 887 347 2;
  • 11) 0.000 000 759 959 220 887 347 2 × 2 = 0 + 0.000 001 519 918 441 774 694 4;
  • 12) 0.000 001 519 918 441 774 694 4 × 2 = 0 + 0.000 003 039 836 883 549 388 8;
  • 13) 0.000 003 039 836 883 549 388 8 × 2 = 0 + 0.000 006 079 673 767 098 777 6;
  • 14) 0.000 006 079 673 767 098 777 6 × 2 = 0 + 0.000 012 159 347 534 197 555 2;
  • 15) 0.000 012 159 347 534 197 555 2 × 2 = 0 + 0.000 024 318 695 068 395 110 4;
  • 16) 0.000 024 318 695 068 395 110 4 × 2 = 0 + 0.000 048 637 390 136 790 220 8;
  • 17) 0.000 048 637 390 136 790 220 8 × 2 = 0 + 0.000 097 274 780 273 580 441 6;
  • 18) 0.000 097 274 780 273 580 441 6 × 2 = 0 + 0.000 194 549 560 547 160 883 2;
  • 19) 0.000 194 549 560 547 160 883 2 × 2 = 0 + 0.000 389 099 121 094 321 766 4;
  • 20) 0.000 389 099 121 094 321 766 4 × 2 = 0 + 0.000 778 198 242 188 643 532 8;
  • 21) 0.000 778 198 242 188 643 532 8 × 2 = 0 + 0.001 556 396 484 377 287 065 6;
  • 22) 0.001 556 396 484 377 287 065 6 × 2 = 0 + 0.003 112 792 968 754 574 131 2;
  • 23) 0.003 112 792 968 754 574 131 2 × 2 = 0 + 0.006 225 585 937 509 148 262 4;
  • 24) 0.006 225 585 937 509 148 262 4 × 2 = 0 + 0.012 451 171 875 018 296 524 8;
  • 25) 0.012 451 171 875 018 296 524 8 × 2 = 0 + 0.024 902 343 750 036 593 049 6;
  • 26) 0.024 902 343 750 036 593 049 6 × 2 = 0 + 0.049 804 687 500 073 186 099 2;
  • 27) 0.049 804 687 500 073 186 099 2 × 2 = 0 + 0.099 609 375 000 146 372 198 4;
  • 28) 0.099 609 375 000 146 372 198 4 × 2 = 0 + 0.199 218 750 000 292 744 396 8;
  • 29) 0.199 218 750 000 292 744 396 8 × 2 = 0 + 0.398 437 500 000 585 488 793 6;
  • 30) 0.398 437 500 000 585 488 793 6 × 2 = 0 + 0.796 875 000 001 170 977 587 2;
  • 31) 0.796 875 000 001 170 977 587 2 × 2 = 1 + 0.593 750 000 002 341 955 174 4;
  • 32) 0.593 750 000 002 341 955 174 4 × 2 = 1 + 0.187 500 000 004 683 910 348 8;
  • 33) 0.187 500 000 004 683 910 348 8 × 2 = 0 + 0.375 000 000 009 367 820 697 6;
  • 34) 0.375 000 000 009 367 820 697 6 × 2 = 0 + 0.750 000 000 018 735 641 395 2;
  • 35) 0.750 000 000 018 735 641 395 2 × 2 = 1 + 0.500 000 000 037 471 282 790 4;
  • 36) 0.500 000 000 037 471 282 790 4 × 2 = 1 + 0.000 000 000 074 942 565 580 8;
  • 37) 0.000 000 000 074 942 565 580 8 × 2 = 0 + 0.000 000 000 149 885 131 161 6;
  • 38) 0.000 000 000 149 885 131 161 6 × 2 = 0 + 0.000 000 000 299 770 262 323 2;
  • 39) 0.000 000 000 299 770 262 323 2 × 2 = 0 + 0.000 000 000 599 540 524 646 4;
  • 40) 0.000 000 000 599 540 524 646 4 × 2 = 0 + 0.000 000 001 199 081 049 292 8;
  • 41) 0.000 000 001 199 081 049 292 8 × 2 = 0 + 0.000 000 002 398 162 098 585 6;
  • 42) 0.000 000 002 398 162 098 585 6 × 2 = 0 + 0.000 000 004 796 324 197 171 2;
  • 43) 0.000 000 004 796 324 197 171 2 × 2 = 0 + 0.000 000 009 592 648 394 342 4;
  • 44) 0.000 000 009 592 648 394 342 4 × 2 = 0 + 0.000 000 019 185 296 788 684 8;
  • 45) 0.000 000 019 185 296 788 684 8 × 2 = 0 + 0.000 000 038 370 593 577 369 6;
  • 46) 0.000 000 038 370 593 577 369 6 × 2 = 0 + 0.000 000 076 741 187 154 739 2;
  • 47) 0.000 000 076 741 187 154 739 2 × 2 = 0 + 0.000 000 153 482 374 309 478 4;
  • 48) 0.000 000 153 482 374 309 478 4 × 2 = 0 + 0.000 000 306 964 748 618 956 8;
  • 49) 0.000 000 306 964 748 618 956 8 × 2 = 0 + 0.000 000 613 929 497 237 913 6;
  • 50) 0.000 000 613 929 497 237 913 6 × 2 = 0 + 0.000 001 227 858 994 475 827 2;
  • 51) 0.000 001 227 858 994 475 827 2 × 2 = 0 + 0.000 002 455 717 988 951 654 4;
  • 52) 0.000 002 455 717 988 951 654 4 × 2 = 0 + 0.000 004 911 435 977 903 308 8;
  • 53) 0.000 004 911 435 977 903 308 8 × 2 = 0 + 0.000 009 822 871 955 806 617 6;
  • 54) 0.000 009 822 871 955 806 617 6 × 2 = 0 + 0.000 019 645 743 911 613 235 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:


0.000 000 000 742 147 676 647 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 647 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 647 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 000


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 0000 =

100 1100 0000 0000 0000 0000


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 0000


Decimal number -0.000 000 000 742 147 676 647 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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